NCERT Solutions Class 12 Maths Chapter 9 Differential Equations: Download Free PDFs, Formulas & Weightage

Ex-9.1

Determine order and degree (if defined) of differential equations given in Questions 1 to 10:

Q1.    $\frac{d^{4}y}{d{x}^4}+sin\left(y\text{'}\right)=0$

A.1. The highest order derivation present in the differential equation (D.E.) is  $\frac{d^{4}y}{d{x}^4}$ , so its order is 4.

As, the given D.E.is not a polynomial equation in its derivative ,its degree is not defined.

 

Q2. y' + 5y = 0

A.2. The highest order derivation present in the D.E. is y , so its order is 1.

As the given D.E. is a polynomial equation in its derivative its degree is 1.

 

Q3.  ${\left(\frac{ds}{dt}\right)}^4+3s\frac{d^{2}s}{d{t}^2}=0$

A.3. The highest order derivation present in the D.E. is  $\frac{d^{2}s}{d{t}^2}$ so its order is 2 .

As the given D.E. is a polynomial equation in its derivative its degree is 1.

 

Q4.  $\frac{d^{2}y}{{\left(d{x}^2\right)}^2}+cos\left(\frac{dy}{dx}\right)=0$

A.4.  $\frac{d^{4}y}{d{x}^4}$ As the given D.E. is not a polynomial equation in its derivative, its degree is not defined.

 

Q5.    $\frac{d^{2}y}{d{x}^2}=cos3x+sin3x$

A.5.  $\frac{d^{4}y}{d{x}^4}$ As the given D.E. is a polynomial equation in its derivative, its degree is 1.

 

Q6. (y′′′) + (y′′) + (y′)⁴ + y⁵ =0

A.6. The highest order derivative present in the D.E. is  $y^{\left|\right||}$ so its order is 3.

As the given D.E. is a polynomial equation in its derivation , its degree is 2.

 

Q7. y''' + 2y'' + y' = 0

A.7. The highest order present in the D.E. is  $y^{\left|\right||}$ so its order is 3.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

 

Q8. y′ + y = e^x

A.8. The given order derivative present in the D.E. is  $y^{|}$ so its order is 1.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

 

Q9. y'′ + (y')²  + 2y = 0

A.9. The highest order derivative present in the D.E. is  $y^{\left|\right|}$ so its order is 2.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

 

Q10.  y'′ + 2y' + sin y = 0

A.10. The highest order derivative present in the D.E. is  $y^{\left|\right|}$ so its order is 2.

As the given D.E. is polynomial equation in its derivative, its degree is 1.

Q11. The degree of the differential equation   ${\left(\frac{d^{2}y}{d{x}^2}\right)}^3+{\left(\frac{dy}{dx}\right)}^2+sin\left(\frac{dy}{dx}\right)+1=0$  is:

(A) 3 

(B) 2 

(C) 1 

(D) Not defined

A.11. In the given D.E,

$sin\frac{dy}{dx}$ is a trigonometric function of derivative  $\frac{dy}{dx}$ . So it is not a polynomial equation so its derivative is not defined.

Hence, Degree of the given D.E. is not defined.

$\therefore$ Option (D) is correct.

 

Q12. The order of the differential equation   $2x^2\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}+y=0$  is:

(A) 2 

(B) 1 

(C) 0 

(D) Not defined

A.12. The highest order derivative present in the given D.E. is  $\frac{d^{2}y}{d{x}^2}$ and its order is 2.

$\therefore$ Option (A) is correct.

 

Ex-9.2

In each of the Questions 1 to 6 verify that the given functions (explicit) is a solution of the corresponding differential equation:

Q1.  y = e^x   + 1  :  y″ – y′ = 0

A.1. Given  $f{x}^n$ is  $y=e^x+1$

Differentiating with  $x$ we get,

$y^{|}=\frac{dy}{dx}=e^x$

Again,

$y^{\left|\right|}=\frac{d^{2}y}{d{x}^2}=e^x$

Substituting value of  $y^{\left|\right|}$ and  $y^{|}$ in the given D.E. we get

$L.H.S.=y^{\left|\right|}-y^{|}=e^x-e^x=0=R.H.S$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

 

Q2. y = x²  + 2x + C  :  y′ – 2x – 2 = 0

A.2. Given ,  $f{x}^n$ is  $y=x^2+2x+c$

So,  $y^{|}=2x+2$

Substituting value of  $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}-2x-2=2x+2-2x-2=0=R.H.S$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

 

Q3. y = cos x + C : y′ + sin x = 0

A.3. Given,  $f{x}^n$ is  $y=cosx+c$

So,  $y^{|}=-sinx$

Putting the value of  $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}+sinx=-sinx+sinx=0=R.H.S$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

Q4.  y = √1 + x² ; y/= $\frac{xy}{1+x^2}$

 

Q5. y = Ax   :  xy′ = y (x ≠ 0)

A.5. Given,  $y=Ax:$

So,  $y^{|}=\frac{Adx}{dx}=A$

Putting value of  $y^{|}$ in L.H.S. of the given D.E.

L.H.S=  $x{y}^{|}=xA=Ax=y$ =R.H.S

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

 

A.6. Given,  $y=xsinx$

So,  $y^{|}=x\frac{d}{dx}sinx+sinx\frac{dx}{dx}=xcosx+sinx$

Now, L.H.S of the given D.E  $=x{y}^{|}$

$=x\left(xcosx+sinx\right)$

$=x^{2}cosx+xsinx$

$y\text{'}=\frac{y^2}{1-xy}\left(xy\ne 1\right)$

A.7. Given,  $xy=logy+c$

Differentiate w.r.t. x we have

$\begin{array}{l}x\frac{dy}{dx}+y\frac{dx}{dx}=\frac{d}{dx}logy+\frac{d}{dx}C\\ \Rightarrow x\frac{dy}{dx}+y=\frac{1}{y}\frac{dy}{dx}+0\\ \Rightarrow x\frac{dy}{dx}-\frac{1}{y}\frac{dy}{dx}=-y\\ \Rightarrow \frac{dy}{dx}\left[x-\frac{1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}\left[\frac{xy-1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}=\frac{-y^2}{xy-1}=\frac{\left(-1\right)\times -y^2}{\left(-1\right)\times \left(xy-1\right)}=\frac{y^2}{1-xy}\\ \therefore y^{|}=\frac{y^2}{1-xy}\end{array}$

Hence, y is a Solution of the given D.E

 

Q8. y – cos y = x :  (y sin y + cos y + x) y′ = y

A.8. Given  $y-cosy=x$

Differentiate w.r.t ‘x’ we get

$\begin{array}{l}\frac{dy}{dx}-\left(-siny\right)\frac{dy}{dx}=\frac{dx}{dx}\\ \Rightarrow \frac{dy}{dx}\left[1+siny\right]=1\\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+siny}=y^{|}\end{array}$

So, L.H.S of given D.E  $=\left(ysiny+cosy+x\right)y^{|}$

$\begin{array}{l}=\left(ysiny+cosy+y-cosx\right)\left[\frac{1}{1+siny}\right]\\ =\frac{y\left(1+siny\right)}{\left(1+siny\right)}=y=R.H.S\end{array}$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E.

 

Q9. x + y = tan^-1 y   :   y²y' + y² + 1 = 0

A.9. Given,  $x+y=ta{n}^{-1}y$

Differentiate with ‘x’ we get

$\begin{array}{l}1+\frac{dy}{dx}=\frac{1}{1+y^2}\frac{dy}{dx}\\ =1+y^{|}=\frac{1}{1+y^2}{y}^{|}\\ =\left(1+y^{|}\right)\left(1+y^2\right)=y^{|}\\ =1+y^2{y}^{|}+y^{|}+y^2=y^{|}\\ =y^2{y}^{|}+y^2+1=0\end{array}$

$\therefore$ The given  $f{x}^n$ is a solution of the given D.E

 

 

Q11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0

(B) 2

(C) 3

(D) 4

A.11. The number of arbitrary constant is general solution of D.E of 4^th order is four.

$\therefore$ Option (D) is correct.

 

Q12. The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3

(B) 2

(C) 1

(D) 0

A.12. In a particular solution, there are no arbitrary constant.

Hence, option (D) is correct.

 

Ex:9.3

In each of the questions 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b

Q1.  $\frac{x}{a}+\frac{y}{b}=1$

A.1. Given: Equation of the family of curves   $\frac{x}{a}+\frac{y}{b}=1..........(i)$

Differentiating both sides of the given equation with respect to x, we get:

$\begin{array}{l}\frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0\\ \Rightarrow \frac{1}{a}+\frac{1}{b}{y}^{\text{'}}=0\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}0+\frac{1}{b}{y}^{"}=0\\ \Rightarrow \frac{1}{b}{y}^{"}=0\\ \Rightarrow y^{"}=0\end{array}$

Hence, the required differential equation of the given curve is  $y^{"}=0$ .

 

Q2.  $y^2=a\left(b^2-x^2\right)$

A.2. Given: Equation of the family of curves   $y^2=a\left(b^2-x^2\right)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}2y\frac{dy}{dx}=a\left(-2x\right)\\ \Rightarrow 2y{y}^{\text{'}}=-2ax\\ \Rightarrow y{y}^{\text{'}}=-ax..........(1)\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}.y^{\text{'}}+y{y}^{"}=-a\\ \Rightarrow {\left(y^{\text{'}}\right)}^2+y{y}^{"}=-a..........(2)\end{array}$

Dividing equation (2) by equation (1), we get:

$\begin{array}{l}\frac{{\left(y^{\text{'}}\right)}^2+y{y}^{"}}{y{y}^{\text{'}}}=\frac{-a}{-ax}\\ \Rightarrow xy{y}^{"}+x{\left(y^{\text{'}}\right)}^2-y{y}^{"}=0\end{array}$

This is the required differential equation of the given curve.

 

Q3.  $y=a{e}^{3x}+b{e}^{-2x}$

A.3. Given: Equation of the family of curves   $y=a{e}^{3x}+b{e}^{-2x}..........(i)$

Differentiating both sides with respect to x, we get:

$y^{\text{'}}=3a{e}^{3x}-2b{e}^{-2x}..........(ii)$

Again, differentiating both sides with respect to x, we get:

$y^{"}=9a{e}^{3x}-4b{e}^{-2x}..........(iii)$

Multiplying equation (i) with (ii) and then adding it to equation (ii), we get:

$\begin{array}{l}\left(2a{e}^{3x}+2b{e}^{-2x}\right)+\left(3a{e}^{3x}-2b{e}^{-2x}\right)=2y+y^{\text{'}}\\ \Rightarrow 5a{e}^{3x}=2y+y^{\text{'}}\\ \Rightarrow a{e}^{3x}=\frac{2y+y^{\text{'}}}{5}\end{array}$

Now, multiplying equation (i) with (iii) and subtracting equation (ii) from it, we get:

$\begin{array}{l}\left(3a{e}^{3x}+2b{e}^{-2x}\right)-\left(3a{e}^{3x}-2b{e}^{-2x}\right)=3y-y^{\text{'}}\\ \Rightarrow 5b{e}^{-2x}=3y-y^{\text{'}}\\ \Rightarrow b{e}^{-2x}=\frac{3y-y^{\text{'}}}{5}\end{array}$

Substituting the values of  $a{e}^{3x}$ and  $b{e}^{-2x}$ in equation (iii), we get:

$\begin{array}{l}{y}^{"}=9.\frac{\left(2y-y^{\text{'}}\right)}{5}+4\frac{\left(3y-y^{\text{'}}\right)}{5}\\ \Rightarrow y^{"}=\frac{18y+9y^{\text{'}}}{5}+\frac{12y-4y^{\text{'}}}{5}\\ \Rightarrow y^{"}=\frac{30y+5y^{\text{'}}}{5}\\ \Rightarrow y^{"}=6y+y^{\text{'}}\\ \Rightarrow y^{"}-y^{\text{'}}-6y=0\end{array}$

This is the required differential equation of the given curve.

 

Q4.  $y=e^{2x}\left(a+bx\right)$

A.4. Given: Equation of the family of curves   $y=e^{2x}\left(a+bx\right)..........(1)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}=2e^{2x}\left(a+bx\right)+e^{2x}.b\\ \Rightarrow y^{\text{'}}=e^{2x}\left(2a+2bx+b\right)..........(2)\end{array}$

Multiplying equation (1) with (2) and then subtracting it from equation (2), we get:

$\begin{array}{l}{y}^{\text{'}}-2y=e^{2x}\left(2a+2bx+b\right)-e^{2x}\left(2a+2bx\right)\\ \Rightarrow y^{\text{'}}-2y=b{e}^{2x}..........(3)\end{array}$

Differentiating both sides with respect to x, we get:

$y^{"}-2y^{\text{'}}=2b{e}^{2x}..........(4)$

Dividing equation (4) by equation (3), we get:

$\begin{array}{l}\frac{y^{"}-2y^{\text{'}}}{y^{\text{'}}-2y}=2\\ \Rightarrow y^{"}-2y^{\text{'}}=2y^{\text{'}}-4y\\ \Rightarrow y^{"}-4y^{\text{'}}+4y=0\end{array}$

This is the required differential equation of the given curve.

 

Q5.  $y=e^x\left(acosx+bsinx\right)$

A.5. Given: Equation of the family of curves  $y=e^x\left(acosx+bsinx\right)..........(i)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}=e^x\left(acosx+bsinx\right)+e^x\left(-acosx+bsinx\right)\\ \Rightarrow y^{\text{'}}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]..........(2)\end{array}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}{y}^{"}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]+e^x\left[-\left(a+b\right)sinx-\left(a-b\right)cosx\right]\\ y^{"}=e^x\left[2bcosx-2asinx\right]\\ y^{"}=2e^x\left(bcosx-asinx\right)\\ \Rightarrow \frac{y^{"}}{2}=e^x\left(bcosx-asinx\right)..........(3)\end{array}$

Adding equations (1) and (3), we get:

$\begin{array}{l}y+\frac{y^{"}}{2}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]\\ \Rightarrow y+\frac{y^{"}}{2}=y^{\text{'}}\\ \Rightarrow 2y+y^{"}=2y^{\text{'}}\\ \Rightarrow y^{"}-2y^{\text{'}}+2y=0\end{array}$

This is the required differential equation of the given curve.

 

Q6. Form the differential equation of the family of circles touching the y-axis at the origin.

A.6. The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is

$\begin{array}{l}{\left(x-a\right)}^2+y^2=a^2.\\ \Rightarrow x^2+y^2=2ax..........(1)\end{array}$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2x+2y{y}^{\text{'}}=2a\\ \Rightarrow x+y{y}^{\text{'}}=a\end{array}$

Now, on substituting the value of a in equation (1), we get:

$\begin{array}{l}{x}^2+y^2=2\left(x+y{y}^{\text{'}}\right)x\\ \Rightarrow x^2+y^2=2x^2+2xy{y}^{\text{'}}\\ \Rightarrow 2xy{y}^{\text{'}}+x^2=y^2\end{array}$

This is the required differential equation.

 

Q7. Find the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

A.7. The equation of the parabola having the vertex at origin and the axis along the positive y-axis is:

$x^2 =4ay$

Differentiating equation (1) with respect to x, we get:

$2x=4ay\text{'}$

Dividing equation (2) by equation (1), we get:

$\begin{array}{l}\frac{2x}{x^2}=\frac{4a{y}^{\text{'}}}{4ay}\\ \Rightarrow \frac{2}{x}=\frac{y^{\text{'}}}{y}\\ \Rightarrow x{y}^{\text{'}}=2y\\ \Rightarrow x{y}^{\text{'}}-2y=0\end{array}$

This is the required differential equation.

 

Q8. Form the differential equation of family of ellipse having foci on y-axis and centre at the origin.

A.8. The equation of the family of ellipses having foci on the y-axis and the centre at origin is as follows:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1..........(1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}\frac{2x}{b^2}+\frac{2y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow \frac{x}{b^2}+\frac{y{y}^{\text{'}}}{a^2}..........(2)\end{array}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}\Rightarrow \frac{1}{b^2}+\frac{y^{\text{'}}.y^{\text{'}}+y.y^{"}}{a^2}=0\\ \Rightarrow \frac{1}{b^2}+\frac{1}{a^2}\left({y^{\text{'}}}^2+y{y}^{"}\right)=0\\ \Rightarrow \frac{1}{b^2}=-\frac{1}{a^2}\left({y^{\text{'}}}^2+y{y}^{"}\right)\end{array}$

Substituting this value in equation (2), we get:

$\begin{array}{l}x\left[-\frac{1}{a^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)\right]+\frac{y{y}^{"}}{a^2}=0\\ \Rightarrow -x{\left(y^{\text{'}}\right)}^2-xy{y}^{"}+y{y}^{\text{'}}=0\\ \Rightarrow xy{y}^{"}+x{\left(y^{\text{'}}\right)}^2=0\end{array}$

This is the required differential equation

 

Q9. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.

