Class 12 Ncert Solutions Maths Differential Equations

                                      Class 12 Ncert Solutions Maths Differential Equations
                                                                     Exercise 9.5


Previous Exercises
Exercise 9.1 &9.2
Exercise 9.4

Next Exercises
Exercise 9.6



In each of the Exercises 1 to 10 , sow that the given differential equation is homogeneous and solve each of them.

1. \((x^2+xy)\) dy = \((x^2+y^2)\) dx

Sol: The given equation can be written as follows

\begin{align}
\frac{dy}{dx}&=\frac{x^2+y^2}{x^2+xy}\\
\\
\frac{dy}{dx}&={x^2(1+\frac{y^2}{x^2})\over x^2(1+\frac{y}{x})}\\
\\
\frac{dy}{dx}&={(1+\frac{y^2}{x^2})\over (1+\frac{y}{x})}\tag 1\\
\\
&=g({y\over x})\\

\end{align}
Thus the given differential equation is a homogeneous equation.

Let y =vx, therefore $$ \frac{dy}{dx} = v+ x\frac{dv}{dx}$$.
Substituting these values in equation (1) we get
\begin{align}
v+x\frac{dv}{dx} &= \frac{1+\frac{v^2x^2}{x^2}}{1+\frac{vx}{x}}\\
\\
v+ x\frac{dv}{dx}&=\frac{1+v^2}{1+v}\\
\\
x\frac{dv}{dx}&=\frac{1+v^2}{1+v}-v\\
\\
x\frac{dv}{dx}&=\frac{1+v^2-v-v^2}{1+v}\\
\\
x\frac{dv}{dx}&=\frac{1-v}{1+v}\\
\\
\frac{1+v}{1-v}\ dv&=\frac{dx}{x}\\
\\
\frac{2-(1-v)}{1-v}\ dv&=\frac{dx}{x}\\
\\
\frac{2}{1-v}dv - dv &=\frac{dx}{x}\\
\\
-2\frac{d(1-v)}{1-v}-dv&=\frac{dx}{x}\\
\\
2\frac{d(1-v)}{1-v}+dv&=-\frac{dx}{x}\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
2\int \frac{d(1-v)}{1-v}\ dv + \int dv&=- \int \frac{dx}{x}\\
2\log |1-v| + v&=-\log x + \log C\\
2\log |1-v| + v&= \log \frac{C}{x}\\
\end{align}
Substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
2\log |1-v| + v &= \log \frac{C}{x}\\
\\
2\log |1-\frac{y}{x}| + \frac{y}{x}\log e &= \log \frac{C}{x}\\
\\
\log |1-\frac{y}{x}|^2+ \log e^{\frac{y}{x}} &= \log \frac{C}{x}\\
\\
\log \left(1-\frac{y^2}{x^2}\right)e^{\frac{y}{x}} &=\log \frac{C}{x}\\
\\
\left(1-\frac{y^2}{x^2}\right)e^{\frac{y}{x}} &=\ \frac{C}{x}\\
\\
x^2-y^2 &=\frac{x^2C}{xe^{\frac{y}{x}}}\\
\\
x^2-y^2 &= xCe^{-\frac{y}{x}}\\
\end{align}
This is the required general solution of the given differential equation.

2. \(y^{'}=\frac{x+y}{x}\)

Sol: The above equation can be written as
\begin{align}
\frac{dy}{dx}&=\frac{x+y}{x}\\
\\
\frac{dy}{dx} &=1+\frac{y}{x} \tag 1\\
&=g(\frac{y}{x})
\end{align}
Thus the given differential equation is  homogeneous.
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx} &= 1+\frac{vx}{x}\\
v+ x\frac{dv}{dx}&=1+v\\
x\frac{dv}{dx}&=1+v-v\\
dv&=\frac{dx}{x}\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
\int \ dv &=\frac{dx}{x}\\
v =\log x + C\\
\end{align}
Substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\frac{y}{x} &= \log x + C\\
y &=x\log x + xC\\
\end{align}
This is the required solution of the differential equation.

