Exercise 9.4 (Revised) - Chapter 9 - Differential Equations - Ncert Solutions class 12 - Maths
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NCERT Class 12 Maths Solutions: Chapter 9 - Differential Equations
In each of the following Questions 1 to 5 , show that the differential equation is homogenous and solve each of them:
Ex 9.4 Question 1.
$\left(x^2+x y\right) d y=\left(x^2+y^2\right) d x$
Answer.
Given: Differential equation $\left(x^2+x y\right) d y=\left(x^2+y^2\right) d x$
Here degree of each coefficients of $d x$ and $d y$ is same therefore, it is homogenous.
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=\frac{x^2+y^2}{x^2+x y} \\
& \Rightarrow \frac{d y}{d x}=\frac{x^2\left(1+\frac{y^2}{x^2}\right)}{x^2\left(1+\frac{y}{x}\right)} \\
& \Rightarrow \frac{d y}{d x}=\frac{1+\left(\frac{y}{x}\right)^2}{1+\left(\frac{y}{x}\right)} \ldots \\
& \Rightarrow \mathrm{F}\left(\frac{y}{x}\right),
\end{aligned}
$
therefore the given differential equation is homogeneous.
$
\text { Putting } \frac{y}{x}=v
$
$
\begin{aligned}
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting value of $\frac{y}{x}$ and $\frac{d y}{d x}$ in eq. (ii), $v+x \frac{d v}{d x}=\frac{1+v^2}{1+v}$
$
\begin{aligned}
& \Rightarrow x \frac{d v}{d x}=\frac{1+v^2}{1+v}-v \\
& \Rightarrow x \frac{d v}{d x}=\frac{1+v^2-v-v^2}{1+v}=\frac{1-v}{1+v} \\
& \Rightarrow x(1+v) d v=(1-v) d x \\
& \Rightarrow \frac{1+v}{1-v} d v=\frac{d x}{x} \text { [Separating variables] }
\end{aligned}
$
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{1+v}{1-v} d v=\int \frac{1}{x} d x \\
& \Rightarrow \int \frac{1+1-1+v}{1-v} d v=\int \frac{1}{x} d x \\
& \Rightarrow \int \frac{2-(1-v)}{1-v} d v=\log x+c \\
& \Rightarrow \int\left(\frac{2}{1-v}-1\right) d v=\log x+c
\end{aligned}
$
$\begin{aligned}
& \Rightarrow \int\left(\frac{2}{1-v}-1\right) d v=\log x+c \\
& \Rightarrow \frac{2 \log (1-v)}{-1}-v=\log x+c \\
& \Rightarrow-2 \log (1-v)-v=\log x+c
\end{aligned}$
$\begin{aligned}
& \text { Putting } v=\frac{y}{x}-2 \log \left(1-\frac{y}{x}\right)-\frac{y}{x}=\log x+c \\
& \Rightarrow 2 \log \left(1-\frac{y}{x}\right)+\frac{y}{x}=-\log x-c \\
& \Rightarrow \log \left(\frac{x-y}{x}\right)^2+\log x=-\frac{y}{x}-c \\
& \Rightarrow \log \left(\frac{x-y}{x}\right)^2 x=-\frac{y}{x}-c \\
& \Rightarrow \frac{(x-y)^2}{x}=e^{\frac{-y}{x}-\varepsilon} \\
& \Rightarrow \frac{(x-y)^2}{x}=e^{\frac{-y}{x}} \cdot e^{-c} \\
& \Rightarrow(x-y)^2=\mathrm{C} x e^{\frac{-y}{x}} \text { where } \mathrm{C}=e^{-c}
\end{aligned}$
Ex 9.4 Question 2.
$y^{\prime}=\frac{x+y}{x}$
Answer.
