NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.4 - Differential Equations

Question 2: Show that the given differential equation is homogeneousand solve each of them. $y' = \frac{x+y}{x}$

Answer:

the above differential eq can be written as,

$\frac{dy}{dx} = F(x,y)=\frac{x+y}{x}$ ............................(i)

Now, $F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x} = \lambda ^{0}F(x,y)$
Thus the given differential eq is a homogeneous equaion
Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= \frac{x+vx}{x} = 1+v$
$\\x\frac{dv}{dx}= 1\\ $

$dv = \frac{dx}{x}$
Integrating on both sides, we get; (and substitute the value of $v =\frac{y}{x}$ )

$\\v =\log x+C\\ $

$\frac{y}{x}=\log x+C\\ y = x\log x +Cx$
this is the required solution

Question 3: Show that the given differential equation is homogeneous and solve each of them.

$(x-y)dy - (x+y)dx = 0$

Answer:

The given differential eq can be written as;

$\frac{dy}{dx}=\frac{x+y}{x-y} = F(x,y)(let\ say)$ ....................................(i)

$F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}= \lambda ^{0}F(x,y)$
Hence it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

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

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

$\frac{1-v}{1+v^{2}}dv = (\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}})dv=\frac{dx}{x}$
Integrating on both sides, we get;

$\tan^{-1}v-1/2 \log(1+v^{2})=\log x+C$
again substitute the value of $v=y/x$
$\\\tan^{-1}(y/x)-1/2 \log(1+(y/x)^{2})=\log x+C\\$

$\tan^{-1}(y/x)-1/2 [\log(x^{2}+y^{2})-\log x^{2}]=\log x+C\\ $

$tan^{-1}(y/x) = 1/2[\log (x^{2}+y^{2})]+C$

This is the required solution.

Question 4: Show that the given differential equation is homogeneous and solve each of them.

$(x^2 - y^2)dx + 2xydy = 0$

Answer:

we can write it as;

$\frac{dy}{dx}= -\frac{(x^{2}-y^{2})}{2xy} = F(x,y)\ (let\ say)$ ...................................(i)

$F(\lambda x,\lambda y) = \frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)} = \lambda ^{0}.F(x,y)$
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx} = \frac{ x^{2}-(vx)^{2}}{2x(vx)} =\frac{v^{2}-1}{2v}$
$\\x\frac{dv}{dx} =\frac{v^{2}+1}{2v}\\ \frac{2v}{1+v^{2}}dv=\frac{dx}{x}$
integrating on both sides, we get

$\log (1+v^{2})= -\log x +\log C = \log C/x$
$\\= 1+v^{2} = C/x\\ = x^2+y^{2}=Cx$ .............[ $v =y/x$ ]
This is the required solution.

Question 5: Show that the given differential equation is homogeneous and solve it.

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

Answer:

$\frac{dy}{dx}= \frac{x^{2}-2y^{2}+xy}{x^{2}} = F(x,y)\ (let\ say)$

$F(\lambda x,\lambda y)= \frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda .\lambda )xy}{(\lambda x)^{2}} = \lambda ^{0}.F(x,y)$ ............(i)
Hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

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

$\frac{dv}{1-2v^{2}}=\frac{dx}{x}$

$1/2[\frac{dv}{(1/\sqrt{2})^{2}-v^{2}}] = \frac{dx}{x}$

On integrating both sides, we get;

$\frac{1}{2\sqrt{2}}\log (\frac{1/\sqrt{2}+v}{1/\sqrt{2}-v}) = \log x +C$
after substituting the value of $v= y/x$

$\frac{1}{2\sqrt{2}}\log (\frac{x+\sqrt{2}y}{x-\sqrt{2}y}) = \log \left | x \right | +C$

This is the required solution

Question 6: Show that the given differential equation is homogeneous and solve it.

$xdy - ydx = \sqrt{x^2 + y^2}dx$

Answer:

$\frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} = F(x,y)$ .................................(i)

$F(\mu x,\mu y)=\frac{\mu y+\sqrt{(\mu x)^{2}+(\mu y)^{2}}}{\mu x} =\mu^{0}.F(x,y)$
henxe it is a homogeneous equation

Now, to solve substitute y = vx

Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

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

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

On integrating both sides,

$\Rightarrow \log \left | v+\sqrt{1+v^{2}} \right | = \log \left | x \right |+\log C$
Substitute the value of v=y/x , we get

$\\\Rightarrow \log \left | \frac{y+\sqrt{x^{2}+y^{2}}}{x} \right | = \log \left | Cx \right |\\$

$y+\sqrt{x^{2}+y^{2}} = Cx^{2}$

Required solution

Question 7: Solve.

$\left\{x\cos\left(\frac{y}{x} \right ) + y\sin\left(\frac{y}{x} \right ) \right \}ydx = \left\{y\sin\left(\frac{y}{x} \right ) - x\cos\left(\frac{y}{x} \right ) \right \}xdy$

Answer:

$\frac{dy}{dx} =\frac{x \cos(y/x)+y\sin(y/x)}{y\sin(y/x)-x\cos(y/x)}.\frac{y}{x} = F(x,y)$ ......................(i)
By looking at the equation we can directly say that it is a homogenous equation.