3.(x-y) dy - (x+y) dx=0

 Sol: The above equation can be written as
\begin{align}
\frac{dy}{dx}&=\frac{x+y}{x-y}\\
\\
\frac{dy}{dx} &=\frac{1+\frac{y}{x}}{1-\frac{y}{x}} \tag 1\\
&=g(\frac{y}{x})
\end{align}
Thus the given differential equation is  homogeneous.
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx} &= \frac{1+\frac{vx}{x}}{1-\frac{vx}{x}}\\
v+ x\frac{dv}{dx}&=\frac{1+v}{1-v}\\
x\frac{dv}{dx}&=\frac{1+v}{1-v}-v\\
x\frac{dv}{dx}&=\frac{1+v-v+v^2}{1-v}\\
x\frac{dv}{dx}&=\frac{1+v^2}{1-v}\\
\frac{1-v}{1+v^2}dv&=\frac{dx}{x}\\
\end{align}
Integrating both sides of the above equation we get
\begin{align}
\int \frac{1-v}{1+v^2}dv&=\int \frac{dx}{x}\\
\int \frac{1}{1+v^2}dv- \int \frac{vdv}{1+v^2}&=\frac{dx}{x}\\
\implies \tan^{-1} v -\frac{1}{2}\log |1+v^2| &= \log x + C\\

\end{align}
Substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\tan^{-1} v -\frac{1}{2}\log |1+\frac{y}{x}^2| &= \log x + C\\
\implies  \tan^{-1} v &=\frac{1}{2}\log |x^2+y^2| - \frac{1}{2}\log x^2+\log x + C\\
\implies \tan^{-1} v &=\frac{1}{2}\log |x^2+y^2| - \log x+\log x + C\\
\implies \tan^{-1} v &=\frac{1}{2}\log |x^2+y^2|  + C\\
\end{align}
This is the required solution of the differential equation.

4. \((x^2 -y^2) dx +2xy\ dy= 0\)

Sol: The given differential equation can be written as
\begin{align}
\frac{dy}{dx}&=\frac{x^2 -y^2}{-2xy}\\
\\
\frac{dy}{dx} &=\frac{1 -\frac{y^2}{x^2}}{-2\frac{y}{x}} \tag 1\\
&=g\left(\frac{y}{x}\right)
\end{align}
Thus the given differential equation is  homogeneous.
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx}&=-\frac{1}{2}\left(\frac{1-v^2}{v}\right)\\
\\
x\frac{dv}{dx}&=\frac{v^2-1}{2v}-v\\
\\
x\frac{dv}{dx}&=\frac{v^2-1-2v^2}{}\\
\\
x\frac{dv}{dx}&=\frac{-1-v^2}{2v}\\
\\
\frac{2v}{1+v^2}\ dv&=-\frac{dx}{x}\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
\int \frac{2v}{1+v^2}\ dv&=-\int \frac{dx}{x}\\
\\
\int\frac{d(1+v^2)}{1+v^2} &=-\int\frac{dx}{x}\\
\\
\log |1+v^2| &=-\log x+\log C\tag 2\\
\end{align}
Substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\log |1+\frac{y^2}{x^2}| &=-\log x+\log C\\
\\
\log |\frac{x^2+y^2}{x^2}| &=\log \frac{C}{x}\\
\\
\frac{x^2+y^2}{x^2} &=\frac{C}{x}\\
\\
x^2 +y^2 &= xC\\
\end{align}
This is the required general solution of the given differential equation.

5.\(x^2\frac{dy}{dx}=\ x^2-2y^2+xy\)

Sol: The above equation can be written as
\begin{align}
\frac{dy}{dx}&=\frac{x^2 -2y^2 +xy}{x}\\
\\
\frac{dy}{dx} &=1-2\frac{y^2}{x^2} + \frac{y}{x} \tag 1\\
&=g\left(\frac{y}{x}\right)
\end{align}
Thus the given differential equation is  homogeneous.
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx} &= 1-2v^2 + v\\
 x\frac{dv}{dx}&=1-2v^2 + v-v\\
x\frac{dv}{dx}&=1-2v^2\\

\frac{dv}{1-2v^2}&=\frac{dx}{x}\\
\end{align}
Integrating both sides of the above equation we get
\begin{align}