Given: Differential equation $y^{\prime}=\frac{x+y}{x}$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=\frac{x}{x}+\frac{y}{x} \\
& \Rightarrow \frac{d y}{d x}=1+\frac{y}{x}=f\left(\frac{y}{x}\right)
\end{aligned}
$
Therefore, eq. (i) is homogeneous.
Putting $\frac{y}{x}=v$
$
\begin{aligned}
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$$
Putting value of $y$ and $\frac{d y}{d x}$ in eq. (i)
$
\begin{aligned}
& \Rightarrow v+x \frac{d v}{d x}=1+v \\
& \Rightarrow x \frac{d v}{d x}=1 \\
& \Rightarrow x d v=d x \\
& \Rightarrow d v=\frac{d x}{x} \text { [Separating variables] }
\end{aligned}
$
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int 1 d v=\int \frac{d x}{x} \\
& \Rightarrow v=\log |x|+c
\end{aligned}
$
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v, \\
& \Rightarrow \frac{y}{x}=\log |x|+c \\
& \Rightarrow y=x \log |x|+x c
\end{aligned}
$
Ex 9.4 Question 3.
$(x-y) d y+(x+y) d x=0$
Answer.
Given: Differential equation $(x-y) d y-(x+y) d x=0$
This given equation is homogeneous because each coefficients of $d x$ and $d y$ is of degree 1.
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=\frac{x+y}{x-y} \\
& \Rightarrow \frac{d y}{d x}=\frac{x\left(1+\frac{y}{x}\right)}{x\left(1-\frac{y}{x}\right)} \\
& \Rightarrow \frac{d y}{d x}=\frac{1+\frac{y}{x}}{1-\frac{y}{x}}=f\left(\frac{y}{x}\right) \\
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting value of $y$ and $\frac{d y}{d x}$ in eq. (ii)
$
\begin{aligned}
& v+x \frac{d v}{d x}=\frac{1+v}{1-v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{1+v}{1-v}-v \\
& =>x \frac{d v}{d x}=\frac{1+v-v+v^2}{1-v} \\
& \Rightarrow=>x \frac{d v}{d x}=\frac{1+v^2}{1-v}
\end{aligned}
$
$\Rightarrow x(1-v) d v=\left(1+v^2\right) d x$
$\Rightarrow \frac{(1-v)}{1+v^2} d v=\frac{d x}{x}$ [Separating variables]
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{(1-v)}{1+v^2} d v=\int \frac{1}{x} d x \\
& \Rightarrow \int \frac{1}{1+v^2} d v-\int \frac{v}{1+v^2} d v=\int \frac{1}{x} d x+c \\
& \Rightarrow \tan ^{-1} v-\frac{1}{2} \int \frac{2 v}{1+v^2} d v=\int \frac{1}{x} d x+c \\
& \Rightarrow \tan ^{-1} v-\frac{1}{2} \log \left(1+v^2\right)=\log x+c
\end{aligned}
$
Putting $v=\frac{y}{x}$,
$\begin{aligned}
& \Rightarrow \tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left(1+\frac{y^2}{x^2}\right)=\log x+c \\
& \Rightarrow \tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left(\frac{x^2+y^2}{x^2}\right)=\log x+c \\
& \Rightarrow \tan ^{-1} \frac{y}{x}-\left[\frac{1}{2} \log \left(x^2+y^2\right)-\log x^2\right]=\log x+c \\
& \Rightarrow \tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left(x^2+y^2\right)+\frac{1}{2} \log x^2=\log x+c \\
& \Rightarrow \tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left(x^2+y^2\right)+\frac{1}{2} \cdot 2 \log x=\log x+c \\
& \Rightarrow \tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left(x^2+y^2\right)=c
\end{aligned}$
$\Rightarrow \tan ^{-1} \frac{y}{x}=\frac{1}{2} \log \left(x^2+y^2\right)+c$
Ex 9.4 Question 4.
$\left(x^2-y^2\right) d x+2 x y d y=0$
Answer.