A.9. The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1..........(1)$

Differentiating both sides of equation (1) with respect to x, we get:

$\begin{array}{l}\frac{2x}{a^2}-\frac{2y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow \frac{x}{a^2}-\frac{y{y}^{\text{'}}}{b^2}=0..........(2)\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}\frac{1}{a^2}-\frac{y^{\text{'}}.y^{\text{'}}+y.y^{"}}{b^2}=0\\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)=0\end{array}$

Substituting the value of  $\frac{1}{a^2}$ in equation (2), we get:

$\begin{array}{l}\frac{x}{b^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)-\frac{y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow x{\left(y^{\text{'}}\right)}^2+xy{y}^{"}-y{y}^{\text{'}}=0\\ \Rightarrow xy{y}^{"}+x{\left(y^{\text{'}}\right)}^2-y{y}^{\text{'}}=0\end{array}$

This is the required differential equation.

 

Q10. Form the differential equation of the family of circles having centres on y-axis and radius 3 units.

A.10. Let the centre of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

$\begin{array}{l}{x}^2+{\left(y-b\right)}^2=3^2\\ \Rightarrow x^2+{\left(y-b\right)}^2=9..........(1)\end{array}$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2x+2(y-b).y^{\text{'}}=0\\ \Rightarrow (y-b).y^{\text{'}}=-x\\ \Rightarrow y-b=\frac{-x}{y^{\text{'}}}\end{array}$

Substituting the value of  $(y-b)$ in equation (1), we get:

$\begin{array}{l}{x}^2+{\left(\frac{-x}{y^{\text{'}}}\right)}^2=9\\ \Rightarrow x^2\left[1+\frac{1}{{\left(y^{\text{'}}\right)}^2}\right]=9\\ \Rightarrow x^2\left({\left(y^{\text{'}}\right)}^2+1\right)=9{\left(y^{\text{'}}\right)}^2\\ \Rightarrow (x^2-9){\left(y^{\text{'}}\right)}^2+x^2=0\end{array}$

This is the required differential equation.

 

Q11. Which of the following differential equation has   $y=c_1{e}^x+c_2{e}^{-x}$  as the general solution:

$\begin{array}{l}(A)\frac{d^{2}y}{d{x}^2}+y=0\\ (B)\frac{d^{2}y}{d{x}^2}-y=0\\ (C)\frac{d^{2}y}{d{x}^2}+1=0\\ (D)\frac{d^{2}y}{d{x}^2}-1=0\end{array}$

A.11. Given:  $y=c_1{e}^x+c_2{e}^{-x}.........(1)$

Differentiating with respect to x, we get:

$\frac{dy}{dx}=c_1{e}^x-c_2{e}^{-x}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=c_1{e}^x+c_2{e}^{-x}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=y\\ \Rightarrow \frac{d^{2}y}{d{x}^2}-y=0\end{array}$

This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.

 

Q12. Which of the following differential equations has y = x as one of its particular solutions:

$\begin{array}{l}(A)\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=x\\ (B)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+xy=x\\ (C)\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0\\ (D)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+xy=0\end{array}$

A12. The given equation of curve is   $y = x$ ..

Differentiating with respect to x, we get:

$\frac{dy}{dx}=1 ..........(1)$

Again, differentiating with respect to x, we get:

$\frac{d^{2}y}{d{x}^2}=0..........(2)$

Now, on substituting the values of y,  $\frac{d^{2}y}{d{x}^2}$ and  $\frac{dy}{dx}$   from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0-x^2.1+x.x\\ =-x^2+x^2\\ =0\end{array}$

Therefore, option (C) is correct.

 

Ex:9.4

For each of the following differential equations in Exercise 1 to 10, find the general solution:

Q1.  $\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$

A.1. The given D.E is

$\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$

By separable of variable,

$\begin{array}{l}\Rightarrow dy=\frac{1-cosx}{1+cosx}dx\left\{\begin{array}{l}cos2x=1-2si{n}^{2}x\\ =2si{n}^{2}x=1-cos2x\\ =2si{n}^2\frac{x}{2}=1-cosx\\ cos2x=2co{s}^{2}x-1\end{array}\right\}\\ \Rightarrow dy=\frac{2si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}dx\\ \Rightarrow dy=ta{n}^2\frac{x}{2}dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int ta{n}^2\frac{x}{2}dx\left\{se{c}^{2}x=1+ta{n}^2\right\}}}\\ \Rightarrow y={\displaystyle \int \left(se{c}^2\frac{x}{2}-1\right)}dx\end{array}$

$\Rightarrow y=\frac{tan\frac{x}{2}}{\frac{1}{2}}-x+c$ c = constant

$\Rightarrow y=2tan\frac{x}{2}-x+c$ is the general solution.

 

 

Q3.  $\frac{dy}{dx}+y=1\left(y\ne 1\right)$

A3. Given,  $\frac{dy}{dx}+y=1$

$\Rightarrow \frac{dy}{dx}=1-y=-\left(y-1\right)$

By separable of variable,

$\frac{dy}{\left(y-1\right)}=-dx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(y-1\right)}=-{\displaystyle \int dx}}\\ \Rightarrow log\left|y-1\right|=-x+c\\ \Rightarrow \left|y-1\right|=e^{-x+c}\\ \Rightarrow y-1=\pm e^{-x}.e^c\end{array}$

$\Rightarrow y=1+Ac$ where  $A=\pm e^c$

Is the general solution.

 

Q4.  $se{c}^{2}xtanydx+se{c}^{2}ytanxdy=0$

A.4. Given,  $se{c}^{2}xtanydx+se{c}^{2}ytanxdy=0$

Dividing throughout by ‘  $tanxtany$ ’ we get,

$\begin{array}{l}\frac{se{c}^{2}xtany}{tanxtany}dx+\frac{se{c}^{2}ytanx}{tanxtany}dy=0\\ \Rightarrow \frac{se{c}^{2}x}{tanx}dx+\frac{se{c}^{2}y}{tany}dy=0\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{se{c}^{2}x}{tanx}dx+{\displaystyle \int \frac{se{c}^{2}y}{tany}dy=logc}}\\ \Rightarrow log\left|tanx\right|+log\left|tany\right|=logc\left\{{\displaystyle \int \frac{f^{|}\left(x\right)}{f\left(x\right)}}dx-log\left|f\left(x\right)\right|\right\}\\ \Rightarrow log\left|\left(tanx+tany\right)\right|=logc\end{array}$

$\Rightarrow tanxtany=\pm c$ is the required general solution.

 

Q5.  $\left(e^x+e^{-x}\right)dy-\left(e^x-e^{-x}\right)dx=0$

A.5. Given,  $\left(e^x+e^{-x}\right)dy-\left(e^x-e^{-x}\right)dx=0$

$\begin{array}{l}\Rightarrow \left(e^x+e^{-x}\right)dy=\left(e^x-e^{-x}\right)dx\\ \Rightarrow dy=\frac{e^x-e^{-x}}{e^x+e^{-x}}dx\end{array}$

Integrating both sides

$\Rightarrow {\displaystyle \int dy=}\frac{e^x-e^{-x}}{e^x+e^{-x}}dx\left\{\because {\displaystyle \int \frac{f^{|}\left(x\right)}{f\left(x\right)}}dx=log\left|x\right|\right\}$

$\Rightarrow y=log\left|e^x+e^{-x}\right|+c$ is the required general solution.

 

Q6.  $\frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)$

A.6. Given,  $\frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)$

$\Rightarrow \frac{dy}{\left(1+y^2\right)}=\left(1+x^2\right)dx$

Integrating both sides

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(1+y^2\right)}dy={\displaystyle \int \left(x^2+1\right)dx}}\\ \Rightarrow ta{n}^{-1}\frac{y}{1}=\frac{x^3}{3}+x+c\end{array}$

$\Rightarrow ta{n}^{-1}y=\frac{x^3}{3}+x+c$ is the general solution.

 

Q7.  $ylogydx-xdy=0$

A.7. Given,  $ylogydx-xdy=0$

$\begin{array}{l}\Rightarrow ylogydx=xdy\\ \Rightarrow \frac{dy}{ylogy}=\frac{dx}{x}\end{array}$

Integration both sides,

${\displaystyle \int \frac{dy}{ylogy}={\displaystyle \int \frac{dx}{x}}}$

Put log  $y=t\Rightarrow \frac{1}{y}=\frac{dt}{dy}\Rightarrow \frac{dy}{y}=dt$

Hence,  ${\displaystyle \int \frac{dt}{t}={\displaystyle \int \frac{dx}{x}}}$

$\begin{array}{l}\Rightarrow log\left|t\right|=log\left|x\right|+log\left|c\right|=log\left|xc\right|\\ \Rightarrow t=\pm xc\end{array}$

$\Rightarrow logy=ax$ where  $a=\pm c$

$\Rightarrow y=e^{ax}$ is the general solution.

 

Q8.  $x^5\frac{dy}{dx}=-y^5$

A.8. Given,  $x^5\frac{dy}{dx}=-y^5$

$\Rightarrow \frac{dy}{y^5}=-\frac{dx}{x^5}$

Integrating both sides

$\begin{array}{l}{\displaystyle \int \frac{dy}{y^5}=-{\displaystyle \int \frac{dx}{x^5}}}\\ \Rightarrow {\displaystyle \int y^{-5}dy=-{\displaystyle \int x^{-5}dx}}\end{array}$

$\begin{array}{l}\Rightarrow \frac{y^{-5+1}}{\left(-5+1\right)}=-\frac{x^{-5+1}}{\left(-5+1\right)}+c\\ \Rightarrow \frac{1}{-4y^4}=\frac{1}{4x^4}+c\end{array}$

$\Rightarrow \frac{1}{y^4}=\frac{1}{x^4}+4c$ is the general solution.

 

Q9.  $\frac{dy}{dx}=si{n}^{-1}x$

A.9. Given,  $\frac{dy}{dx}=si{n}^{-1}x$

$\Rightarrow dy=si{n}^{-1}xdx$

Integrating

 

Q.10.  $e^{x}tanydx+\left(1-e^x\right)se{c}^{2}ydy=0$

A.10. Given,  $e^{x}tanydx+\left(1-e^x\right)se{c}^{2}ydy=0$

Dividing throughout by  $\left(1-e^x\right)tany$ we get,

$\begin{array}{l}\frac{e^{x}tany}{\left(1-e^x\right)tany}dx+\frac{\left(1-e^x\right)se{c}^{2}y}{\left(1-e^x\right)tany}dy=0\\ =\frac{e^x}{1-e^x}dx+\frac{se{c}^{2}y}{tany}dy=0\end{array}$

Integrating both sides

$\begin{array}{l}={\displaystyle \int \frac{-e^x}{1-e^x}dx+{\displaystyle \int \frac{se{c}^{2}y}{tany}dy=clogc}}\\ =-log\left|1-e^x\right|+log\left|tany\right|=clogc\\ =log\frac{tany}{1-e^x}=logc\\ =\frac{tany}{1-e^x}=c\end{array}$

$=tany=\left(1-e^x\right)c$ is the general solution.

 

For each of the differential equations in Question 11 to 12, find a particular solution satisfying the given condition:

Q11.  $\left(x^3+x^2+x+1\right)\frac{dy}{dx}=2x^2+x$

A.11. The given D.E is  $\left(x^3+x^2+x+1\right)\frac{dy}{dx}=2x^2+x$

$\begin{array}{l}\Rightarrow dy=\left(\frac{2x^2+x}{x^3+x^2+x+1}\right)dx\\ \Rightarrow dy=\frac{2x^2+x}{x^2\left(x+1\right)+\left(x+1\right)}dx=\frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}dx\end{array}$

Integrating both sides we get,

${\displaystyle \int dy={\displaystyle \int \frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}dx}}$

Let,  $\frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{Bx+c}{x^2+1}$

$\begin{array}{l}\Rightarrow 2x^2+2=A\left(x^2+1\right)+\left(Bx+c\right)\left(x+1\right)\\ =A{x}^2+A+B{x}^2+Bx+Cx+C\\ =\left(A+B\right)x^2+\left(B+C\right)x^2+\left(A+C\right)\end{array}$

Comparing the co-efficient we get,

$\begin{array}{l}A+B=2------(1)\\ B+C=1------(2)\\ A+C=0------(3)\end{array}$

Subtracting equation (1) – (2), we get

$\begin{array}{l}A+B-\left(B+C\right)=2-1\\ \to A-C=1\end{array}$

But from equation (3)  $A=-C$ so, we get,

$\begin{array}{l}A\left(-C\right)-C=1\\ \to -2C=1\\ \to C=-\frac{1}{2}\\ \&A=-\left(-\frac{1}{2}\right)=\frac{1}{2}\end{array}$

And putting value of A in equation (1),

$\begin{array}{l}\frac{1}{2}+B=2\\ \Rightarrow B=2-\frac{1}{2}=\frac{4-1}{2}=\frac{3}{2}\end{array}$

Putting value of A,B and C in

$\begin{array}{l}\frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}=\frac{\frac{1}{2}}{x+1}+\frac{\frac{3}{2}x-\frac{1}{2}}{x^2+1}\\ =\frac{1}{2\left(x+1\right)}+\frac{3}{2}\left(\frac{x}{x^2+1}\right)-\frac{1}{2}\left(\frac{1}{x^2+1}\right)\end{array}$

Hence, the integration becomes

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int \frac{1}{2\left(x+1\right)}dx+{\displaystyle \int \frac{3}{4}\left(\frac{2x}{x^2+1}\right)dx-{\displaystyle \int \frac{1}{2}\left(\frac{1}{x^2+1}\right)dx}}}}\\ \Rightarrow y=\frac{1}{2}log\left(x+1\right)+\frac{3}{4}log\left(x^2+1\right)-\frac{1}{2}ta{n}^{-1}\frac{x}{1}+c\end{array}$

Given, At  $x=0,y=1$

Then,  $1=\frac{1}{2}log1+\frac{3}{4}log1-\frac{1}{2}ta{n}^{-1}\left(0\right)+C$

$\begin{array}{l}\Rightarrow 1=0+0-0+C\left\{\begin{array}{l}\because log1=0\\ ta{n}^{-1}0-0\end{array}\right\}\\ \Rightarrow c=1\end{array}$

$\therefore$ The required particular solution is:

$\begin{array}{l}y=\frac{1}{2}log\left(x+1\right)+\frac{3}{4}log\left(x^2+1\right)-\frac{1}{2}ta{n}^{-1}x+1\\ =\frac{1}{4}\left[2log\left(x+1\right)+3log\left(x^2+1\right)\right]-\frac{1}{2}ta{n}^{-1}x+1\\ =\frac{1}{4}\left[log{\left(x+1\right)}^2+log{\left(x^2+1\right)}^3\right]-\frac{1}{2}ta{n}^{-1}x+1\\ =\frac{1}{4}\left[log{\left(x+1\right)}^2{\left(x^2+1\right)}^3\right]-\frac{1}{2}ta{n}^{-1}x+1\end{array}$

 

Q12.  $x\left(x^2-1\right)\frac{dy}{dx}=1;y=0whenx=2$

A.12. The given D.E. is

$\begin{array}{l}x\left(x^2-1\right)\frac{dy}{dx}=1\\ \Rightarrow dy=\frac{dx}{x\left(x^2-1\right)}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy}={\displaystyle \int \frac{dx}{x\left(x^2-1\right)}}\\ \Rightarrow y={\displaystyle \int \frac{dx}{x\left(x^2-1\right)\left(x+1\right)}dx+c.}\end{array}$

Let,  $\frac{1}{x\left(x-1\right)\left(x+1\right)}=\frac{A}{x}+\frac{B}{x-1}+\frac{c}{x+1}$