\int \frac{dv}{1-2v^2}&=\int \frac{dx}{x}\\
\\
\implies \frac{dv}{1^2-(\sqrt2v)^2} &=\frac{dx}{x}\\
\\
\implies \frac{1}{2\sqrt2}\log\left|\frac{1+\sqrt2v}{1-\sqrt2v}\right| &= \log x + C\\


\end{align}
The following formula has been used for integration
$$ \int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right |+ C $$
Substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\frac{1}{2\sqrt2}\log\left|\frac{1+\sqrt2{y\over x}}{1-\sqrt2{y\over x}}\right| &= \log x + C\\
\\
\frac{1}{2\sqrt2}\log\left|\frac{x+\sqrt2y}{x-\sqrt2y}\right| &= \log x + C\\
\end{align}
This is the required solution of the differential equation.

6. \(x\ dy -y\ dx\ = \sqrt{x^2 + y^2}\ dx\)

Sol: The above equation can be written as 

\begin{align}
\frac{dy}{dx}&=\frac{\sqrt{x^2 +y^2} +y}{x}\\
\\
\frac{dy}{dx} &=\sqrt{1 + \frac{y^2}{x^2}} +\frac{y}{x} \tag 1\\
&=g\left(\frac{y}{x}\right)
\end{align}
Thus the given differential equation is  homogeneous.
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v + x\frac{dv}{dx} &= \sqrt{1+\frac{v^2x^2}{x^2}} + \frac{y}{x} \\
v+ x \frac{dv}{dx} &= \sqrt{1+v^2} + v \\
x \frac{dv}{dx} &= \sqrt{1+v^2} \\
\frac{dv}{\sqrt{1+ v^2}} &= \frac{dx}{x} \\
\end{align}
Integrating both sides of the above equation we get
\begin{align}

\int \frac{dv}{\sqrt{1+ v^2}} &= \int \frac{dx}{x}\\
\\
\implies \log |v+ \sqrt{v^2+1}| &= \log x + \log C\tag 2\\
\end{align}
The following formula has been used for integration
$$ \int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right |+ C $$
Substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\log \left|\frac{y}{x}+ \sqrt{\frac{y^2}{x^2}+1}\right| &= \log x + \log C\\
\\
\log \left|\frac{y+ \sqrt{y^2+x^2}}{x}\right| &= \log xC\\
\\
\frac{y+\sqrt{x^2+y^2}}{x}&=xC\\
\\
y+\sqrt{x^2+y^2}&=x^2C\\
\end{align}
This is the required solution of the differential equation.

7. \(\left[x\cos\left(\frac{y}{x}\right)+y\sin\left(\frac{y}{x}\right)\right]y\ dx= \left[y\sin\left(\frac{y}{x}\right)-x\cos\left(\frac{y}{x}\right)\right]x\ dy\)

Sol: The given differential equation can be written as
\begin{align}
\frac{dy}{dx} &= \frac{y\left[x\cos\left({y\over x}\right) + y\sin\left({y\over x}\right)\right]}{x\left[y\sin\left({y\over x}\right) - x\cos\left({y\over x}\right)\right]}\\
\\
\frac{dy}{dx} &=\frac{y}{x} \frac{\left[\cos\left({y\over x}\right) + \frac{y}{x}\sin\left({y\over x}\right)\right]}{\left[\frac{y}{x}\sin\left({y\over x}\right) - \cos\left({y\over x}\right)\right]}\tag 1\\
&=g\left(\frac{y}{x}\right)\\
\end{align}
Thus the given differential equation is  homogeneous.
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx} &=v\frac{\left[\cos v + v\sin v\right]}{\left[v\sin v - \cos v\right]}\\
\\
v+x\frac{dv}{dx}&=\frac{v\cos v + v^2\sin v}{v\sin v - \cos v}\\
\\
x\frac{dv}{dx}&=\frac{v\cos v + v^2\sin v}{v\sin v - \cos v} -v\\
\\
x\frac{dv}{dx}&=\frac{v\cos v + v^2\sin v - v^2\sin v + v\cos v}{v\sin v - \cos v} \\
\\
x\frac{dv}{dx}&=\frac{2v\cos v }{v\sin v - \cos v} \\
\\
\frac{v\sin v - \cos v}{2v\cos v }\ dv &=\frac{dx}{x}\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
\int \frac{v\sin v - \cos v}{2v\cos v }\ dv &=\int \frac{dx}{x}\\
\\
\int  \frac{\sin v} {2\cos v}\ dv - \frac{dv}{2v} &=\int \frac{dx}{x}\\
\\
\int \tan v\ dv - \int \frac{dv}{v} &= 2\int \frac{dx}{x}\\
\\
\log|\sec v| -\log v&=2\log x + \log A\\
\\
\log\left|\frac{\sec v}{v}\right| &=\log Ax^2\\
\\
\implies |\frac{\sec v}{v}|&=Ax^2\\
\\
\sec v =vAx^2\\
\end{align}
now, substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\sec \left(\frac{y}{x}\right) &= \frac{y}{x}Ax^2\\
\sec \left(\frac{y}{x}\right) &= Axy\\
\cos \left(\frac{y}{x}\right) &= \frac{1}{Axy}\\
\implies xy \cos \left(\frac{y}{x}\right) &= \frac{1}{A}\\
\implies xy \cos \left(\frac{y}{x}\right) &= C\\
\end{align}
This is the required solution of the given differential equation.