Given: Differential equation $\left(x^2-y^2\right) d x+2 x y d y=0$ -
This equation is homogeneous because degree of each coefficient of $d x$ and $d y$ is same i.e.,
$
\begin{aligned}
& 2 \\
& \Rightarrow 2 x y d y=-\left(x^2-y^2\right) d x \\
& \Rightarrow \frac{d y}{d x}=\frac{-\left(x^2-y^2\right)}{2 x y}=\frac{y^2-x^2}{2 x y} \\
& \Rightarrow \frac{d y}{d x}=\frac{\left(\frac{y}{x}\right)^2-1}{2 \frac{y}{x}}=f\left(\frac{y}{x}\right) \ldots
\end{aligned}
$
Therefore, the given equation is homogeneous.
Put $\frac{y}{x}=v$
$
\begin{aligned}
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in eq. (ii), we get
$
v+x \frac{d v}{d x}=\frac{v^2-1}{2 v}
$
$
\begin{aligned}
& \Rightarrow x \frac{d v}{d x}=\frac{v^2-1}{2 v}-v \\
& \Rightarrow x \frac{d v}{d x}=\frac{v^2-1-2 v^2}{2 v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{-v^2-1}{2 v}=\frac{-\left(v^2+1\right)}{2 v} \\
& \Rightarrow x 2 v d v=-\left(v^2+1\right) d x \\
& \Rightarrow \frac{2 v d v}{v^2+1}=\frac{-d x}{x}
\end{aligned}
$
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{2 v}{v^2+1} d v=-\int \frac{1}{x} d x \\
& \Rightarrow \log \left(v^2+1\right)+\log x=\log c \\
& \Rightarrow \log \left(v^2+1\right) x=\log c \\
& \Rightarrow\left(v^2+1\right) x=c
\end{aligned}
$
$
\begin{aligned}
& \text { Put } \frac{y}{x}=v, \\
& \Rightarrow\left(\frac{y^2}{x^2}+1\right) x=c \\
& \Rightarrow\left(\frac{y^2+x^2}{x^2}\right) x=c
\end{aligned}
$
$\Rightarrow \frac{y^2+x^2}{x}=c$
$\Rightarrow x^2+y^2=c x$
Ex 9.4 Question 5.
$x^2\left(\frac{d y}{d x}\right)=x^2-2 y^2+x y$
Answer.
Given: Differential equation $x^2 \frac{d y}{d x}=x^2-2 y^2+x y$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=\frac{x^2}{x^2}-\frac{2 y^2}{x^2}+\frac{x y}{x^2} \\
& \Rightarrow \frac{d y}{d x}=1-2\left(\frac{y}{x}\right)^2+\left(\frac{y}{x}\right)=f\left(\frac{y}{x}\right)
\end{aligned}
$
Therefore, the given differential equation is homogeneous as all terms of $x$ and $y$ are of same degree i.e., degree 2.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in eq. (i), we get
$
\begin{aligned}
& v+x \frac{d v}{d x}=1-2 v^2+v \\
& \Rightarrow x \frac{d v}{d x}=1-2 v^2 \\
& \Rightarrow x d v=\left(1-2 v^2\right) d x
\end{aligned}
$
$\Rightarrow \frac{d v}{1-2 v^2}=\frac{d x}{x}$ [Separating variables]
Integrating both sides,
$
\begin{aligned}
& {\left[\because \int \frac{1}{a^2-x^2} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\right]} \\
&
\end{aligned}
$
Putting $\frac{y}{x}=v$,
$
\begin{aligned}
& \Rightarrow \frac{1}{2 \sqrt{2}} \log \left|\frac{1+\sqrt{2} \frac{y}{x}}{1-\sqrt{2} \frac{y}{x}}\right|=\log |x|+c \\
& \Rightarrow \frac{1}{2 \sqrt{2}} \log \left|\frac{x+\sqrt{2} y}{x-\sqrt{2} y}\right|=\log |x|+c
\end{aligned}
$
In each of the Questions 6 to 10 , show that the given differential equation is homogeneous and solve each of them:
Ex 9.4 Question 6.