$\begin{array}{l}1=A\left(x-1\right)\left(x+1\right)+B\left(x\right)\left(x+1\right)+C\left(x\right)\left(x-1\right)\\ =A\left(x^2-1\right)+B{x}^2+Bx+C{x}^2-Cx\\ =A{x}^2-A+B{x}^2+Bx+C{x}^2-Cx\\ =\left(A+B+C\right)x^2+\left(B-C\right)x-A\end{array}$

Comparing the coefficient,

$\begin{array}{l}-A=1\\ \Rightarrow A=-1------------(1)\\ \Rightarrow A+B+C=0---------(2)\\ \Rightarrow B-C=0\\ \Rightarrow B=C-------------(3)\end{array}$

Putting equation (1) & (2) in (1) we get,

$\begin{array}{l}-1+B+B=0\\ \Rightarrow -1+2B=0\\ \Rightarrow B=\frac{1}{2}=C\end{array}$

So,  $\frac{1}{x\left(x-1\right)\left(x+1\right)}=-\frac{1}{x}+\frac{\frac{1}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}$

$=-\frac{1}{x}+\frac{1}{2\left(x-1\right)}+\frac{1}{2\left(x+1\right)}$

Integrating becomes,

$\begin{array}{l}y={\displaystyle \int -\frac{1}{x}dx+{\displaystyle \int \frac{1}{2\left(x-1\right)}dx+{\displaystyle \int \frac{1}{2\left(x+1\right)}dx+c}}}\\ =-log\left(x\right)+\frac{1}{2}log\left(x-1\right)+\frac{1}{2}log\left(x+1\right)+c\\ =\frac{1}{2}\left[-2log\left(x\right)+log\left(x-1\right)+log\left(x+1\right)\right]+c\\ =\frac{1}{2}\left[-log{x}^2+log\left(x+1\right)\left(x-1\right)\right]+c\\ =\frac{1}{2}log\frac{x^2-1}{x^2}+c\end{array}$

Given,  $y=0whenx=2.$

Then,  $0=\frac{1}{2}log\frac{2^2-1}{2^2}+c$

$\begin{array}{l}\Rightarrow 0=\frac{1}{2}log\frac{3}{4}+c\\ \Rightarrow c=-\frac{1}{2}log\frac{3}{4}\end{array}$

$\therefore$ The required particular solution is

$y=\frac{1}{2}log\frac{x^2-1}{x^2}-\frac{1}{2}log\frac{3}{4}$

 

Q13.  $cos\left(\frac{dx}{dy}\right)=a\left(a\in R\right);y=1whenx=0$

A13. Given, D.E. is

$\begin{array}{l}cos\frac{dy}{dx}=a\\ \Rightarrow \frac{dy}{dx}=co{s}^{-1}\left(a\right)\\ \Rightarrow dy=co{s}^{-1}\left(a\right)dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int co{s}^{-1}\left(a\right)dx}}\\ \Rightarrow y=co{s}^{-1}\left(a\right)\times x+c\\ \Rightarrow y=xco{s}^{-1}\left(a\right)dx\end{array}$

Given,  $y=1,atx=0$

Then,  $1=0co{s}^{-1}\left(a\right)+c$

$\Rightarrow c=1$

$\therefore$ The required particular solution is

 

Q14.  $\frac{dy}{dx}=ytanx;y=1whenx=0$

A14. Given,  $\frac{dy}{dx}=ytanx$

$\Rightarrow \frac{dy}{y}=tanxdx$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{dy}{y}}={\displaystyle \int tanxdx}\\ \Rightarrow logy=log\left|secx\right|+logc\\ \Rightarrow logy=log\left|csecx\right|\\ \Rightarrow y=c_{1}secx\left(where,c_1=\pm c\right)\end{array}$

As,  $y=1,at,x=0$ we have,

$1=c_{1}sec\left(0\right)=c\Rightarrow c=1$

$\therefore$ The required particular solution is  $y=secx$ .

 

Q15. Find the equation of the curve passing through the point (0, 0) and whose differential equation is y' = e^x sin x

A.15. The given D.E. is  $y^1=e^{x}sinx$

$dy=e^{x}sinxdx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int e^{x}sinxdx}}\\ \Rightarrow y=I+c\end{array}$

Where,  $I={\displaystyle \int e^{x}sinxdx}$

$\begin{array}{l}=sinx{\displaystyle \int e^{x}dx-{\displaystyle \int \frac{d}{dx}sinx{\displaystyle \int e^{x}dx.dx}}}\\ =sinx.e^x-{\displaystyle \int cosx{e}^{x}dx}\\ =sinx{e}^x-\left\{cosx{\displaystyle \int e^{x}dx}-{\displaystyle \int \frac{d}{dx}\left(cosx\right).{\displaystyle \int I={\displaystyle \int e^{x}x}dx}}\right\}\\ =sinx.e^x-\left\{cosx{e}^x+{\displaystyle \int sinx{e}^{x}dx}\right\}\\ =sinx.e^x-cosx{e}^x-I\\ \Rightarrow I+I=e^x\left(sinx-cosx\right)\\ \Rightarrow I=\frac{e^x}{2}\left(sinx-cosx\right)+c\end{array}$

Hence,  $y=\frac{e^x}{2}\left(sinx-cosx\right)+c$

When the curve passed point (0,0),

$\begin{array}{l}y=0,at,x=0\\ \Rightarrow 0=\frac{e^x}{2}\left(sin0-cos0\right)+c\\ \Rightarrow \frac{e^0}{2}\left(0-1\right)=c\\ \Rightarrow c=\frac{1}{2}\end{array}$

$\therefore$ The required equation of the curve is  $y=\frac{e^x}{2}\left(sinx-cosx\right)+\frac{1}{2}$

$\begin{array}{l}\Rightarrow 2y=e^x\left(sinx-cosx\right)+1\\ \Rightarrow 2y-1=e^x\left(sinx-cosx\right)\end{array}$

 

Q.16. For the differential equation   $xy\frac{dy}{dx}=\left(x+2\right)\left(y+2\right)$ find the solution curve passing through the point (1,-1)

A16.The Given D.E is

$\begin{array}{l}xy\frac{dy}{dx}=\left(x+2\right)\left(y+2\right)\\ \Rightarrow y\frac{dy}{y+2}=\frac{\left(x+2\right)}{2}dx\\ \Rightarrow \frac{y+2-2}{y+2}dy=\left(\frac{x}{x}+\frac{2}{x}\right)dx\\ \Rightarrow \left(1-\frac{2}{y+2}\right)dy=\left(1+\frac{2}{x}\right)dydx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \left(1-\frac{2}{y+2}\right)dy={\displaystyle \int \left(1+\frac{2}{x}\right)dydx}}\\ \Rightarrow y-2log\left|y+2\right|=x+2log\left|x\right|+c\\ \Rightarrow y-log{\left(y+2\right)}^2=x+log{x}^2+c\\ \Rightarrow y-x=log{\left(y+2\right)}^2+log{x}^2+c\\ \Rightarrow y-x=log\left[{\left(y+2\right)}^2.x^2\right]+c\end{array}$

A the curve passes through (-1,1) then  $y=-2,at,x=1$

So,  $-1-1=log{\left(-1+2\right)}^2.{\left(1\right)}^2+c$

$\begin{array}{l}\Rightarrow -2=log1+c\\ \Rightarrow c=-2\end{array}$

$\therefore$ The required equation of curve is,

$y-x=log\left[{\left(y+2\right)}^2{x}^2\right]-2$

 

Q.17 Find the equation of the curve passing through the point (0,-2) given that at any point (x,y) on the curve the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.

A.17. The slope of the tangent to then curve is  $\frac{dy}{dx}$

$\begin{array}{l}\frac{dy}{dx}.y=x\\ \Rightarrow y.dy=xdx\end{array}$

So,

Integrating both sides,

$\begin{array}{l}{\displaystyle \int y.dy={\displaystyle \int xdx}}\\ \Rightarrow {\displaystyle \int \frac{y^2}{2}=\frac{x^2}{2}+c}\\ \Rightarrow y^2=x^2+A,Where,A=2c\end{array}$

As the curve passes through (0,-2) we have,

$\begin{array}{l}{\left(-2\right)}^2=0^2+A\\ \Rightarrow A=4\end{array}$

$\therefore$ The equation of the curve is

$y^2=x^2+4$

 

Q18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (–4, –3). Find the equation of the curve given that it passes through (–2, 1).

A18. The slope of tangent is  $\frac{dy}{dx}$ and slope of line joining line (-4,-3) and point say P(x,y)

$\frac{y-\left(-3\right)}{x-\left(-4\right)}=\frac{y+3}{x+4}$

So,  $\frac{dy}{dx}=2\left(\frac{y+3}{x+4}\right)$

$\Rightarrow \frac{dy}{y+3}=\frac{2}{x+4}dx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{y+3}={\displaystyle \int \frac{2}{x+4}dx}}\\ \Rightarrow log\left|y+3\right|=2log\left|x+4\right|+log\left|c\right|\\ \Rightarrow log\left|y+3\right|=log{\left(x+4\right)}^2+log\left|c\right|\\ \Rightarrow log\left|y+3\right|=log\left|c{\left(x+4\right)}^2\right|\\ \Rightarrow y+3=c_1{\left(x+4\right)}^2,where,c_1=\pm c\end{array}$

Since, the curve passes through (-2,1) we get,

$\begin{array}{l}y=1,at,x=-2\\ \Rightarrow 1+3=c{\left(-2+4\right)}^2\\ \Rightarrow 4=c\times 4\\ \Rightarrow c=1\end{array}$

$\therefore$ The equation of the curve is  $y+3={\left(x+4\right)}^2$

 

Q19. The volume of the spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

A.19. Let ‘r’ and U be the radius and volume of the spherical balloon.

Then,  $\frac{dU}{dt}=k,$ k = constant

$\begin{array}{l}\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=k\\ \Rightarrow 4\pi r^2\frac{dr}{dt}=k\\ \Rightarrow 4\pi r^{2}dr=kdt\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int 4\pi r^{2}dr={\displaystyle \int kdt}}\\ \Rightarrow \frac{4}{3}\pi r^3=kt+c\end{array}$

Given at t = 0, r = 3

So, 4π(3)³ = c

C = 36π

And, at t=3, r=6

So,  $\frac{4}{3}\pi {\left(6\right)}^3=3k+36\pi \left(c=36\pi \right)$

$\begin{array}{l}\Rightarrow 288\pi -36\pi =3k\\ \Rightarrow k=\frac{252\pi }{3}=84\pi \end{array}$

Hence, putting value of c and k in,

$\frac{4}{3}\pi r^3=kt+c$ , we get,

$\begin{array}{l}\frac{4}{3}\pi r^3=84\pi .t+36\pi \\ \Rightarrow r^3=\frac{3}{4\pi }\left(84\pi .t+36\pi \right)\\ \Rightarrow r^3=63t+27\\ \Rightarrow r={\left[63t+27\right]}^{\frac{1}{3}}\end{array}$

 

Q20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).

A.20. Let P, r and t be the principal rate and time respectively.

Then, increase in principal  $\frac{dP}{dt}=P\times r\mathrm{\%}$

$\begin{array}{l}\Rightarrow \frac{dP}{dt}=P.\frac{r}{100}\\ \Rightarrow \frac{dP}{P}=\frac{r}{100}dt\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dP}{P}={\displaystyle \int \frac{r}{100}dt}}\\ \Rightarrow logP=\frac{rt}{100}+c\\ \Rightarrow P=e^{\frac{rt}{100}+c}\end{array}$

Given at t=0,P=100

So,  $100=e^{\frac{r\times 0}{100}+c}$

$\begin{array}{l}\Rightarrow 100=e^0\times e^c\\ \Rightarrow e^c=100\left(\because e^0=1\right)\end{array}$

And at  $t=10,P=2\times 100=200$

So,  $200=e^{\frac{r10}{100}+c}$

$\begin{array}{l}\Rightarrow 200=e^{\frac{r}{10}}.e^e\\ \Rightarrow e^{\frac{r}{10}}=\frac{200}{100}=2\end{array}$

$\begin{array}{l}\Rightarrow \frac{r}{10}=log2\\ \Rightarrow \frac{r}{10}=0.6931\\ \Rightarrow r=6.931\end{array}$

Hence, the rate is 6.931%

 

Q21.  In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years

(e^0.5  = 1.645).

A.21. Let P and t the principal and time respectively.

Then, increase in principal  $\frac{dP}{dt}=P\times 5\mathrm{\%}$

$\Rightarrow \frac{dP}{P}=\frac{5}{100}dt$

Integrating both sides,

$\begin{array}{l}\Rightarrow {\displaystyle \int \frac{dP}{P}={\displaystyle \int \frac{1}{20}dt}}\\ \Rightarrow logP=\frac{t}{20}+c\\ \Rightarrow P=e^{\frac{t}{20}+c}\end{array}$

At, t=0,P=1000

So,  $\begin{array}{l}1000=e^{\frac{0}{20}+c}\\ \Rightarrow e^c=1000\end{array}$

And at t=10,

$\begin{array}{l}P=e^{\frac{10}{100}+c}=e^{0.5}.e^c\\ \Rightarrow P=1.648\times 1000=1648\end{array}$

P = ₹1648

Q22. In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present.

A.22. Let ‘x’ be the number of bacteria present in instantaneous time t.

Then,  $\frac{dx}{dt}\propto x$

$\Rightarrow \frac{dx}{dt}=kx,where,k=$ constant of proportionality.

$\Rightarrow \frac{dx}{x}=kdt$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dx}{x}={\displaystyle \int kdt}}\\ \Rightarrow logx=kt+c\end{array}$

Given, at  $t=0,x=x_0(say)then,$

$log{x}_0=c\left(Initial,x_0=100000\right)$

So, the differential equation is

$\begin{array}{l}logx=kt+log{x}_0\\ \Rightarrow logx-log{x}_0=kt\\ \Rightarrow log\frac{x}{x_0}=kt\end{array}$

As the bacteria number increased by 10% in 2 hours.

The number of bacteria increased in 2hours  $=10\mathrm{\%}\times 100000=10000$

Hence, at t=2,

$x=100000+10000=110000$

So,  $\begin{array}{l}log\frac{110000}{100000}=2k\\ \Rightarrow k=\frac{1}{2}log\left(\frac{11}{10}\right)\end{array}$

Hence,  $log\frac{x}{x_0}=\left[\frac{1}{2}log\frac{11}{10}\right]\times t$

$when,x=200000,$ then we get,

$log\frac{200000}{100000}=\frac{1}{2}log\frac{11}{10}\times t$

$\begin{array}{l}\Rightarrow 2log2=log\left(\frac{11}{10}\right)\times t\\ \Rightarrow t=\frac{2log2}{log\left(\frac{11}{10}\right)}hours\end{array}$

 

Q23. The general solution of the differential equation   $\frac{dy}{dx}=e^{x+y}$ is:

(A) e^x + e^-y = C

(B) e^x + e^y = C

(C) e^-x + e^y = C

(D) e ^x + e^-y = C

A.23. The given D.E. is

$\begin{array}{l}\frac{dy}{dx}=e^{x+y}\\ \Rightarrow \frac{dy}{dx}=e^x.e^y\\ \Rightarrow \frac{dy}{e^y}=e^{x}dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{e^y}={\displaystyle \int e^{x}dx}}\\ \Rightarrow \frac{e^{-y}}{-1}=e^x+c_1\\ \Rightarrow e^{-y}=-e^x-c_1\\ \Rightarrow e^{-y}+e^x=c,where,c=-c_1\end{array}$

$\therefore$ Option (A) is correct.

 

EX-9.5

In each of the following Questions 1 to 5, show that the differential equation is homogenous and solve each of them:

Q1.  $\left(x^2+xy\right)dy=\left(x^2+y^2\right)dx$

A.1. The given D.E. is

$\begin{array}{l}\frac{dy}{dx}=\frac{x^2+y^2}{x^2+x+y}=\frac{x^2\left(1+\frac{y^2}{x^2}\right)}{x^2\left(1+\frac{y}{x}\right)}=F\left(x,y\right)\\ Then,F\left(\lambda x,\lambda y\right)=\frac{{\lambda }^2{x}^2}{{\lambda }^2{x}^2}\left[\frac{1+\frac{{\lambda }^2{y}^2}{{\lambda }^2{x}^2}}{1+\frac{\lambda y}{\lambda x}}\right]={\lambda }^{0}F\left(x,y\right)\\ ={\lambda }^{2}F\left(x,y\right)\end{array}$

Hence,  $F\left(x,y\right)$ is a homogenous  $f{x}^n$ of degree 2.