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

Sol: The given differential equation can be written as
\begin{align}
\frac{dy}{dx} &= \frac{y - x\sin\left(\frac{y}{x}\right)}{x}\\
&= \frac{y}{x} - \sin\left(\frac{y}{x}\right)\tag (1)\\
&= g\left(\frac{y}{x}\right)\\
\end{align}
Thus the given differential equation is  homogeneous.
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx} &= \frac{vx}{x} - \sin\left(\frac{vx}{x}\right)\\
\\
v+x\frac{dv}{dx} &= v- \sin v \\
\\
x{dv}{dx} &= -sinv\\
\\
\frac{dv}{sinv}&=-\frac{dx}{x}\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
\int \frac{dv}{sinv}&=-\int \frac{dx}{x}\\
\\
\int cosec \ v\ dv&=-\frac{dx}{x}\\
\\
\log |cosec\ v - \cot v| &= -\log |x| + \log C\\
\\
\log |cosec\ v - \cot v| &= \log\frac{C}{x}\\
cosec\ v -\cot v =\frac{C}{x}\\
\frac{1}{\sin v} - \frac{ \sin v}{\cos v} &=\frac{C}{x}\\
\\
\frac{1-\cos v}{\sin v}&=\frac{C}{x}\\
\\
x[1- \cos v] &= C\sin v\\
\end{align}
Now, substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
x[1- \cos\left(\frac{y}{x}\right)] &= C\sin \left(\frac{y}{x}\right)\\
\end{align}
This is the required solution of the given differential equation.

9. \(ydx + x\log\left( \frac{y}{x}\right) dy - 2x dy = 0\)

Sol:  The given differential equation can be written as
\begin{align}
\frac{dy}{dx}&= \frac{-y}{x\log \frac{y}{x}-2x}\\
\\
&= \frac{\frac{-y}{x}}{\log \frac{y}{x}-2}\tag 1\\
&=g\left(\frac{y}{x}\right)
\end{align}
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx}&=\frac{-v}{\log v -2}\\
\\
x\frac{dv}{dx}&=\frac{-v}{\log v -2}-v\\
\\
x\frac{dv}{dx}&=\frac{-v-v\log v + 2v}{\log v -2}\\
\\
x\frac{dv}{dx}&=\frac{v-v\log v }{\log v -2}\\
\\
\frac{\log v - 2}{v-v\log v}&=\frac{dx}{x}\\
\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
\int \frac{\log v - 2}{v-v\log v}&=\int \frac{dx}{x}\\
\\
 \int \frac{\log v - 2}{v(1-\log v)}&=\int \frac{dx}{x}\\
\\
 \int \frac{(\log v - 1) -1}{v-v\log v}&=\int \frac{dx}{x}\\
\\
 -\int \frac{dv}{v} - \int \frac{dv}{v(1-\log v)}&=\int \frac{dv}{dx}\\
\\
 -\int \frac{dv}{v} + \int \frac{dv}{v(\log v-1)}&=\int \frac{dv}{dx}\\
\\
 -\int \frac{dv}{v} + \int \frac{d(\log v)}{\log v-1}&=\int \frac{dv}{dx}\\
\\
-\log |v| + \log|\log |v| -1| &=\log |x| + \log C\\
\\
\log \frac{\log v -1}{v}&=\log xC\\
\\
\frac{\log v -1}{v}&= xC\\
\end{align}
Now, substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\frac{\log \left(\frac{y}{x}\right) -1}{\frac{y}{x}}&= xC\\
\\
\log \left(\frac{y}{x}\right) -1&=Cx\times\frac{y}{x}\\
\\
\log \left(\frac{y}{x}\right) -1&=Cy\\
\end{align}
This is the required solution of the given differential equation.