Now, to solve substitute y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx} =\frac{v \cos v+v^{2}\sin v}{v\sin v-\cos v}\\$

$=x\frac{dv}{dx} = \frac{2v\cos v}{v\sin v-\cos v}\\$

$=(\tan v-1/v)dv = \frac{2dx}{x}$

integrating on both sides, we get

$\\=\log(\frac{\sec v}{v})= \log (Cx^{2})\\$

$=\sec v/v =Cx^{2}$
substitute the value of v= y/x , we get

$\\\sec(y/x) =Cxy \\ $

$xy \cos (y/x) = k$

Required solution

Question 8: Solve.

$x\frac{dy}{dx} - y + x\sin\left(\frac{y}{x}\right ) = 0$

Answer:

$\frac{dy}{dx}=\frac{y-x \sin(y/x)}{x} = F(x,y)$ ...............................(i)

$F(\mu x, \mu y)=\frac{\mu y-\mu x \sin(\mu y/\mu x)}{\mu x} = \mu^{0}.F(x,y)$
it is a homogeneous equation

Now, to solve substitute y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= v- \sin v = -\sin v$
$\Rightarrow -\frac{dv}{\sin v} = -(cosec\ v)dv=\frac{dx}{x}$

On integrating both sides we get;

$\\\Rightarrow \log \left | cosec\ v-\cot v \right |=-\log x+ \log C\\ $

$\Rightarrow cosec (y/x) - \cot (y/x) = C/x$

$= x[1-\cos (y/x)] = C \sin (y/x)$ Required solution

Question 9: Solve.

$ydx + x\log\left(\frac{y}{x} \right ) -2xdy = 0$

Answer:

$\frac{dy}{dx}= \frac{y}{2x-x \log(y/x)} = F(x,y)$ ..................(i)

$\frac{\mu y}{2\mu x-\mu x \log(\mu y/\mu x)} = F(\mu x,\mu y) = \mu^{0}.F(x,y)$

hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}= \frac{v}{2-\log v}\\ =x\frac{dv}{dx} = \frac{v\log v-v}{2-\log v}\\$

$=[\frac{1}{v(\log v-1)}-\frac{1}{v}]dv=\frac{dx}{x}$
integrating on both sides, we get; ( substituting v =y/x)

$\\\Rightarrow \log[\log(y/x)-1]-\log(y/x)=\log(Cx)\\\Rightarrow \frac{x}{y}[\log(y/x)-1]=Cx\\$

$\Rightarrow \log (y/x)-1=Cy$

This is the required solution of the given differential eq

Question 10: Solve.

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

Answer:

$\frac{dx}{dy}=\frac{-e^{x/y}(1-x/y)}{1+e^{x/y}} = F(x,y)$ .......................................(i)

$= F(\mu x,\mu y)=\frac{-e^{\mu x/\mu y}(1-\mu x/\mu y)}{1+e^{\mu x/\mu y}} =\mu^{0}.F(x,y)$
Hence it is a homogeneous equation.

Now, to solve substitute x = yv

Diff erentiating on both sides wrt $x$
$\frac{dx}{dy}= v +y\frac{dv}{dy}$

Substitute this value in equation (i)

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

$=y\frac{dv}{dy} = -\frac{v+e^{v}}{1+e^{v}}\\$

$=\frac{1+e^{v}}{v+e^{v}}dv=-\frac{dy}{y}$

Integrating on both sides, we get;

${100} \log(v+e^{v})=-\log y+ \log c =\log (c/y)\\$

$=[\frac{x}{y}+e^{x/y}]= \frac{c}{y}\\\Rightarrow x+ye^{x/y}=c$
This is the required solution of the diff equation.

Question 11: Solve for particular solution.

$(x + y)dy + (x -y)dx = 0;\ y =1\ when \ x =1$

Answer:

$\frac{dy}{dx}=\frac{-(x-y)}{x+y} =F(x,y)$ ..........................(i)

We can clearly say that it is a homogeneous equation.

Now, to solve substitute y = vx

Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

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

$\Rightarrow x\frac{dv}{dx} = -\frac{(1+v^{2})}{1+v}$

$\frac{1+v}{1+v^{2}}dv = [\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}]dv=-\frac{dx}{x}$

On integrating both sides

$\\=\frac{1}{2}[\log (1+v^{2})]+\tan^{-1}v = -\log x +k\\$

$=\log(1+v^{2})+2\tan^{-1}v=-2\log x +2k\\$

$=\log[(1+(y/x)^{2}).x^{2}]+2\tan^{-1}(y/x)=2k\\$]

$=\log(x^{2}+:y^{2})+2\tan^{-1}(y/x) = 2k$ ......................(ii)

Now, y=1 and x= 1

$\\=\log 2 +2\tan^{-1}1=2k\\ $

$=\pi/2+\log 2 = 2k\\$

After substituting the value of 2k in eq. (ii)

$\log(x^{2}+y^2)+2\tan^{-1}(y/x)=\pi/2+\log 2$

This is the required solution.

Question 12: Solve for particular solution.