To solve it, let

$\begin{array}{l}y=vx,sothat,\frac{dy}{dx}=v.\frac{dx}{dx}+x\frac{dv}{dx}\\ v=\frac{y}{x}\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}$

The D.E. now becomes,

$\begin{array}{l}v+x\frac{dv}{dx}=\frac{1+v^2}{1+v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v=\frac{1+v^2-v-v^2}{1+v}=\frac{1-v}{1+v}\\ \Rightarrow \left(\frac{1+v}{1-v}\right)dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{1+v}{1-v}dv={\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow {\displaystyle \int \frac{1+v}{1-v}dv=logx+c}\\ \Rightarrow {\displaystyle \int \frac{2-\left(1-v\right)}{1-v}dv=logx+c}\\ \Rightarrow {\displaystyle \int \frac{2}{1-v}dv-{\displaystyle \int 1dv=logx+c}}\\ \Rightarrow \frac{2log\left|1-v\right|}{-1}-v=logx+c\\ \Rightarrow log{\left(1-v\right)}^2+v=-logx-c\end{array}$

Put  $v=\frac{y}{x},$

$\begin{array}{l}log{\left(1-\frac{y}{x}\right)}^2+\frac{y}{2}=-logx+c\\ \Rightarrow log{\left(\frac{x-y}{x}\right)}^2+logx=-\frac{y}{x}-c\end{array}$

$\begin{array}{l}\Rightarrow log\left[{\left(\frac{x-y}{x}\right)}^2\times x\right]=-\frac{y}{x}-c\\ \Rightarrow \frac{x-y}{x}=e^{-\frac{y}{x}-c}=c_1{e}^{-\frac{y}{x}-c},where,c_{1}e\end{array}$

$\Rightarrow {\left(x-y\right)}^2=c_1.x{e}^{-\frac{y}{x}}$ is the required solution of the D.E.

 

Q2.  $y\text{'}=\frac{x+y}{x}$

A.2. The given D.E. is

$\begin{array}{l}{y}^1=\frac{x+y}{x}\\ =\frac{dy}{dx}=1+\frac{y}{x}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E is homogenous.

Let,  $y=vx\to \frac{y}{x}=v,sothat,\frac{dy}{dx}=v+\frac{xdv}{dx}$

So, the D.E. becomes

$\begin{array}{l}v+\frac{xdv}{dx}=1+v\\ \Rightarrow \frac{xdv}{dx}=1\\ \Rightarrow dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dv={\displaystyle \int \frac{dx}{x}}}\\ v=log\left|x\right|+c\end{array}$

Putting  $v=\frac{y}{x}$ back we get,

$\frac{y}{x}=log\left|x\right|+c$

 

Q3.  $\left(x-y\right)dy-\left(x+y\right)dx=0$

A.3. The Given D.E. is  $\left(x-y\right)dy-\left(x+y\right)dx=0$

$\begin{array}{l}\Rightarrow \left(x-y\right)dy=\left(x+y\right)dx\\ \Rightarrow \frac{dy}{dx}=\frac{x+y}{x-y}=\frac{x\left(1+\frac{y}{x}\right)}{x\left(1-\frac{y}{x}\right)}=\frac{1+\frac{y}{x}}{1-\frac{y}{x}}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let,  $y=vx=\frac{y}{x}=v,so,\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

Then,  $\begin{array}{l}v+x\frac{dv}{dx}=\frac{1+v}{1-v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-1+v^2}{1-v}=\frac{1+v^2}{1-v}\\ \Rightarrow \left[\frac{1-v}{1+v^2}\right]dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{1-v}{1+v^2}dv={\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow {\displaystyle \int \frac{1}{1+v^2}dv-\frac{1}{2}{\displaystyle \int \frac{2v}{1+v^2}dv=logx+c}}\\ \Rightarrow ta{n}^{-1}v-\frac{1}{2}log\left(1+v^2\right)=logx+c\end{array}$

Putting back  $v=\frac{y}{x},weget,$

$\begin{array}{l}=ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left|1+\frac{y^2}{x^2}\right|=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left|\frac{x^2+y^2}{x^2}\right|=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}\left[log\left(x^2+y^2\right)-log{x}^2\right]=logx+c\end{array}$

$\begin{array}{l}=ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left(x^2+y^2\right)+\frac{1}{2}log{x}^2=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left(x^2+y^2\right)+log{\left(x^2\right)}^{\frac{1}{2}}=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left(x^2+y^2\right)+logx=log+c\\ =ta{n}^{-1}\frac{y}{x}=\frac{1}{2}log\left(x^2+y^2\right)+c\end{array}$

 

Q4.  $\left(x^2-y^2\right)dx+2xydy=0$

A.4. The Given D.E. is

$\begin{array}{l}\left(x^2-y^2\right)dx+2xydy=0\\ \Rightarrow 2\times ydy=-\left(x^2-y^2\right)dx\\ =\frac{dy}{dx}=-\frac{\left(x^2-y^2\right)}{2xy}=\frac{\left(y^2-x^2\right)}{2xy}\\ =\frac{1}{2}\frac{\left(\frac{y^2-x^2}{x^2}\right)}{\left(\frac{xy}{x^2}\right)}\\ =\frac{1}{2}\frac{\left[{\left(\frac{y}{x}\right)}^2-1\right]}{\left(\frac{y}{x}\right)}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let,  $y=vx\Rightarrow \frac{y}{x}=v,sothat,\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

$\begin{array}{l}\Rightarrow v+x\frac{dv}{dx}=\frac{1}{2}\left[\frac{v^2-1}{v}\right]\\ \Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}=\frac{-1-v^2}{2v}\\ \Rightarrow \left(\frac{2v}{1+v^2}\right)dv=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

Putting back  $v=\frac{y}{x}$ we get,

$\begin{array}{l}\Rightarrow x\left[\frac{y^2}{x^2}+1\right]=c\\ \Rightarrow \frac{y^2+x^2}{x}=c\end{array}$

$\Rightarrow y^2+x^2=xc$ is the required solution.

 

Q5.  $x^2\frac{dy}{dx}=x^2-2y^2+xy$

A5. The given D.E. is

$\begin{array}{l}{x}^2\frac{dy}{dx}=x^2-2y^2++xy\\ \Rightarrow \frac{dy}{dx}=\frac{x^2-2y^2++xy}{x^2}=1-\frac{2y^2}{x^2}+\frac{y}{x}\\ \Rightarrow \frac{dy}{dx}=1-2{\left(\frac{y}{x}\right)}^2+\frac{y}{x}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the D.E. is homogenous  $f{x}^n$

Let,  $y=vx.=\frac{y}{x}=v$ so that,  $\frac{dy}{dx}=v+x\frac{dv}{dx}$ is the D.E.

Thus,  $v+x\frac{dv}{dx}=1-2v^2+v$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=1-2v^2\\ \Rightarrow \frac{dv}{1-2v^2}=\frac{dx}{x}\end{array}$

Integrating both sides,

Integrating both sides,

 

Q7.  $\left\{xcos\left(\frac{y}{x}\right)+ysin\left(\frac{y}{x}\right)\right\}ydx=\left\{ysin\left(\frac{y}{x}\right)+xcos\left(\frac{y}{x}\right)\right\}xdy$

A.7. The given D.E is

.  $\begin{array}{l}\left\{xcos\left(\frac{y}{x}\right)+ysin\left(\frac{y}{x}\right)\right\}ydx=\left\{ysin\left(\frac{y}{x}\right)-xcos\left(\frac{y}{x}\right)\right\}xdy\\ \Rightarrow \frac{dy}{dx}=\frac{\left\{xcos\frac{y}{x}+ysin\frac{y}{x}\right\}y}{\left\{ysin\frac{y}{x}-xcos\frac{y}{x}\right\}x}=\frac{xycos\frac{y}{x}+y^{2}sin\frac{y}{x}}{xysin\frac{y}{x}-x^{2}cos\frac{y}{x}}\end{array}$

$\Rightarrow \frac{\frac{y}{x}cos\frac{y}{x}+{\left(\frac{y}{x}\right)}^2-sin\frac{y}{x}}{\left(\frac{y}{x}\right)sin\frac{y}{x}-cos\frac{y}{x}}$ {Dividing numerator and denominator by  $x^2$ }

$=f\left(\frac{y}{x}\right)$

Hence, the given D.E is homogenous.

Let,  $y=vx=\frac{y}{x}=v$ so that  $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E.

Then,  $v+x\frac{dv}{dx}=\frac{vcosv+v^{2}sinv}{vsinv-cosv}$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=\frac{vcosv+v^{2}sinv}{vsinv-cosv}-v=\frac{vcosv+v^{2}sinv-v^{2}sinv+vcosv}{vsinv-cosv}\\ \Rightarrow \frac{vcosv}{vsinv-cosv}\\ \Rightarrow \left(\frac{vsinv-cosv}{vcosv}\right)dv=\frac{2dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{vsinv-cosv}{vcosv}dv=2{\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow {\displaystyle \int tanvdv-{\displaystyle \int \frac{1}{v}dv=2log\left|x\right|+}}log\left|c\right|\\ \Rightarrow log\left|secv\right|-log\left|v\right|=log{x}^2+logc\\ \Rightarrow log\left|\frac{secv}{v}\right|=logc{x}^2\\ \Rightarrow \frac{secv}{v}=c{x}^2\\ \Rightarrow secv=c{x}^{2}v\end{array}$

Putting back  $\Rightarrow v=\frac{y}{x}=c{x}^2\frac{y}{x}=cxy$

$\begin{array}{l}sec\frac{y}{x}=c{x}^2\frac{y}{x}=cxy\\ \Rightarrow \frac{1}{cos\frac{y}{x}}=cxy\\ \Rightarrow \frac{1}{c}=xycos\frac{y}{x}\end{array}$

$\Rightarrow xycos\frac{y}{x}=c_1$ where  $c_1=\frac{1}{c}$ .

 

Q8.  $x\frac{dy}{dx}-y=xsin\left(\frac{y}{x}\right)=0$

A.8. The given D.E. is  $\frac{xdy}{dx}-y+xsin\left(\frac{y}{x}\right)=0$

$\begin{array}{l}\Rightarrow \frac{xdy}{dx}=y-xsin\left(\frac{y}{x}\right)\\ \Rightarrow \frac{dy}{dx}=\frac{y-xsin\left(\frac{y}{x}\right)}{x}=\frac{y}{x}-sin\frac{y}{x}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let,  $y=vx=\frac{y}{x}=v$ so that  $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E.

Then,  $v+x\frac{dv}{dx}=v-sinv$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=-sinv\\ \Rightarrow \frac{dv}{sinv}=-\frac{dx}{x}\\ \Rightarrow cosecvdv=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int cosecvdv=}{\displaystyle \int -\frac{dx}{x}}\\ \Rightarrow log\left|cosecv-cotv\right|=-logx+logc\\ \Rightarrow log\left|cosecv-cotv\right|=log\frac{c}{x}\\ \Rightarrow cosecv-cotv=\frac{c}{x}\end{array}$

Putting back  $v=\frac{y}{x}$ we get,

$\begin{array}{l}cosec\frac{y}{x}-cot\frac{y}{x}=\frac{c}{x}\\ \Rightarrow \frac{1}{sin\frac{y}{x}}-\frac{cos\frac{y}{x}}{sin\frac{y}{x}}=\frac{c}{x}\end{array}$

$\Rightarrow x\left[1-cos\frac{y}{x}\right]=csin\left(\frac{y}{x}\right)$ is the solution of the D.E.

 

Q9.  $ydx=xlog\left(\frac{y}{x}\right)dy-2xdy=0$

A.9. The given D.E is

$\begin{array}{l}ydx+xlog\left(\frac{y}{x}\right).dy-2xdy=0\\ \Rightarrow ydx=\left[2xdy-xlog\left(\frac{y}{x}\right)dy\right]\\ \Rightarrow ydx=\left[2x-xlog\left(\frac{y}{x}\right)\right]dy\\ \Rightarrow \frac{dy}{dx}=\frac{y}{2x-xlog\left(\frac{y}{x}\right)}=\frac{y}{x\left(2-log\frac{y}{x}\right)}\\ =\frac{\frac{y}{x}}{2-log\frac{y}{x}}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E is homogenous.

Let,  $y=vx=\frac{y}{x}=v$ so that  $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E.

Then,  $v+x\frac{dv}{dx}=\frac{v}{2-logv}$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=\frac{v}{2-logv}-v=\frac{v-2v+vlogv}{2-logv}=\frac{vlogv-v}{2-logv}\\ \Rightarrow \frac{2logv}{v\left[logv-1\right]}dv=\frac{dx}{x}\\ \Rightarrow \frac{1+1-logv}{v\left[logv-1\right]}dv=\frac{dx}{x}\\ \Rightarrow \frac{1-\left[logv-1\right]}{v\left[logv-1\right]}dv=\frac{dx}{x}\end{array}$

$\Rightarrow \left[\frac{1}{v\left(logv-1\right)}-\frac{1}{v}\right]dv=\frac{dx}{x}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \left[\frac{1}{v\left(logv-1\right)}-\frac{1}{v}\right]dv={\displaystyle \int \frac{dx}{x}}}\\ {\displaystyle \int \frac{dv}{v\left(logv-1\right)}}-logv=logx+logc\end{array}$

Let,  $logv-1=t,so,\frac{d}{dv}\left(logv-1\right)=\frac{dt}{dv}$

$\begin{array}{l}\Rightarrow \frac{1}{v}=\frac{dt}{dv}\\ \Rightarrow \frac{dv}{v}=dt\\ \Rightarrow {\displaystyle \int \frac{dt}{t}-logv=logx+logc}\\ \Rightarrow logt-logv=logx+logc\\ \Rightarrow log\left|logv-1\right|-logv=logcx\end{array}$

Putting back  $v=\frac{y}{x}$ we get,

$\begin{array}{l}log\left|log\left(\frac{y}{x}\right)-1\right|-log\frac{y}{x}=log\frac{c}{x}\\ =log\left[\frac{log\left(\frac{y}{x}\right)-1}{\frac{y}{x}}\right]=log\frac{c}{x}\\ =\frac{y}{x}\left[log\left(\frac{y}{x}\right)-1\right]=cx\end{array}$

$=log\left(\frac{y}{x}\right)-1=cy$ is the required solution.

 

Q10.  $\left(1+e^{\frac{x}{y}}\right)dx+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)dy=0$

A10. The given D.E. is

$\begin{array}{l}\left(1+e^{\frac{x}{y}}\right)dx+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)dy=0\\ \Rightarrow \left(1+e^{\frac{x}{y}}\right)dx=-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)dy\\ \Rightarrow \frac{dx}{dy}=-\frac{e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}=f\left(\frac{x}{y}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let,  $x=yv=\frac{y}{x}$ so that  $\frac{dy}{dx}=v+y\frac{dx}{dy}$ in the D.E.

Then,  $v+y\frac{dv}{dy}=\frac{-e^v\left(1-v\right)}{1+e^v}$

$\begin{array}{l}\Rightarrow y\frac{dv}{dy}=\frac{v{e}^v-e^v}{1+e^v}-v=\frac{v{e}^v-e^v-v-v{e}^v}{1+e^v}\\ \Rightarrow y\frac{dv}{dy}=-\frac{\left(e^v+v\right)}{1+e^v}\\ \Rightarrow \left(\frac{1+e^v}{e^v+v}\right)dv=-\frac{dy}{y}\end{array}$

Integrating both sides we get,

$log\left|e^v+v\right|=-log\left|y\right|+log\left|c\right|$

Putting back  $v=\frac{x}{y}$ we get,

$\begin{array}{l}log\left|e^{\frac{x}{y}}+\frac{x}{y}\right|=log\left|\frac{c}{y}\right|\\ =e^{\frac{x}{y}}+\frac{x}{y}=\frac{c}{y}\end{array}$

$=x+y{e}^{\frac{x}{y}}=c$ is the general solution.