10. \((1+e^{\frac{x}{y}}) dx + e^{\frac{x}{y}}(1-\frac{x}{y})dy=0\)

Sol: The given differential equation can be written as 

\begin{align}
\frac{dx}{dy}&= -\frac{e^{\frac{x}{y}}(1-\frac{x}{y})}{(1+e^{\frac{x}{y}})}\tag 1\\
\\
&=h\left(\frac{x}{y}\right)\\
\end{align}
Thus the differential equation is homogeneous.
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+y\frac{dv}{dy}&= -\frac{e^{v}(1-v)}{(1+e^v)}\\
\\
y\frac{dv}{dy}&= -\frac{e^{v}(1-v)}{(1+e^v)}-v\\
\\
y\frac{dv}{dy}&= \frac{v+e^v}{(1+e^v)}\\
\\
\frac{(1+e^v)}{v+e^v}\ dv &=-\frac{dy}{y}\\
\\
\frac{d(v+e^v)}{v+e^v}\ dv &=-\frac{dy}{y}\\
\\
\log |v+e^v| &=-\log y + \log C\\
\\
\log |v+e^v| &=-\log \frac{C}{y}\\
\\
v+e^v &=\frac{C}{y}\\
\\
\end{align}
Now, substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\frac{x}{y}+e^{\frac{x}{y}} &=\frac{C}{y}\\
\\
x+ye^{\frac{x}{y}}&=C\\
\end{align}
This is the required solution of the given differential equation.

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

11. (x+y)dy + (x-y)dx =0

Sol: The given differential equation can be written as 

\begin{align}
\frac{dy}{dx}&=-\frac{(x-y)}{(x+y)}\\
\\
\frac{dy}{dx}&=-\frac{x(1-\frac{y}{x})}{x(1+\frac{y}{x})}\\
\\
\frac{dy}{dx}&=-\frac{(1-\frac{y}{x})}{(1+\frac{y}{x})}\\
\\
&=g\left(\frac{y}{x}\right)\\
\end{align}
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx}&= -\frac{(1-v)}{(1+ v)}\\
\\
x\frac{dv}{dx}&= -\frac{(1-v)}{(1+ v)}-v\\
\\
x\frac{dv}{dx}&= \frac{-1+v-v-v^2}{(1+ v)}\\
\\
x\frac{dv}{dx}&= -\frac{(1+v^2)}{(1+ v)}\\
\\
\frac{1+v}{1+v^2}&= \frac{-dx}{x}\\
\\
\frac{dv}{1+v^2} + \frac{v}{1+v^2}\ dv &= -\frac{dx}{x}\\
\end{align}

Integrating both sides we get
\begin{align}
\int \frac{dv}{1+v^2} +\int  \frac{v}{1+v^2}\ dv &= -\int \frac{dx}{x}\\
\\
\tan^{-1}v +\frac{1}{2}\log |1+v^2|&= -\log x + A\\
\\
2\tan^{-1}v +\log |1+v^2|&= -2\log x + 2A\\
\\
2\tan^{-1}v +\log |1+v^2|&= -2\log x + C\\
\end{align}
Now, substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
2\tan^{-1}\frac{y}{x} +\log |1+\frac{y^2}{x^2}|&= -2\log x + C\\
\\
2\tan^{-1}\frac{y}{x} +\log |x^2+ y^2| - \log x^2&= -\log x^2 + C\\
\\
2\tan^{-1}\frac{y}{x} +\log |x^2+ y^2| &=  C\tag 2\\
\\
\end{align}
When y =1 then x=1. Putting these values in the baove equation we get
\begin{align}
2\tan^{-1}\frac{1}{1} +\log |1^2+ 1^2| &=  C\\
\\
C&=\log 2 + \tan^{-1}1\\
\\
C&=\log 2 + \frac{\pi}{2}\\
\end{align}
Putting this value of C in eq (2) we get
\begin{align}
2\tan^{-1}\frac{y}{x} +\log |x^2+ y^2| &=  \log 2 + \frac{\pi}{2}
\end{align}
This is the required particular solution of the differential equation.

