NCERT Solutions for Exercise 9.5 Class 12 Maths Chapter 9- Differential Equations

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:

$$\frac{dy}{dx}=\frac{-[x\sin^{2}(y/x)-y]}{x} = F(x,y)$$ ..................(i)

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

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)

on integrating both sides, we get;

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

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

On substituting v =y/x

$$=\cot (y/x) = \log\left | Cx \right |$$ ............................(ii)

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

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

put this value of C in eq (ii)

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

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