 

For each of the differential equations in Questions from 11 to 15, find the particular solution satisfying the given condition

Q11. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1

A.11. The given D.E. is

$\begin{array}{l}\left(x+y\right)dy+\left(x-y\right)dx=0\\ =\left(x+y\right)dy=-\left(x-y\right)dx\\ =\frac{dy}{dx}=\frac{y-x}{x+y}=\frac{\frac{y-x}{x}}{\frac{x+y}{y}}=\frac{\frac{y}{x}-1}{1+\frac{y}{x}}=f\left(\frac{y}{x}\right)\end{array}$

i.e, homogenous

Let,  $y=vx=v=\frac{y}{x}$ so that  $\frac{dy}{dx}=v+y\frac{dy}{dx}$ in the D.E.

Then,  $v+x\frac{dv}{dx}=\frac{v-1}{v+1}$

$\begin{array}{l}=x\frac{dv}{dx}=\frac{v-1}{v+1}-v=\frac{v-1-v^2-v}{v+1}=-\frac{\left(v^2+1\right)}{v+1}\\ =\left[\frac{v+1}{v^2+1}\right]dv=-\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{v+1}{v^2+1}dv={\displaystyle \int -\frac{dx}{x}}}\\ =\frac{1}{2}{\displaystyle \int \frac{2v}{v^2+1}}dv+{\displaystyle \int \frac{1}{v^2+1}dv=-logx+c}\\ =\frac{log\left|v^2+1\right|}{2}+ta{n}^{-1}v=-logx+c\end{array}$

Putting back  $v=\frac{y}{x}$ we get,

$\begin{array}{l}=\frac{1}{2}log\left|\frac{y^2}{x^2}+1\right|+ta{n}^{-1}\frac{y}{x}=-logx+c\\ =\frac{1}{2}\left[log\left(y^2+x^2\right)-log{x}^2\right]+ta{n}^{-1}\frac{y}{x}+logx=c\\ =\frac{1}{2}log\left(y^2+x^2\right)-\frac{1}{2}log{x}^2+logx+ta{n}^{-1}\frac{y}{x}=c\\ =\frac{1}{2}log\left(y^2+x^2\right)-logx+logx+ta{n}^{-1}\frac{y}{x}=c\\ =\frac{1}{2}log\left(y^2+x^2\right)+ta{n}^{-1}\frac{y}{x}=c\end{array}$

Given,  $y=1,when,x=1$

So,  $=\frac{1}{2}log\left(1^2+1^2\right)+ta{n}^{-1}\frac{1}{1}=c$

$=\frac{1}{2}log2+\frac{\pi }{4}=c$

Hence, the particular solution is

$\frac{1}{2}log\left(1^2+1^2\right)+ta{n}^{-1}\frac{y}{x}=\frac{1}{2}log2+\frac{\pi }{4}$

 

Q12. x² dy + (xy + y²  ) dx = 0; y = 1 when x = 1

A.12. The given D.E. is

.  $\begin{array}{l}{x}^{2}dy+\left(xy+y^2\right)dx=0\\ =x^{2}dy=-\left(xy+y^2\right)dx\\ =\frac{dy}{dx}=-\left(\frac{xy+y^2}{x^2}\right)=-\left[\frac{y}{x}+\left(\frac{y^2}{x}\right)\right]=f\left(\frac{y}{x}\right)\end{array}$

i.e, the D.E is homogenous.

Let,  $y=vx=v=\frac{y}{x}$ so that  $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the given D.E.

Then,  $v+x\frac{dv}{dx}=-\left[v+v^2\right]=-v-v^2$

$\begin{array}{l}=x\frac{dv}{dx}=-2v-v^2=-v\left(2+v\right)\\ =\frac{dv}{v\left(2+v\right)}=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{dv}{v\left(2+v\right)}=-{\displaystyle \int \frac{dx}{x}}}\\ =\frac{1}{2}{\displaystyle \int \frac{2dv}{2\left(v+2\right)}=}-{\displaystyle \int \frac{dx}{x}}\\ =\frac{1}{2}{\displaystyle \int \frac{v+2-v}{v\left(v+2\right)}}dv={\displaystyle \int -\frac{dx}{x}}\\ =\frac{1}{2}\left\{{\displaystyle \int \frac{1}{v}dv-\frac{1}{v+2}dv}\right\}={\displaystyle \int -\frac{dx}{x}}\end{array}$

$\begin{array}{l}=\frac{1}{2}\left[logv-log\left|v+2\right|\right]=-logx+logc\\ =\frac{1}{2}log\left(\frac{v}{v+2}\right)=log\frac{c}{x}\\ =log{\left(\frac{v}{v+2}\right)}^{\frac{1}{2}}=log\frac{c}{x}\\ ={\left(\frac{v}{v+2}\right)}^{\frac{1}{2}}=\frac{c}{x}\\ =\frac{v}{v+2}={\left(\frac{c}{x}\right)}^2\end{array}$

Putting back  $v=\frac{y}{x}$ we get,

$\begin{array}{l}=\frac{\frac{y}{x}}{\frac{y}{x}+2}={\left(\frac{c}{x}\right)}^2\\ =\frac{y}{y+2x}={\left(\frac{c}{x}\right)}^2\\ =\frac{x^{2}y}{y+2x}=c^2\end{array}$

Given, y = 1 when x = 1

So,  $=\frac{1}{1+2}=c^2=c^2=\frac{1}{3}$

Hence, the required particular solution is,

$\begin{array}{l}=\frac{x^{2}y}{y+2x}=\frac{1}{3}\\ =3x^{2}y=y+2x\end{array}$

 

Q13.  $\left[xsi{n}^2\left(\frac{y}{x}-y\right)\right]dx+xdy=0;y=\frac{\pi }{4}whenx=1$

A.13. The given D.E.is

$\begin{array}{l}\left[xsi{n}^2\left(\frac{y}{x}\right)-y\right]dx+xdy=0\\ =\left[xsi{n}^2\left(\frac{y}{x}\right)-y\right]dx=-xdy\\ =\frac{dy}{dx}=\frac{\left[xsi{n}^2\left(\frac{y}{x}\right)-y\right]}{-x}\\ =-\left[si{n}^2\left(\frac{y}{x}\right)-\frac{y}{x}\right]=f\left(\frac{y}{x}\right)\end{array}$

i.e, the given D.E is homogenous.

Let,  $y=vx=\frac{y}{x}=v$ so that,  $\frac{dy}{dx}=vx\frac{dv}{dx}$ in the D.E.

$\begin{array}{l}=v+\frac{dv}{dx}=-\left[si{n}^{2}v-v\right]=v-si{n}^{2}v\\ =\frac{dv}{dx}=-si{n}^{2}v\\ =\frac{dv}{si{n}^{2}v}=-dx\end{array}$

Integrating both sides we get,

$\begin{array}{l}={\displaystyle \int cose{c}^{2}vdv={\displaystyle \int -dx}}\\ =-cotv=-log\left|x\right|+c\\ =cotv=log\left|x\right|-c\end{array}$

Putting back  $v=\frac{y}{x}$ we have,

$cot\frac{y}{x}=log\left|x\right|-c$

Then,  $y=\frac{\pi }{4}$ when,  $x=1$

$\begin{array}{l}cot\frac{\pi }{4}=log\left|1\right|-c\\ =c=-1\end{array}$

$\therefore$ The required particular solution is,

$\begin{array}{l}cot\frac{y}{x}=log\left|x\right|+1=log\left|x\right|+logc\left\{\because loge=1\right\}\\ =cot\frac{y}{x}=log\left|ex\right|\end{array}$

 

Q14.  $\frac{dy}{dx}-\frac{y}{x}+cosec\left(\frac{y}{x}\right)=0;y=0whenx=1$

A.14.The given D.E.is

$\begin{array}{l}\frac{dy}{dx}-\frac{y}{x}+cosec\left(\frac{y}{x}\right)=0\\ =\frac{dy}{dx}=\frac{y}{x}-cosec\left(\frac{y}{x}\right)=f\left(\frac{y}{x}\right)\end{array}$

i.e, the given D.E. is homogenous.

Let,  $y=vx=\frac{y}{x}=v$ So that,  $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

Then,  $v+x\frac{dv}{dx}=v-cosecv$

$\begin{array}{l}=x\frac{dv}{dx}=-cosecv\\ =\frac{dv}{cosecv}=-\frac{dx}{x}\\ =sinvdv=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int sinvdv=-{\displaystyle \int \frac{dx}{x}}}\\ =-cosv=-log\left|x\right|+c\\ =cosv=log\left|x\right|-c\end{array}$

Putting back  $v=\frac{y}{x}$ we get,

$=cos\frac{y}{x}=log\left|x\right|-c$

Given,  $y=0,when,x=1$

$\begin{array}{l}=cos0=log1-c\\ =c=-1\end{array}$

$\therefore$ The required particular solution is

$\begin{array}{l}cos\left(\frac{y}{x}\right)=log\left|x\right|+1=log\left|x\right|+log\left|c\right|\\ =cos\left(\frac{y}{x}\right)=log\left|cx\right|\end{array}$

 

Q15.  $2xy+y^2-2x^2\frac{dy}{dx}=0;y=2whenx=1$

A.15. The given D.E. is

$\begin{array}{l}2xy+y^2-2x^2\frac{dy}{dx}=0\\ =2x^2\frac{dy}{dx}=2xy+y^2\\ =\frac{dy}{dx}=\frac{2xy+y^2}{2x^2}=\frac{y}{x}+\frac{1}{2}{\left(\frac{y}{x}\right)}^2=f\left(\frac{y}{x}\right)\end{array}$

i.e, the given is homogenous.

Let,  $y=vx$  $=\frac{y}{x}=v$ so that  $\frac{dy}{dx}=v+x\frac{dv}{dx}$ is the D.E.

Then,  $v+x\frac{dv}{dx}=v+\frac{1}{2}{v}^2$

$\begin{array}{l}=x\frac{dv}{dx}=\frac{1}{2}{v}^2\\ =\frac{dv}{v^2}=\frac{dx}{2x}\end{array}$

Now,  $={\displaystyle \int \frac{dv}{v^2}={\displaystyle \int \frac{dx}{2x}}}$

$\begin{array}{l}=\frac{v-2+1}{-2+1}=\frac{1}{2}log\left|x\right|+c\\ =\frac{-1}{v}=\frac{1}{2}log\left|x\right|+c\end{array}$

Putting back  $\frac{y}{x}=v$ we get,

$=-\frac{x}{y}=\frac{1}{2}log\left|x\right|+c$

$Given\frac{y}{x}=vwhenx=1$ and y= 2

$\begin{array}{l}=-\frac{1}{2}=\frac{1}{2}log\left|1\right|+c\\ =c=-\frac{1}{2}\end{array}$

$\therefore$ The particular solution is,

$\begin{array}{l}=-\frac{x}{y}=\frac{1}{2}log\left|x\right|-\frac{1}{2}\\ =-\frac{2x}{y}=log\left|x\right|-1\\ =y=\frac{-2x}{log\left|x\right|-1}=\frac{2x}{1-log\left|x\right|}\end{array}$

 

Q16. A homogeneous differential equation of the form  $\frac{dx}{dy}=h\left(\frac{x}{y}\right)$ can be solved by making the substitution:

(A) y = vx

(B) v = yx

(C) x = vy

(D) x = v

A.16. For a homogenous D.E. of the formula  $f\left(\frac{y}{x}\right)$

We put,  $\frac{x}{y}=0=x=vy$

$\therefore$ Option(c) is correct.

 

Q17. Which of the following is a homogeneous differential equation:

(A) (4x +6y +5) dy – (3y + 2x + 4) dx = 0

(B) (xy) dx – x³ + y³) dy = 0

(C) (x² + 2y²) dx + (2xy + dy) = 0

(D) y² dx + (x² – xy + y²) dy = 0

A.17. In (D), each of the terms has a degree 2.

Hence, (D) is homogenous

$\therefore$ Option (D) is correct.

 

Ex.9.6

In each of the following differential equations given in each Questions 1 to 4, find the general solution:

Q1.  $\frac{dy}{dx}+2y=2sinx$

A.1. The given D.E. is

$\frac{dy}{dx}+2y=2sinx$ which is of form  $\frac{dy}{dx}+Px=Q$

We have, P = 2

$Q=sinx$

So, I.F.  $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 2dx}}=e^{2x}$

The solution is  $y\times I.F={\displaystyle \int Q\times \left(I.F\right)dx+c}$

$\begin{array}{l}y{e}^{2x}={\displaystyle \int sinx{e}^{2x}dx+c}\\ =y{e}^{2x}=I+c---------(1)\\ Where,I={\displaystyle \int sinx{e}^{2x}}\\ =sinx{\displaystyle \int e^{2x}-{\displaystyle \int \left(\frac{d}{dx}sinx{\displaystyle \int e^{2x}dx}\right)dx}}\\ =sinx\frac{e^{2x}}{2}-\frac{1}{2}\left[{\displaystyle \int cosx{e}^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cos{\displaystyle \int e^{2x}dx-{\displaystyle \int \left(\frac{d}{dx}cosx{\displaystyle \int e^{2x}dx}\right)dx}}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cosx\frac{e^{2x}}{2}\frac{1}{2}{\displaystyle \int \left(-cosx\right)e^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{4}{\displaystyle \int sinx{e}^{2x}dx}\end{array}$

$\begin{array}{l}=I=e^{2x}\frac{sinx}{2}-e^{2x}\frac{cosx}{4}-\frac{1}{4}1\\ =I+\frac{1}{4}I=2e^{2x}\frac{sinx-e^{2x}cosx}{4}\\ =\frac{5}{4}I=e^{2x}\frac{\left(2sinx-cosx\right)}{4}\\ =I=e^{2x}\frac{\left(2sinx-cosx\right)}{5}\end{array}$

Hence, equation, (1) becomes,

$\begin{array}{l}y{e}^{2x}=\frac{e^{2x}}{5}\left(2sinx-cosx\right)+c\\ =y=\frac{\left(2sinx-cosx\right)}{5}+c{e}^{2x}\end{array}$

Is the required solution.

 

Q2.  $\frac{dy}{dx}+3y=e^{-2x}$

A.2. Given, D.E. is

$\frac{dy}{dx}+3y=e^{-2x}$ which is of the form

$\frac{dy}{dx}+Py=Q$

Where  $\begin{array}{l}P=3\\ \&Q=e^{-2x}\end{array}$

So, I.F  $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 3dx}}=e^{3x}$

So, the solution is  $=y\times I.F={\displaystyle \int e^{-2x}\left(I.F\right).dx+c}$

$\begin{array}{l}=y\times e^{3x}={\displaystyle \int e^{-2x}.}{e}^{3x}dx+c\\ =e^{3x}y={\displaystyle \int e^{x}dx+c}\\ =e^{3x}y=e^x+c\\ =y=\frac{e^x}{e^{3x}}+\frac{c}{e^{3x}}\\ =y=e^{-2x}+c{e}^{-3x}\end{array}$

Is the required general solution.

 

Q3.  $\frac{dy}{dx}+\frac{y}{x}=x^2$

A.3. The given D.E. is

$\frac{dy}{dx}+\frac{y}{x}=x^2$ Which is in the form  $\frac{dy}{dx}+Py=Q$

So,  $P=\frac{1}{x}=\&Q=x^2$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{1}{2}dx}}=e^{logx}=x\left\{\because e^{logx}=x\right\}$

Thus, the general solution is

$\begin{array}{l}y\times I.F={\displaystyle \int Q\times I.Fdx+c}\\ y.x={\displaystyle \int x^2.xdx+c}\\ =xy={\displaystyle \int x^{3}dx+c}\\ =xy=\frac{x^4}{4}+c\end{array}$

 

Q4.  $\frac{dy}{dx}+secxy=tanx\left(0\le x\le \frac{\pi }{2}\right)$

A.4. The given D.E.is

$\frac{dy}{dx}+\left(secx\right)y=tanx$ Which is in the form  $\frac{dy}{dx}+Py=Q$

So,  $P=secx\&Q=tanx$

$\therefore$  $\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int secxdx}}=e^{log\left|secx+tanx\right|}=secx+tanx$

Thus, the general solution is ,

$\begin{array}{l}y\times I.F={\displaystyle \int Q\times I.Fdx+c}\\ =y\times \left(secx+tanx\right)={\displaystyle \int tanx}\left(secx+tanx\right)dx+c\end{array}$

$\begin{array}{l}={\displaystyle \int \left(tanxsecx+ta{n}^{2}x\right)dx+c}\\ ={\displaystyle \int \left(tanxsecx+se{c}^2-1\right)}dx+c\\ =sec+tanx-x+c\end{array}$

$=\left(secx+tanx\right)y=secx+tanx-x+c\left\{\because se{c}^{2}x=ta{n}^{2}x+1\right\}$

 

Q5.  $co{s}^{2}x\frac{dy}{dx}+y=tanx\left(0\le x\le \frac{\pi }{2}\right)$

A.5. The given D.E.is

$\begin{array}{l}co{s}^{2}x\frac{dy}{dx}+y=tanx\\ =\frac{dy}{dx}+\frac{1}{co{s}^{2}x}y=\frac{tanx}{co{s}^{2}x}\end{array}$

$=\frac{dy}{dx}+se{c}^{2}xy=se{c}^{2}xtanx$ Which is of form  $\frac{dy}{dx}+Py=Q$

So,  $P=se{c}^{2}x\&Q=se{c}^{2}xtanx$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int se{c}^{2}dx}}=e^{tanx}$

Thus, the general solution is of the form.