12. \(x^2dy + y + y^2)dx=0\)

Sol: The given equation can be written as
\begin{align}
\frac{dy}{dx}&=-\frac{xy+y^2}{x^2}\\
&=-x^2\frac{\frac{y}{x}+\frac{y^2}{x^2}}{x^2}\\
&=-\left(\frac{y}{x} +\frac{y^2}{x^2}\right)\tag 1\\
&=g\left(\frac{y}{x}\right)\\
\end{align}
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx}&=-(v+v^2)\\
\\
v+x\frac{dv}{dx}&=-v-v^2\\
\\
x\frac{dv}{dx}&=-2v-v^2\\
\\
\frac{dv}{2v+v^2}&=-\frac{dx}{x}\\
\\
\frac{dv}{v(v+2}&=-\frac{dx}{x}\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
\int \frac{dv}{v(v+2)} &=-\int \frac{dx}{x}\\
\\
\int \frac{1}{2}\left[\frac{1}{v}-\frac{1}{v+2}\right]\ dv &=-\int \frac{dx}{x}\\
\\
\frac{1}{2}\left[\log v -\log |v+2|\right] &= -\log x + \log C\\
\\
\frac{1}{2}\log \left(\frac{v}{v+2}\right)&=\log\frac{C}{x}\\
\\
\log \left(\frac{v}{v+2}\right)&=2\log\frac{C}{x}\\
\\
\log \left(\frac{v}{v+2}\right)&=\log\left(\frac{C}{x}\right)^2\\
\\
\frac{v}{v+2}&= \frac{C^2}{x^2}\\
\end{align}
Now, substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\frac{\frac{y}{x}}{\frac{y}{x}+2}&=\frac{C^2}{x^2}\\
\\
\frac{x^2y}{y+2x}&=C^2\tag 2\\
\end{align}
given that when y=1 then x=1. Putting these values in above equation we get
\begin{align}
C^2 &= \frac{1\times1}{1+2}\\
&=\frac{1}{3}\\
\end{align}
Substituting the value of C in eq (2) we get
\begin{align}
\frac{x^2y}{y+2x}&=\frac{1}{3}\\
\\
y+2x =3x^2y\\
\end{align}
This is the required particular solution of the given differential equation.

13. \(\left[x\sin ^2\left(\frac{y}{x}\right)-y\right]dx +xdy=0\)

Sol: The given equation can be written as follows
\begin{align}
\frac{dy}{dx}&=-\frac{\left[x\sin ^2\left(\frac{y}{x}\right)-y\right]}{x}\\
\\
\frac{dy}{dx}&=\frac{y}{x}-\sin ^2\left(\frac{y}{x}\right)\tag 1\\
&=g\left(\frac{y}{x}\right)\\
\end{align}
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx} &= v - \sin ^2 v\\
\\
\frac{dv}{dx}&=-\sin ^2 v\\
\\
\frac{dv}{\sin ^2 v}&=-\frac{dx}{x}\\
\end{align}
Integrating both sides  of the equation we get
\begin{align}
\int \frac{dv}{\sin ^2 v}&=-\int \frac{dx}{x}\\
\\
\int cosec^2v\ dv &=-\int\frac{dx}{x}\\
\\
-\cot v&= -\log x-\log C\\
\cot v&= \log Cx\\
\end{align}
Now, substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\cot \frac{y}{x}&=\log (xC)\tag 2\\
\end{align}
given that y=\(\frac{\pi}{4}\) then x=1.Putting these values in eq (2) we get
\begin{align}
\cot \frac{\frac{\pi}{4}}{1}&=\log C\\
\\
\cot \frac{\pi}{4}&=\log C\\
\log C&=1\\
C&=e^1=e\\
\end{align}
Substituting the value of C in equation 2 we get
$$ \cot \frac{y}{x}=\log (xe) $$
This is the required particular solution of given differential equation.