$x^2dy + (xy + y^2)dx = 0; y =1\ \textup{when}\ x = 1$

Answer:

$\frac{dy}{dx}= \frac{-(xy+y^{2})}{x^{2}} = F(x,y)$ ...............................(i)

$F(\mu x, \mu y)=\frac{-\mu^{2}(xy+(\mu y)^{2})}{(\mu x)^{2}} =\mu ^{0}. F(x,y)$
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i), we get

$\\=v+\frac{xdv}{dx}= -v- v^{2}\\ $

$=\frac{xdv}{dx}=-v(v+2)\\ $

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

$=1/2[\frac{1}{v}-\frac{1}{v+2}]dv=-\frac{dx}{x}$

Integrating on both sides, we get;

$\\=\frac{1}{2}[\log v -\log(v+2)]= -\log x+\log C\\$

$=\frac{v}{v+2}=(C/x)^{2}$

replace the value of v=y/x

$\frac{x^{2}y}{y+2x}=C^{2}$ .............................(ii)

Now y =1 and x = 1

$C = 1/\sqrt{3}$
therefore,

$\frac{x^{2}y}{y+2x}=1/3$

Required solution

Question 13: Solve for particular solution.

$\left [x\sin^2\left(\frac{y}{x} \right ) - y \right ]dx + xdy = 0;\ y =\frac{\pi}{4}\ when \ x = 1$

Answer:

$
\begin{aligned}
& \frac{d y}{d x}=\frac{-\left[x \sin ^2(y / x)-y\right]}{x}=F(x, y) \cdots \cdots \cdots \cdots \cdots \cdots . . . . . . .(i) \\
& F(\mu x, \mu y)=\frac{-\left[\mu x \sin ^2(\mu y / \mu x)-\mu y\right]}{\mu x}=\mu^0 . F(x, y)
\end{aligned}
$

Hence it is a homogeneous eq
Now, to solve substitute $\mathrm{y}=\mathrm{vx}$
Differentiating on both sides wrt $x$

$
\frac{d y}{d x}=v+x \frac{d v}{d x}
$

Substitute this value in equation (i)
on integrating both sides, we get;

$
\begin{aligned}
& -\cot v=\log |x|-C \\
& =\cot v=\log |x|+\log C
\end{aligned}
$

On substituting $\mathrm{v}=\mathrm{y} / \mathrm{x}$

$
=\cot (y / x)=\log |C x| .
$

Now, $y=\pi / 4 @ x=1$

$
\begin{aligned}
& \cot (\pi / 4)=\log C \\
& =C=e^1
\end{aligned}
$

put this value of C in eq (ii)

$
\cot (y / x)=\log |e x|
$

Required solution.

Question 14: Solve for particular solution.

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

Answer:

$\frac{dy}{dx} = \frac{y}{x} -cosec(y/x) =F(x,y)$ ....................................(i)

the above eq is homogeneous. So,
Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}=v- cosec\ v\\ $

$=x\frac{dv}{dx} = -cosec\ v\\$

$=-\frac{dv}{cosec\ v}= \frac{dx}{x}\\$

$=-\sin v dv = \frac{dx}{x}$

on integrating both sides, we get;

$\\=cos\ v = \log x +\log C =\log Cx\\$

$=\cos(y/x)= \log Cx$ .................................(ii)

now y = 0 and x =1 , we get

$C =e^{1}$

put the value of C in eq 2

$\cos(y/x)=\log \left | ex \right |$

Question 15: Solve for particular solution.

$2xy + y^2 - 2x^2\frac{dy}{dx} = 0 ;\ y = 2\ \textup{when}\ x = 1$

Answer:

The above eq can be written as;

$\frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}} = F(x,y)$
By looking, we can say that it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

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

$=x\frac{dv}{dx} = v^{2}/2\\ $

$= \frac{2dv}{v^{2}}=\frac{dx}{x}$

integrating on both sides, we get;

$\\=-2/v=\log \left | x \right |+C\\$

$=-\frac{2x}{y}=\log \left | x \right |+C$ .............................(ii)

Now, y = 2 and x =1, we get

C =-1
put this value in equation(ii)

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

$\Rightarrow y = \frac{2x}{1- \log x}$

Question 16: A homogeneous differential equation of the from $\frac{dx}{dy}= h\left(\frac{x}{y} \right )$ can be solved by making the substitution.

(A) $y = vx$

(B) $v = yx$

(C) $x = vy$

(D) $x =v$

Answer:

$\frac{dx}{dy}= h\left(\frac{x}{y} \right )$
for solving this type of equation put x/y = v
x = vy

option C is correct

Question 17: Which of the following is a homogeneous differential equation?

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

(B) $(xy)dx - (x^3 + y^3)dy = 0$

(C) $(x^3 +2y^2)dx + 2xydy =0$

(D) $y^2dx + (x^2 -xy -y^2)dy = 0$

Answer:

Option D is the right answer.

$y^2dx + (x^2 -xy -y^2)dy = 0$
$\frac{dy}{dx}=\frac{y^{2}}{x^{2}-xy-y^{2}} = F(x,y)$
we can take out lambda as a common factor and it can be cancelled out