$y.e^{tanx}={\displaystyle \int se{c}^{2}xtanx.}{e}^{tanx}dx+c$

Let,  $tanx=t=se{c}^{2}xdx=dt$

$\begin{array}{l}=y{e}^t={\displaystyle \int t.}{e}^{t}dt+c\\ =t{\displaystyle \int e^{t}dt-{\displaystyle \int \frac{d}{dt}t{\displaystyle \int e^{t}dt.dt+c}}}\\ =t{e}^t-{\displaystyle \int e^{t}dt+c}\\ =t{e}^t-e^t+c\\ =e^t\left(t-1\right)+c\end{array}$

$\begin{array}{l}\Rightarrow y{e}^{tanx}=e^{tanx}\left(tanx-1\right)+c\\ y=\left(tanx-1\right)+c{e}^{-tanx}\end{array}$

 

Q6.  $x\frac{dy}{dx}+2y=x^{2}logx$

A.6. The given D.E. is

$x\frac{dy}{dx}+2y=x^{2}logx$

$\Rightarrow \frac{dy}{dx}+\frac{2}{x}.y=xlogx$ which is of form  $\frac{dy}{dx}+Py=Q$

So,  $P=\frac{2}{x}\&Q=xlogx$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2}{x}dx}}=e^{2logx}=e^{log{x}^2}=x^2$

Thus, the general solution is of the form.

$y\times x^2={\displaystyle \int xlogx.x^{2}dx+c}$

$={\displaystyle \int logx}{x}^{3}dx+c$

$\begin{array}{l}=logx{\displaystyle \int x^{3}dx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int x^{3}dxdx+c}}}\\ =\frac{logx.x^4}{4}-{\displaystyle \int \frac{1}{4}\times \frac{x^4}{4}dx+c}\end{array}$

$\begin{array}{l}=y{x}^2=\frac{x^4}{4}logx-\frac{x^4}{16}+c\\ =y=\frac{x^2}{4}logx-\frac{x^2}{16}+\frac{c}{x^2}\end{array}$

 

Q7.  $xlogx\frac{dy}{dx}+y=\frac{2}{x}logx$

A.7. The given D.E. is

$xlogx\frac{dy}{dx}+y=\frac{2}{x}logx$

$=\frac{dy}{dx}+\frac{y}{xlogx}=\frac{2}{x^2}$ Which is of form  $\frac{dy}{dx}+Py=Q$

So,  $P=\frac{1}{xlogx}\&Q=\frac{2}{x^2}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{1}{x}logxdx}}=e^{{\displaystyle \int \frac{\frac{1}{x}}{logx}}dx}=e^{log\left|logx\right|}=logx$

Thus, the general solution is of the form

$\begin{array}{l}y\times logx={\displaystyle \int \frac{2}{x^2}}\times logxdx+c\\ y.logx=2\left[logx{\displaystyle \int \frac{1}{x^2}dx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int \frac{1}{x^2}dx.dx}}}\right]+c\\ =2\left[logx\times \left(\frac{x^{-1}}{-1}\right)-{\displaystyle \int \frac{1}{x}\left(\frac{x^{-1}}{-1}\right)dx}\right]+c\\ =2\left[-\frac{logx}{x}+{\displaystyle \int x^{-2}dx}\right]+c\\ =2\left[-\frac{logx}{x}-\left(\frac{x^{-1}}{-1}\right)\right]+c\end{array}$

$=ylogx=\frac{-2}{x}\left[logx+1+c\right]$

 

Q8.  $\left(1+x^2\right)dy+2xydx=cotxdx\left(x\ne 0\right)$

A.8. The given D.E. is

$\begin{array}{l}{\left(1+x\right)}^{2}dy+2xydx=cotxdx\\ ={\left(1+x\right)}^2\frac{dy}{dx}+2xy=cotx\end{array}$

$=\frac{dy}{dx}+\frac{2x}{1+x^2}\times y=\frac{cotx}{1+x^2}$ Which is of form  $\frac{dy}{dx}+Py=Q$

So ,  $P=\frac{2x}{1+x^2}\&Q=\frac{cotx}{1+x^2}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2x}{1+x^2}dx}}=e^{log\left|1+x^2\right|}=1+x^2$

Thus the solution is of the form,

$\begin{array}{l}\Rightarrow y\left(1+x^2\right)={\displaystyle \int \frac{cotx}{1+x^2}\times }\left(1+x^2\right)dx+c\\ \Rightarrow y\left(1+x^2\right)={\displaystyle \int cotxdx+c}\end{array}$

$=log\left|sinx\right|+c$

$\begin{array}{l}\Rightarrow y=\frac{log\left|sinx\right|}{1+x^2}+\frac{c}{1+x^2}\\ ={\left(1+x^2\right)}^{-1}log\left|sinx\right|+{\left(1+x^2\right)}^{-1}c\end{array}$

 

Q9.  $x\frac{dy}{dx}+y-x+xycotx=0\left(x\ne 0\right)$

A.9. The given D.E is

$=\frac{dy}{dx}+y\frac{\left(1+xcotx\right)}{x}=1$ Which is of form  $\frac{dy}{dx}+Py=Q$

So,  $P=\frac{\left(1+xcotx\right)}{x}\&Q=1$

${\displaystyle \int Pdx={\displaystyle \int \left(\frac{1}{x}+\frac{xcotx}{x}\right)}}dx=log\left|x\right|+log\left|sinx\right|=log\left|xsinx\right|$

$\therefore I.F=e^{{\displaystyle \int Pdx}}-e^{{\displaystyle \int log\left|xsinx\right|}}xsinx$

Thus the solution is of the form.

$\begin{array}{l}y\times xsinx={\displaystyle \int 1.xsinxdx+c}\\ =x{\displaystyle \int sinx-{\displaystyle \int \frac{d}{dx}{\displaystyle \int sinxdxdx+c}}}\\ =-2cosx+{\displaystyle \int cosxdx+c}\\ =y\times xsinx=-xcosx+sinx+c\\ =y=\frac{-xcosx}{sinx}+\frac{sinx}{xsinx}+\frac{c}{xsinx}\\ =y=-cotx+\frac{1}{x}+\frac{c}{xsinx}\end{array}$

 

Q10.  $\left(x+y\right)\frac{dy}{dx}=1$

A10. The given D.E is

$\begin{array}{l}\left(x+y\right)\frac{dy}{dx}=1\\ =x+y=\frac{dx}{dy}\end{array}$

$=\frac{dx}{dy}-x=y$ Which is of form  $=\frac{dx}{dy}+Px=Q$

So,  $P=-1\&Q=y$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int -1dy}}=e^y$

Thus the general solution is of the form,  $x\times \left(I.F\right)={\displaystyle \int Q}\left(I.F\right)dy+c$

 

Q11.  $ydx+\left(x-y^2\right)dy=0$

A.11. The given D.E. is

$\begin{array}{l}ydx+\left(x-y^2\right)dy=0\\ =y\frac{dx}{dy}+x-y^2=0\\ =\frac{dx}{dy}+\frac{x}{y}-y=0\end{array}$

$=\frac{dx}{dy}+\frac{1}{y}.x=y$ Which is of form.

$\frac{dx}{dy}+Px=Q$

So,  $P=\frac{1}{y}\&Q=y$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int \frac{1}{y}dy}}=e^{logy}=y$

Thus the general solution is of form,  $x\times \left(I.F\right)={\displaystyle \int Q\times }\left(I.F\right)dy+c$

$\begin{array}{l}x.y={\displaystyle \int y.ydy+c}\\ =xy={\displaystyle \int y^{2}dy+c}\\ =xy=\frac{y^3}{3}+c\\ =x=\frac{y^2}{3}+\frac{c}{y}\end{array}$

 

Q12.  $\left(x+3y^2\right)\frac{dy}{dx}=y\left(y>0\right)$

A.12. The given D.E. is

$\begin{array}{l}\left(x+3y^2\right)\frac{dy}{dx}=y\\ \Rightarrow \left(x+3y^2\right)dy=ydx\\ \Rightarrow y\frac{dx}{dy}=x+3y^2\\ \Rightarrow \frac{dx}{dy}=\frac{x}{y}+3y\end{array}$

$\Rightarrow \frac{dx}{dy}-\frac{1}{y}.x=3y$ Which is form  $\frac{dx}{dy}+Px=Q$

So,  $P=-\frac{1}{y}\&Q=3y$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int -\frac{1}{y}dy}}=e^{-log\left|y\right|}=e^{log{y}^{-1}}=y^{-1}=\frac{1}{y}$

Thus the solution is of the form.

$\begin{array}{l}x\times \frac{1}{y}={\displaystyle \int 3y.\frac{1}{y}dy+c}\\ =\frac{x}{y}={\displaystyle \int 3dy+c}\\ =\frac{x}{y}=3y+c\\ =x=3y^2+cy\end{array}$

For each of the differential equations given in Questions 13 to 15, find a particular solution satisfying the given condition:

 

Q13.  $\frac{dy}{dx}+2ytanx=sinx;y=0whenx=\frac{\pi }{3}$

A.13. The given D.E. is

$\frac{dy}{dx}+2ytanx=sinx$

$\Rightarrow \frac{dy}{dx}+\left(2tanx\right)y=sinx$ Which is of form  $\frac{dy}{dx}+Px=Q$

So,  $P=2tanx\&Q=sinx$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int 2tanxdx}}=e^{2log\left|secx\right|}=e^{logse{c}^2}=se{c}^2$

Thus the solution is of the form  $y\times \left(I.F\right)={\displaystyle \int Q}.\left(I.F\right)dx+c$

$\begin{array}{l}\Rightarrow y.se{c}^{2}x={\displaystyle \int sinx.se{c}^{2}xdx+c}\\ =\frac{sinx}{co{s}^2}dx+c\\ ={\displaystyle \int tanx.secxdx+c}\\ =yse{c}^2=secx+c\\ =y=\frac{1}{secx}+\frac{c}{se{c}^{2}x}=cosx+cco{s}^2\\ =y=cosx+cco{s}^{2}x\end{array}$

Given,  $y=0,Whenx=\frac{\pi }{3}$

$\begin{array}{l}=0=cos\frac{\pi }{3}+cco{s}^2\frac{\pi }{3}\left\{\frac{c}{4}=-\frac{1}{2},c=-\frac{4}{2},c=-2\right\}\\ =0=\frac{1}{2}+c{\left(\frac{1}{2}\right)}^2\\ =0=\frac{1}{2}+\frac{c}{4}\end{array}$

C = -2

$\therefore$ The particular solution is

$y=cosx-2co{s}^{2}x$

 

Q14.  $\left(1+x^2\right)\frac{dy}{dx}+2xy=\frac{1}{1+x^2};y=0whenx=1$

A.14. The given D.E. is

$\left(1+x^2\right)\frac{dy}{dx}+2xy=\frac{1}{1+x^2}$

$\frac{dy}{dx}+Px=Q$

So,  $P=\frac{2x}{1+x^2}\&Q=\frac{1}{{\left(1+x^2\right)}^2}$

${\displaystyle \int Pdx={\displaystyle \int \frac{2x}{1+x^2}dx=log\left|1+x^2\right|}}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{log\left|1+x^2\right|}=1+x^2$

Thus the solution is if form,

$y\times \left(I.F\right)={\displaystyle \int Q}.\left(I.F\right)dx+c$

$\begin{array}{l}y\left(1+x^2\right)={\displaystyle \int \frac{1}{{\left(1+x^2\right)}^2}\times \left(1+x^2\right)dx+c}\\ ={\displaystyle \int \frac{1}{\left(1+x^2\right)}dx+c}\\ y\times \left(1+x^2\right)=ta{n}^{-1}x+c\end{array}$

Given,  $y=0,when,x=1$

$\begin{array}{l}0=ta{n}^{-1}1+c\\ c=ta{n}^{-1}1=-\frac{\pi }{4}\end{array}$

$\therefore$ The particular solution is

$y\left(1+x^2\right)=ta{n}^{-1}x-\frac{\pi }{4}$

 

Q15.  $\frac{dy}{dx}-3ycotx=sin2x;y=2whenx=\frac{\pi }{2}$

A.15. The given D.E. is

$\frac{dy}{dx}-3ycotx=sin2x$

$\frac{dy}{dx}-3cotx.y=sin2x$ Which is of form  $\frac{dy}{dx}+Py=Q$

So,  $P=-3cotx\&Q=sin2x$

$\therefore I.F=e^{-{\displaystyle \int Pdx}}=e^{{\displaystyle \int -3cotxdx}}=e^{-3{\displaystyle \int cotxdx}}=e^{-3log\left|cotxdx\right|}=e^{log{\left(sin\right)}^{-3}}=\frac{1}{si{n}^{3}x}$

Thus the solution is of the form.

$\begin{array}{l}y\times \frac{1}{si{n}^{3}x}={\displaystyle \int sin2x.\frac{1}{si{n}^{3}x}dx+c}\\ ={\displaystyle \int \frac{2sinxcosx}{si{n}^{3}x}}dx+c\left\{\because sin2x=2sinxcosx\right\}\\ ={\displaystyle \int 2cosecxcotxdx+c}\\ =-2cosecx+c\\ =\frac{y}{si{n}^{3}x}=\frac{-2}{sinx}+c\\ =-2y=-2si{n}^{2}x+csi{n}^{3}x\end{array}$

Given,  $y=2,when,x=\frac{\pi }{2}$

$\begin{array}{l}2=-2si{n}^2\frac{1}{2}+csi{n}^3\frac{\pi }{2}\\ =2=-2+c\\ =e=2+2=4\end{array}$

$\therefore$ The particular solution is,  $y=-2si{n}^{2}x+4si{n}^{3}x$

 

Q16. Find the equation of the curve passing through the origin, given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of coordinates of that point.

A.16. We know the slope of tangent to curve is  $\frac{dy}{dx}$ .

$\therefore$  $\frac{dy}{dx}=x+y$

$=\frac{dy}{dx}-y=x$ which has form  $\frac{dy}{dx}+Py=Q$

So,  $P=-1\&Q=x$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -dx}}=e^{-x}$

Thus the solution is of the form .

$\begin{array}{l}y\times e^{-x}={\displaystyle \int x.e^{-x}dx+c}\\ =x{\displaystyle \int e^{-x}dx-{\displaystyle \int \frac{dx}{dx}{\displaystyle \int e^{-x}dxdx+c}}}\\ =-x{e}^{-x}+{\displaystyle \int e^{-x}dx+c}\\ =y{e}^{-x}=-x{e}^{-x}-e^{-x}+c\\ =y=-x-1+\frac{c}{e^{-x}}\\ =y+x+1=c{e}^x\end{array}$

Given, the curve passes through origin (0,0) i.e,  $y=0,when,x=0$

$0+0+1=c{e}^0=c=1$

$\therefore$ Thus, equation of the curve is

$y+x+1=e^x$

 

Q17. Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangents to the curve at that point by 5.