14. \(\frac{dy}{dx} - \frac{y}{x}+cosec\left(\frac{y}{x}\right)=0\)

Sol: The given differential equation can be written as
\begin{align}
\frac{dy}{dx}&=\frac{y}{x}-cosec\left(\frac{y}{x}\right)\\
\end{align}
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx}&=v-cosec\ v\\
\\
x\frac{dv}{dx}&=-cosec\ v\\
\\
\frac{dv}{cosec\ v}&=-\frac{dx}{x}\\
 \end{align}
Integrating both sides of the equation we get
\begin{align}
\int \frac{dv}{cosec\ v}&=-\int \frac{dx}{x}\\
\int \sin v\ dv&=-\int \frac{dx}{x}\\
\\
-\cos v &=-\log x- \log C\\
\cos v&=\log xC\\
\end{align}
Now, substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
\cos\frac{y}{x}&=\log xC\tag 2\\
\end{align}
Given that when y=0 then x=1.Substituting these values in above equation we get
\begin{align}
\cos 0 &=\log (1\times C)\\
\log C&=1\\
C&=e^1\\
C&=e\\
\end{align}
Putting this value of C in eq(2) we get
\begin{align}
\cos \frac{y}{x}&=\log |xe|\\
\end{align}
This is the required particular solution of the given differential equation


15. \(2xy +y^2-2x^2\frac{dy}{dx}=0\)

Sol: The given differential equation can be written as
\begin{align}
\frac{dy}{dx} &=\frac{2xy+y^2}{2x^2}\\
&=\frac{y}{x}+\frac{1}{2}\frac{y^2}{x^2}\tag 1\\
&=g\left(\frac{y}{x}\right)\\
\end{align}
Let y =vx. Therefore, $$ \frac{dy}{dx} = v+x\frac{dv}{dx}$$
Substituting these values in eq (1) we get
\begin{align}
v+x\frac{dv}{dx}&=v+\frac{1}{2}v^2\\
x\frac{dv}{dx}&=\frac{1}{2}v^2\\
2\frac{dv}{v^2}&=\frac{dx}{x}\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
2\int \frac{dv}{v^2}&=\int \frac{dx}{x}\\
\\
2\frac{v^{-2+1}}{-2+1}&=\log x+ C\\
\\
-2\frac{1}{v}&=\log x + C
\end{align}
Now, substituitng \(v=\frac{y}{x}\) in the above equation we get
\begin{align}
-2\frac{1}{\frac{y}{x}}&=\log x + C\tag 2\\
\\
-2\frac{x}{y}&=\log x + C\\
\end{align}
Given that y=2 when x=1. Putting these values in above equation we get
\begin{align}
-2\frac{1}{2}&=\log 1 + C\\
C&=-1\\
\end{align}
Substituting the value of C in equation 2 we get
\begin{align}
-2\frac{x}{y} &=\log x -1\\
\frac{-2x}{\log x -1}&=y\\
y&=\frac{2x}{1-log|x|}\\
\end{align}
This is the required particular solution of the given differential equation.

16. A homogeneous differential equation of the form \(\frac{dy}{dx}=h\left(\frac{x}{y}\right)\) can be solved by making the substitution.
(A) y = vx  (B)   v=yx   (C) x=vy   (D) x=v.

Ans: The correct answer is option (C) beacuse only when we substitute x=vy , the function in \(\frac{x}{y}\) turns into a function of v.

17. Which of the following is a homogeneous differential equation?
(A) (4x+6y+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

Sol: The correct answer is option  (D) because (D) can be written as
\begin{align}
\frac{dy}{dx}&=-\frac{y^2}{x^2-xy-y^2}\\
&=-\frac{y^2}{x^2(1-\frac{y}{x}-\frac{y^2}{x^2})}\\
&=-\frac{\frac{y^2}{x^2}}{1-\frac{y}{x}-\frac{y^2}{x^2}}\\
&=g\left(\frac{y}{x}\right)\\
\end{align}


                                           Class 12 Ncert Solution Maths Differential Equations

Previous Exercises
Exercise 9.1 &9.2
Exercise 9.4

Next Exercises
Exercise 9.6

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Chapter 11 Three dimensional geometry
Chapter 13 Probability

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