A.17. We know that slope of tangent to the curve is  $\frac{dy}{dx}$

$\therefore x+y=\frac{dy}{dx}+5$

$\frac{dy}{dx}-y=x-5$ Which has form  $\frac{dy}{dx}+Px=Q$

$where,P=1\&Q=x-5$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -1dx}}=e^{-x}$

Thus the solution has the form

$\begin{array}{l}y{e}^{-x}={\displaystyle \int \left(x-5\right)e^{-x}dx+c}\\ ={\displaystyle \int x{e}^{-x}dx-{\displaystyle \int 5e^{-x}dx+c}}\\ =y{e}^{-x}=I+5e^{-x}+c\\ where,I={\displaystyle \int x{e}^{-x}dx}\\ =x{\displaystyle \int e^{-x}dx-{\displaystyle \int \frac{d}{dx}x{\displaystyle \int e^{-x}dxdx}}}\\ =-x{e}^{-x}+{\displaystyle \int e^{-x}dx}\\ =-x{e}^{-x}-e^{-x}\end{array}$

$\begin{array}{l}\therefore y{e}^{-x}=-x{e}^{-x}-e^{-x}+5e^{-x}+c\\ =y{e}^{-x}=-x{e}^{-x}+4e^{-x}+c\\ =y=-x+4+\frac{c}{e^{-x}}\\ =y+x=4+c{e}^x\end{array}$

Given, the curve passes through (0,2) so y=2 when x=0

$\begin{array}{l}2+0=4c{e}^0\\ 2-4=c\\ c=-2\end{array}$

$\therefore$ The particular solution is

$y+x=4-2e^x$

 

Q18. Choose the correct answer:

The integrating factor of the differential equation   $\frac{dy}{dx}-y=2x^2$ is:

(A) e^–x

(B) e^–y

(C)  $\frac{1}{x}$

(D) x

A.18. The given D.E is

$\begin{array}{l}x\frac{dy}{dx}-y=2x^2\\ \frac{dy}{dx}-\frac{1}{x}y=2x\end{array}$

Which is of form  $\frac{dy}{dx}+Py=Q$

So,  $P=-\frac{1}{x}$

$\therefore I.E=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -\frac{1}{x}dx}}=e^{log-x}=e^{log-x^{-1}}=x^{-1}=\frac{1}{x}$

$\therefore$ Option (c) is correct

 

Q19. Choose the correct answer:

The integrating factor of the differential equation   $\left(1-y^2\right)\frac{dx}{dy}+yx=ay\left(-1<<1\right)$ .

A.19. The given D.E. is

$\left(1-y^2\right)\frac{dx}{dy}+yx=\propto y\left(-1<y<1\right)$

$\frac{dx}{dy}+\frac{y}{1-y^2}\times x=\frac{\propto y}{1-y^2}$ which is of form

$\begin{array}{l}\frac{dx}{dy}+Px=Q\&P=\frac{y}{1-y^2}\&Q=\frac{\propto y}{1-y^2}\\ {\displaystyle \int pdx={\displaystyle \int \frac{y}{1-y^2}dx=-\frac{1}{2}{\displaystyle \int \frac{-2y}{1-y^{2}dx}}}}\end{array}$

$\begin{array}{l}=-\frac{1}{2}log\left|1-y^2\right|\\ =log{\left[1-y^2\right]}^{-\frac{1}{2}}\end{array}$

$\begin{array}{l}\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{log{\left[1-y^2\right]}^{-\frac{1}{2}}}\\ ={\left[1-y^2\right]}^{-\frac{1}{2}}\\ <br>\frac{\mathrm{<br>}}{}\end{array}$

$\therefore$ option (D ) is correct.

 

Miscellaneous Exercise

Q1. For each of the differential equations given below, indicate its order and degree (if defined):

$\begin{array}{l}(i)\frac{d^{2}y}{d{x}^2}+5x{\left(\frac{dy}{dx}\right)}^2-6y=logx\\ (ii){\left(\frac{dy}{dx}\right)}^3-4{\left(\frac{dy}{dx}\right)}^2+7y=sinx\\ (iii)\frac{d^{4}y}{d{x}^4}-sin\frac{d^{3}y}{d{x}^3}=0\end{array}$

A.1. (i) Given: Differential equation    $\frac{d^{2}y}{d{x}^2}+5x{\left(\frac{dy}{dx}\right)}^2-6y=logx$

The highest order derivative present in this differential equation is  $\frac{d^{2}y}{d{x}^2}$ and hence order of this differential equation if 2.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative  $\frac{d^{2}y}{d{x}^2}$ is 1.

Therefore, Order = 2, Degree = 1

(ii) Given: Differential equation  ${\left(\frac{dy}{dx}\right)}^3-4{\left(\frac{dy}{dx}\right)}^2+7y=sinx$

The highest order derivative present in this differential equation is  $\frac{dy}{dx}$ and hence order of this differential equation if 1.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative  $\frac{dy}{dx}$   is 3.

Therefore, Order = 1, Degree = 3

(iii) Given: Differential equation    $\frac{d^{4}y}{d{x}^4}-sin\frac{d^{3}y}{d{x}^3}=0$

The highest order derivative present in this differential equation is  $\frac{d^{4}y}{d{x}^4}$ and hence order of this differential equation if 4.

The given differential equation is not a polynomial equation in derivatives therefore, degree of this differential equation is not defined.

Therefore, Order = 4, Degree not defined

 

Q2. For each of the exercises given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation:

$\begin{array}{l}(i)ya{e}^2+b{e}^{-x}+x^2:x\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}-xy+x^2-2=0\\ (ii)y=e^x(acosx+bsinx):\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+2y=0\\ (iii)y=xsin3x:\frac{d^{2}y}{d{x}^2}+9y-6cos3x=0\\ (iv)x^2=2y^{2}logy:(x^2+y^2)\frac{dy}{dx}-xy=0\end{array}$

A.2. (i)  $ya{e}^2+b{e}^{-x}+x^2$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}\frac{dy}{dx}=a\frac{d}{dx}\left(e^x\right)+b\frac{d}{dx}\left(e^{-x}\right)+\frac{d}{dx}\left(x^2\right)\\ \Rightarrow \frac{dy}{dx}=a{e}^x-b{e}^{-x}+2x\end{array}$

Again, differentiating both sides with respect to x, we get:

$\frac{d^{2}y}{d{x}^2}=a{e}^x-b{e}^{-x}+2x$

Now, on substituting the values of  $\frac{dy}{dx}$ and  $\frac{d^{2}y}{d{x}^2}$ in the differential equation, we get:

L.H.S

$\begin{array}{l}x\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}-xy+x^2-2\\ =x\left(a{e}^x-b{e}^{-x}+2\right)+2\left(a{e}^x-b{e}^{-x}+2x\right)-x\left(a{e}^x+b{e}^{-x}+x^2\right)+x^2-2\\ =\left(xa{e}^x-bx{e}^{-x}+2x\right)+\left(2a{e}^x-2b{e}^{-x}+4x\right)-\left(ax{e}^x+bx{e}^{-x}+x^3\right)+x^2-2\\ =2a{e}^x-2b{e}^{-x}+x^2+6x-2\\ \ne 0\end{array}$

Therefore, Function given by equation (i) is a solution of differential equation. (ii).

(ii)  $y=e^x(acosx+bsinx)=a{e}^{x}cosx+b{e}^{x}sinx$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}\frac{dy}{dx}=a.\frac{d}{dx}(e^{x}cosx)+b.\frac{d}{dx}(e^{x}sinx)\\ \Rightarrow \frac{dy}{dx}=a\left(e^{x}cosx-e^{x}sinx\right)+b.\left(e^{x}sinx+e^{x}cosx\right)\\ \Rightarrow \frac{dy}{dx}=\left(a+b\right)e^{x}cosx+(b-a)e^{x}sinx\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\left(a+b\right).\frac{d}{dx}(e^{x}cosx)(b-a)\frac{d}{dx}(e^{x}sinx)\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\left(a+b\right).\left[e^{x}cosx-e^{x}sinx\right]+(b-a)\left[e^{x}sinx+e^{x}cosx\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=e^x\left[\left(a+b\right)(cosx-sinx)+(b-a)(sinx+cosx)\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=e^x\left[acosx-asinx+bcosx-bsinx+bsinx+bcosx-asinx-acosx\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\left[2e^x(bcosx-asinx)\right]\end{array}$

Now, on substituting the values of  $\frac{d^{2}y}{d{x}^2}$ and  $\frac{dy}{dx}$ in the L.H.S of the given differential equation, we get:

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+2y\\ =2e^x(bcosx-asinx)-2e^x\left[\left(a+b\right)cosx+(b-a)sinx\right]+2e^x(acosx+bsinx)\\ =e^x\left[(2bcosx-2asinx)-(2acosx+2bcosx)-(2bsinx-2asinx)+(2acosx+2bsinx)\right]\\ =e^x\left[\left(2b-2a-2b+2a\right)cosx\right]+e^x\left[\left(-2a-2b+2a+2b\right)sinx\right]\\ =0\end{array}$

Therefore, Function given by equation (i) is solution of differential equation (ii)

(iii)  $y=xsin3x$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}(xsin3x)=sin3x+x.cos3x.3\\ \Rightarrow \frac{dy}{dx}=sin3x+3xcos3x\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(sin3x)+3\frac{d}{dx}(xcos3x)\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=3cos3x+3\left[cos3x+x(-sin3x).3\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}6cos3x-9xsin3x\end{array}$

Substituting the value of  $\frac{d^{2}y}{d{x}^2}$ in the L.H.S. of the given differential equation, we get:

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}+9y-6cos3x\\ =(6.cos3x-9xsin3x)+9xsin3x-6cos3x\\ =0\end{array}$

Therefore, Function given by equation (i) is a solution of differential equation (ii).

(iv)  $x^2=2y^{2}logy$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}2x=2.\frac{d}{dx}=\left[y^{2}logy\right]\\ \Rightarrow x=\left[2y.logy.\frac{dy}{dx}+y^2.\frac{1}{y}.\frac{dy}{dx}\right]\\ \Rightarrow x=\frac{dy}{dx}(2ylogy+y)\\ \Rightarrow \frac{dy}{dx}=\frac{x}{y(1+2logy)}\end{array}$

Substituting the value of  $\frac{dy}{dx}$ in the L.H.S. of the given differential equation, we get:

$\begin{array}{l}\left(x^2+y^2\right)\frac{dy}{dx}-xy\\ =\left(2y^{2}logy+y^2\right).\frac{x}{y\left(1+2logy\right)}-xy\\ =y^2(1+2logy).\frac{x}{y\left(1+2logy\right)}-xy\\ =xy-xy\\ =0\end{array}$

Therefore, Function given by equation (i) is a solution of differential equation (ii).

 

Q3. Form the differential equation representing the family of curves   ${(x-a)}^2+2y^2=a^2$ where a an arbitrary constant.

A.3. Equation of the given family of curves is   ${(x-a)}^2+2y^2=a^2$

$\begin{array}{l}{(x-a)}^2+2y^2=a^2\\ \Rightarrow x^2+a^2-2ax+2y^2=a^2\\ \Rightarrow 2y^2=2ax-x^2..........(1)\end{array}$

Differentiating with respect to x, we get:

$\begin{array}{l}2y\frac{dy}{dx}=\frac{2a-2x}{2}\\ \Rightarrow \frac{dy}{dx}=\frac{a-x}{2y}\\ \Rightarrow \frac{dy}{dx}=\frac{2a-2x^2}{4xy}..........(2)\end{array}$

From equation (*1), we get:

$2ax=2y^2+x^2$

On substituting this value in equation (3), we get:

$\begin{array}{l}\frac{dy}{dx}=\frac{2y^2+x^2-2x^2}{4xy}\\ \Rightarrow \frac{dy}{dx}=\frac{2y^2-x^2}{4xy}\end{array}$

Hence, the differential equation of the family of curves is given as  $\frac{dy}{dx}=\frac{2y^2-x^2}{4xy}$ .

Q4. Prove that  $x^2-y^2=c{\left(x^2+y^2\right)}^2$  is the general equation of the differential equation   $\left(x^3-3x{y}^2\right)dx=\left(y^3-3x^{2}y\right)dy$  where c is a parameter.

A.4. $\Rightarrow \frac{dy}{dx}=\frac{x^3-3x{y}^2}{y^3-3x^{2}y}..........(1)$

This is a homogenous equation. To simplify it, we need to make the substitution as:

$\begin{array}{l}y=vx\\ \Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(vx)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}$

Substituting the values of y and  $\frac{dv}{dx}$ in equation (1), we get:

$\begin{array}{l}v+x\frac{dv}{dx}=\frac{x^3-3x{(vx)}^2}{{(vx)}^3-3x^2(vx)}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}-v\\ \Rightarrow x\frac{dv}{dx}=\frac{1-3v^2-v(v^3-3v)}{v^3-3v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1-v^4}{v^3-3v}\\ \Rightarrow \left(\frac{v^3-3v}{1-v^4}\right)dv=\frac{dx}{x}\end{array}$

Integrating both sides, we get:

$\begin{array}{l}{\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)dv=logx+log{C}^{\text{'}}}.........(2)\\ Now,{\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)dv={\displaystyle \int \frac{v^{3}dv}{1-v^4}}}-3{\displaystyle \int \frac{vdv}{1-v^4}}\\ \Rightarrow {\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)}dv=I_1-3I_2,Where,I_1={\displaystyle \int \frac{v^{3}dv}{1-v^4}}and{I}_2={\displaystyle \int \frac{vdv}{1-v^4}...........(3)}\end{array}$

$\begin{array}{l}Let,1-v^4=t.\\ \therefore \frac{d}{dv}(1-v^4)=\frac{dt}{dv}\\ \Rightarrow -4v^3=\frac{dt}{dv}\\ \Rightarrow v^{3}dv=-\frac{dt}{4}\\ Now,I_1={\displaystyle \int -\frac{dt}{4}=-logt=-\frac{1}{4}}log(1-v^4)\end{array}$

$\begin{array}{l}And,I_2={\displaystyle \int \frac{vdv}{1-v^4}}={\displaystyle \int \frac{vdv}{1-(v^2){}^2}}\\ Let,v^2=p.\\ \therefore \frac{d}{dv}\left(v^2\right)=\frac{dp}{dv}\\ \Rightarrow 2v=\frac{dp}{dv}\\ \Rightarrow vdv=\frac{p}{2}\\ \Rightarrow I_2=\frac{1}{2}{\displaystyle \int \frac{dp}{1-p^2}=\frac{1}{2\times 2}log\left|\frac{1+p}{1-p}\right|=\frac{1}{4}log}\left|\frac{1+v^2}{1-v^2}\right|\end{array}$

Substituting the values of  $I_1$ and  $I_2$ in equation (3), we get:

${\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)}dv=-\frac{1}{4}log(1-v^4)-\frac{3}{4}log\left|\frac{1+v^2}{1-v^2}\right|$

Therefore, equation (2) becomes:

$\begin{array}{l}\frac{1}{4}log(1-v^4)-\frac{3}{4}log\left|\frac{1+v^2}{1-v^2}\right|=logx+log{C}^{\text{'}}\\ \Rightarrow -\frac{1}{4}log\left[(1-v^4)\left(\frac{1+v^2}{1-v^2}\right)\right]=log{C}^{\text{'}}x\\ \Rightarrow \frac{{\left(1+v^2\right)}^4}{{\left(1-v^2\right)}^2}={\left(C^{\text{'}}x\right)}^{-4}\\ \Rightarrow \frac{{\left(1+\frac{y^2}{x^2}\right)}^4}{{\left(1-\frac{y^2}{x^2}\right)}^2}=\frac{1}{C^{\text{'}4}{x}^4}\\ \Rightarrow \frac{{\left(x^2+y^2\right)}^4}{x^4{\left(x^2-y^2\right)}^2}=\frac{1}{C^{\text{'}4}{x}^4}\\ \Rightarrow {\left(x^2-y^2\right)}^2=C^{\text{'}4}{\left(x^2+y^2\right)}^4\\ \Rightarrow \left(x^2-y^2\right)=C^{\text{'}2}{\left(x^2+y^2\right)}^2\\ \Rightarrow x^2-y^2=C{\left(x^2+y^2\right)}^2,whereC=C^{\text{'}2}\end{array}$

Hence, the given result is proved.

 

Q5. For the differential equation of the family of the circles in the first quadrant which touch the coordinate axes.

A.5. The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:

${(x-a)}^2+{(y-a)}^2=a^2..........(1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2(x-a)+2(y-a)\frac{dy}{dx}=0\\ \Rightarrow (x-a)+(y-a)y^{\text{'}}=0\\ \Rightarrow x-a+y{y}^{\text{'}}-a{y}^{\text{'}}=0\\ \Rightarrow x+y{y}^{\text{'}}-a(1+y^{\text{'}})=0\\ \Rightarrow a=\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\end{array}$

Substituting the value of a in equation (1), we get:

$\begin{array}{l}{\left[x-\left(\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right)\right]}^2+{\left[y-\left(\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right)\right]}^2={\left(\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right)}^2\\ \Rightarrow {\left[\frac{(x-a)y^{\text{'}}}{(1+y^{\text{'}})}\right]}^2+{\left[\frac{y-x}{1+y^{\text{'}}}\right]}^2={\left[\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right]}^2\\ \Rightarrow {(x-y)}^2.y^{\text{'}2}+{(x-y)}^2=(x+y{y}^{\text{'}}){}^2\\ \Rightarrow {(x-y)}^2\left[1+(y^{\text{'}}){}^2\right]=(x+y{y}^{\text{'}}){}^2\end{array}$

Hence, the required differential equation of the family of circles is  ${(x-y)}^2\left[1+(y^{\text{'}}){}^2\right]=(x+y{y}^{\text{'}}){}^2$

 

$\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0$ is given by  $(x+y+1)A(1-x-y-2xy),$ where A is parameter.

A.7. Given: Differential equation  $\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0$

$\begin{array}{l}\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0\\ \Rightarrow \frac{dy}{dx}=-\frac{(y^2+y+1)}{x^2+x+1}\\ \Rightarrow \frac{dy}{y^2+y+1}=\frac{-dx}{x^2+x+1}\\ \Rightarrow \frac{dy}{y^2+y+1}+\frac{dx}{x^2+x+1}=0\end{array}$

Integrating both sides,

A.8. The differential equation of the given curve is:

$\begin{array}{l}sinxcosydx+cosxsinydy=0\\ \Rightarrow \frac{sinxcosydx+cosxsinydy}{cosxcosy}=0\\ \Rightarrow tanxdx+tanydy=0\end{array}$

Integrating both sides, we get:

$\begin{array}{l}log(secx)+log(secy)=logC\\ log(secx.secy)=logC\\ \Rightarrow secx.secy=C..........(1)\end{array}$

The curve passes through point  $\left(0,\frac{\pi }{4}\right)$

$\begin{array}{l}\therefore 1\times \surd 2=C\\ \Rightarrow C=\surd 2\end{array}$

On subtracting  $C=\surd 2$ in equation (10, we get:

$\begin{array}{l}secx.secy=\surd 2\\ \Rightarrow secx.\frac{1}{cosy}=\surd 2\\ \Rightarrow cosy=\frac{sec\mathrm{x/\surd 2}}{}\end{array}$

 

Q9. Find the particular solution of the differential equation (1 + e2x ) dy + (1 + y2 ) ex dx = 0, given that y = 1 when x = 0.

A.9.  $\begin{array}{l}\left(1+e^{2x}\right)dy+\left(1+y^2\right)e^{x}dx=0\\ \Rightarrow \frac{dy}{1+y^2}+\frac{e^{x}dx}{1+e^{2x}}=0\end{array}$

Integrating both sides, we get:

$\begin{array}{l}ta{n}^{-1}y+{\displaystyle \int \frac{e^{x}dx}{1+e^{2x}}=C..........(1)}\\ Let,e^x=t\Rightarrow e^{2x}=t^2\\ \Rightarrow \frac{d}{dx}\left(e^x\right)=\frac{dt}{dx}\\ \Rightarrow e^x=\frac{dt}{dx}\\ \Rightarrow e^{x}dx=dt\end{array}$

Substituting these values in equation (1), we get:

$\begin{array}{l}ta{n}^{-1}y+{\displaystyle \int \frac{dt}{1+t^2}}=C\\ \Rightarrow ta{n}^{-1}y+ta{n}^{-1}t=C\\ \Rightarrow ta{n}^{-1}y+ta{n}^{-1}(e^x)=C..........(2)\\ Now,y=1,at,x=0\end{array}$

Therefore, equation (2) becomes:

$\begin{array}{l}ta{n}^{-1}1+ta{n}^{-1}1=C\\ \Rightarrow \frac{\pi }{4}+\frac{\pi }{4}=C\\ \Rightarrow C=\frac{\pi }{2}\end{array}$

Substituting  $C=\frac{\pi }{2}$ in equation (2), we get:

$ta{n}^{-1}y+ta{n}^{-1}(e^x)=\frac{\pi }{2}$

This is the required solution of the given differential equation.

 

Q10. Solve the differential equation:  $y{e}^{\frac{x}{y}}dx=\left(x{e}^{\frac{x}{y}}+y^2\right)dy(y\ne 0)$

A.10.

$\begin{array}{l}y{e}^{\frac{x}{y}}dx=\left(x{e}^{\frac{x}{y}}+y^2\right)dy\\ \Rightarrow y{e}^{\frac{x}{y}}\frac{dx}{dy}=x{e}^{\frac{x}{y}}+y^2\\ \\ \Rightarrow e^{\frac{x}{y}}\left[y.\frac{dx}{dy}-x\right]=y^2\\ \Rightarrow e^{\frac{x}{y}}.\frac{\left[y.\frac{dx}{dy}-x\right]}{y^2}=1..........(1)\end{array}$

$Let,e^{\frac{x}{y}}=z$

Differentiating it with respect to y, we get:

$\begin{array}{l}\left(e^{\frac{x}{y}}\right)=\frac{dz}{dy}\\ \Rightarrow e^{\frac{x}{y}}.\frac{d}{dy}\left(\frac{x}{y}\right)=\frac{dz}{dy}\\ \Rightarrow e^{\frac{x}{y}}.\left[\frac{y.\frac{dx}{dy}-x}{y^2}\right]=\frac{dz}{dy}..........(2)\end{array}$

From equation (1) and equation (2), we get:

$\begin{array}{l}\frac{dz}{dy}=1\\ \Rightarrow dz=dy\end{array}$

Integration both sides, we get:

$\begin{array}{l}z=y+C\\ \Rightarrow e^{\frac{x}{y}}y+C\end{array}$

 

Q11. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)

A.11.

$\begin{array}{l}(x-y)(dx+dy)=dx-dy\\ \Rightarrow (x-y+1)dy=(1-x+y)dx\\ \Rightarrow \frac{dy}{dx}=\frac{1-x+y}{x-y+1}\\ \Rightarrow \frac{dy}{dx}=\frac{1-(x-y)}{1+(x-y)}..........(1)\\ Let,x-y=t\\ \Rightarrow \frac{d}{dx}(x-y)=\frac{dt}{dx}\\ \Rightarrow 1-\frac{dy}{dx}=\frac{dt}{dx}\\ \Rightarrow 1-\frac{dt}{dx}=\frac{dy}{dx}\end{array}$

Substituting the values of  $x-y$ and  $\frac{dy}{dx}$ in equation (1), we get:

$\begin{array}{l}1-\frac{dt}{dx}=\frac{1-t}{1+t}\\ \Rightarrow \frac{dt}{dx}=1-\left(\frac{1-t}{1+t}\right)\\ \Rightarrow \frac{dt}{dx}=\frac{(1+t)-(1-t)}{1+t}\\ \Rightarrow \frac{dt}{dx}=\frac{2t}{1+t}\end{array}$

$\begin{array}{l}\Rightarrow \left(\frac{1+t}{t}dt\right)=2dx\\ \Rightarrow \left(1+\frac{1}{t}\right)dt=2dx..........(2)\end{array}$

Integrating both sides, we get:

$\begin{array}{l}t+log\left|t\right|=2x+C\\ \Rightarrow (x-y)+log\left|x-y\right|=2x+C\\ \Rightarrow log\left|x-y\right|=x+y+C..........(3)\end{array}$

$Now,y=-1,at,x=0$

Therefore, equation (3) becomes:

$log1=0-1+C$

$\Rightarrow C=1$

Substituting  $C=1$ in equation (3), we get:

$og\left|x-y\right|=x+y+1$

This is the required particular solution of the given differential equation .

 

Q13. Find the particular solution of the differential equation  $\frac{dy}{dx}+ycotx=4xcosecx(x\ne 0)$

given that y = 0 when x = π/2

A.13. The given differential equation is:

$\frac{dy}{dx}+ycotx=4xcosecx$

This equation is a linear equation of the form

$\begin{array}{l}\frac{dy}{dx}+Py=Q,where,p=cotx\&Q=4xcosecx\\ Now,I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int cotxdx}}=e^{log\left|sinx\right|}=sinx\end{array}$

The general solution of the given differential equation is given by,

$y(I.F)={\displaystyle \int (Q\times I.F.)dx+C}$

$\begin{array}{l}\Rightarrow ysinx={\displaystyle \int (4xcosecx.sinx)dx+C}\\ \Rightarrow ysinx=4{\displaystyle \int xdx+C}\\ \Rightarrow ysinx=4.\frac{x^2}{2}+C\\ \Rightarrow ysinx=2x^2+C..........(1)\\ Now,y=0at,x=\frac{\pi }{2}\end{array}$

Therefore, equation (1) becomes:

$\begin{array}{l}0=2\times \frac{\pi }{2}+C\\ \Rightarrow C=-\frac{{\pi }^2}{2}\end{array}$

Substituting  $C=-\frac{{\pi }^2}{2}$ in equation (1), we get:

$ysinx=2x^2-\frac{{\pi }^2}{2}$

This is the required particular solution of the given differential equation.

 

Q14. Find the particular solution of the differential equation  $(x+1)\frac{dy}{dx}=2e^{-y}-1$ given that y = 0 when x = 0

A.14.

$\begin{array}{l}(x+1)\frac{dy}{dx}=2e^{-y}-1\\ \Rightarrow \frac{dy}{2e^{-y}-1}=\frac{dx}{x+1}\\ \Rightarrow \frac{e^{y}dy}{2-e^y}=\frac{dx}{x+1}\end{array}$

Integrating both sides, we get:

$\begin{array}{l}{\displaystyle \int \frac{e^{y}dy}{2-e^y}=log\left|x+1\right|+logC..........(1)}\\ Let2-e^y=t\\ \therefore \frac{d}{dy}(2-e^y)=\frac{dt}{dy}\\ \Rightarrow -e^y=\frac{dt}{dy}\\ \Rightarrow e^{y}dy=-dt\end{array}$

Substituting this value in equation (1), we get:

$\begin{array}{l}{\displaystyle \int \frac{-dt}{t}=log}og\left|x+1\right|+logC\\ \Rightarrow -log\left|t\right|=log\left|C(x+1)\right|\\ \Rightarrow -log\left|2-e^y\right|=log\left|C(x+1)\right|\\ \Rightarrow \frac{1}{2-e^y}=C(x+1)\\ \Rightarrow 2-e^y=\frac{1}{C(x+1)}..........(2)\end{array}$

Now, at x=0& y=0, equation (2) becomes:

$\begin{array}{l}\Rightarrow 2-1=\frac{1}{C}\\ \Rightarrow C=1\end{array}$

Substituting  $C=1$ in equation (2), we get:

$\begin{array}{l}2-e^y=\frac{1}{x+1}\\ \Rightarrow e^y=2-\frac{1}{x+1}\\ \Rightarrow e^y=\frac{2x+2-1}{x+1}\\ \Rightarrow e^y=\frac{2x+1}{x+1}\\ \Rightarrow y=log\left|\frac{2x+1}{x+1}\right|,(x\ne -1)\end{array}$

This is the required particular solution of the given differential equation.

Q15. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009?

A.15. Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

$\therefore \frac{dy}{dt}\alpha y$

$\Rightarrow \frac{dy}{dt}=ky$ (k is constant)

$\Rightarrow \frac{dy}{y}=kdt$

Integration both sides, we get:

$logy=kt+C..........(1)$

In the year  $1999,t=0\&y=20000.$

Therefore, we get:

$log20000=C..........(2)$

In the year  $2004,t=5\&y=25000.$

Therefore, we get:

$\begin{array}{l}\Rightarrow log25000=5k+log20000\\ \Rightarrow 5k=log\left(\frac{25000}{20000}\right)=log\left(\frac{5}{4}\right)\\ \Rightarrow k=\frac{1}{5}log\left(\frac{5}{4}\right)..........(3)\end{array}$

In the year  $2009,t=10years$

Now, on substituting the values of t, k, and C in equation (1), we get:

$\begin{array}{l}logy=10\times \frac{1}{5}log\left(\frac{5}{4}\right)+log(20000)\\ \Rightarrow logy=log\left[20000\times {\left(\frac{5}{4}\right)}^2\right]\\ \Rightarrow y=20000\times \frac{5}{4}\times \frac{5}{4}\\ \Rightarrow y=31250\end{array}$

Hence, the population of the village in 2009 will be 31250.

Choose the correct answer:

Q16. The general solution of the differential equation   $\frac{ydx-xdy}{y}=0$ is:

$\begin{array}{l}(A)xy=c\\ (B)x=c{y}^2\\ (C)y=cx\\ (D)y=c{x}^2\end{array}$

A.16.

The given differential equation is:

$\begin{array}{l}\frac{ydx-xdy}{y}=0\\ \Rightarrow \frac{ydx-xdy}{xy}=0\\ \Rightarrow \frac{1}{x}dx-\frac{1}{y}dy=0\end{array}$

Integration both sides, we get:

$\begin{array}{l}log\left|x\right|-log\left|y\right|=logk\\ \Rightarrow log\left|\frac{x}{y}\right|=logk\\ \Rightarrow \frac{x}{y}=k\\ \Rightarrow y=\frac{1}{k}x\\ \Rightarrow y=Cx,where,C=\frac{1}{k}\end{array}$

Therefore, option (C) is correct.

 

Q17. The general equation of a differential equation of the type  $\frac{dx}{dy}+P_{1}x=Q,is:$

$\begin{array}{l}(A)y{e}^{{\displaystyle \int P_{1}dy}}={\displaystyle \int \left(Q_1{e}^{{\displaystyle \int P_{1}dy}}\right)}dy+C\\ (B)y.e^{{\displaystyle \int P_{1}dx}}={\displaystyle \int \left(Q_1{e}^{{\displaystyle \int P_{1}dx}}\right)}dx+C\\ (C)x{e}^{{\displaystyle \int P_{1}dy}}={\displaystyle \int \left(Q_1{e}^{{\displaystyle \int P_{1}dy}}\right)}dy+C\\ (D)x.e^{{\displaystyle \int P_{1}dx}}={\displaystyle \int \left(Q_1{e}^{{\displaystyle \int P_{1}dx}}\right)}dx+C\end{array}$

A.17.

The integrating factor of the given differential equation   $\frac{dx}{dy}+P_{1}x=Q_1$

The general solution of the differential equation is given by,

$\begin{array}{l}x(I.F.)={\displaystyle \int \left(Q\times I.F.\right)dy+C}\\ \Rightarrow x.e^{{\displaystyle \int P_{1}dy}}={\displaystyle \int \left(Q_1{e}^{{\displaystyle \int P_{1}dy}}\right)}dy+C\end{array}$

Hence, the correct answer is C.

 

Q18. The general solution of the differential equation   $e^{x}dy+(y{e}^x+2x)dx=0$  is:

$\begin{array}{l}(A)x{e}^y+x^2=C\\ (B)x{e}^y+y^2=C\\ (C)y{e}^x+x^2=C\\ (D)y{e}^y+x^2=C\end{array}$

A.18.

The given differential equation is:

$\begin{array}{l}{e}^{x}dy+(y{e}^x+2x)dx=0\\ \Rightarrow e^x\frac{dy}{dx}+y{e}^x+2x=0\\ \Rightarrow \frac{dy}{dx}+y=-2x{e}^{-x}\end{array}$

This is a linear differential equation of the form

$\begin{array}{l}\frac{dy}{dx}+Py=Q,whereP=1\&Q=-2x{e}^{-x}\\ Now,I.F.=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int dx}}=e^x\end{array}$

The general solution of the given differential equation is given by,

$\begin{array}{l}y(I.F.)={\displaystyle \int \left(Q\times I.F.\right)dx+C}\\ \Rightarrow y{e}^x={\displaystyle \int (-2x{e}^{-x}.e^x)dx+C}\\ \Rightarrow y{e}^x=-{\displaystyle \int 2xdx+C}\\ \Rightarrow y{e}^x=-x^2+C\\ \Rightarrow y{e}^x+x^2=C\end{array}$

Therefore, option (c) is correct.