$x d y-y d x=\sqrt{x^2+y^2} d x$
Answer.
Given: Differential equation $x d y-y d x=\sqrt{x^2+y^2} d x$
$
\begin{aligned}
& \Rightarrow x d y=y d x+\sqrt{x^2+y^2} d x \\
& \Rightarrow x \frac{d y}{d x}=y+\sqrt{x^2+y^2} \\
& \Rightarrow x \frac{d y}{d x}=y+x \sqrt{1+\left(\frac{y}{x}\right)^2}
\end{aligned}
$
[Dividing by $x$ ]
$
\Rightarrow \frac{d y}{d x}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}=f\left(\frac{y}{x}\right)
$
Therefore given differential equation is homogeneous.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in eq. (i), we get
$
\begin{aligned}
& \Rightarrow v+x \frac{d v}{d x}=v+\sqrt{1+v^2} \\
& \Rightarrow x \frac{d v}{d x}=\sqrt{1+v^2} \\
& \Rightarrow \frac{d v}{\sqrt{1+v^2}}=\frac{d x}{x}
\end{aligned}
$
Integrating both sides,
$\begin{aligned}
& \Rightarrow \int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x} \\
& \Rightarrow \log \left(v+\sqrt{1+v^2}\right)=\log x+\log c \\
& \text { Putting } \frac{y}{x}=v, \\
& \Rightarrow \log \left(\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}\right)=\log x c \\
& \Rightarrow \log \left(\frac{y+\sqrt{x^2+y^2}}{x}\right)=\log c x \\
& \Rightarrow y+\sqrt{x^2+y^2}=c x^2
\end{aligned}$
Ex 9.4 Question 7.
$\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y$
Answer.
Given: Differential equation
$\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y$
$\Rightarrow \frac{d y}{d x}=\frac{\left(x \cos \frac{y}{x}+y \sin \frac{y}{x}\right) y}{\left(y \sin \frac{y}{x}-x \cos \frac{y}{x}\right) x}=\frac{x y \cos \frac{y}{x}+y^2 \sin \frac{y}{x}}{x y \sin \frac{y}{x}-x^2 \cos \frac{y}{x}}$
$\Rightarrow \frac{d y}{d x}=\frac{\frac{y}{x} \cos \frac{y}{x}+\left(\frac{y}{x}\right)^2 \sin \frac{y}{x}}{\frac{y}{x} \sin \frac{y}{x}-\cos \frac{y}{x}}=f\left(\frac{y}{x}\right)$.
Therefore, the given differential equation is homogeneous.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x} \\
& \text { Putting these values of } \frac{y}{x} \text { and } \frac{d y}{d x} \text { in eq. (i), we get } \\
& v+x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v}-v \\
& \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v-v^2 \sin v+v \cos v}{v \sin v-\cos v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v} \\
& \Rightarrow x(v \sin v-\cos v) d v=2 v \cos v d x \\
& \Rightarrow \frac{v \sin v-\cos v}{v \cos v} d v=2 \frac{d x}{x}[\text { Separating variables] }
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in eq. (i), we get
$
\begin{aligned}
& v+x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v}-v \\
& \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v-v^2 \sin v+v \cos v}{v \sin v-\cos v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v} \\
& \Rightarrow x(v \sin v-\cos v) d v=2 v \cos v d x \\
& \Rightarrow \frac{v \sin v-\cos v}{v \cos v} d v=2 \frac{d x}{x} \text { [Separating variables] }
\end{aligned}
$
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{v \sin v-\cos v}{v \cos v} d v=2 \int \frac{1}{x} d x \\
& \Rightarrow \int\left(\frac{v \sin v}{v \cos v}-\frac{\cos v}{v \cos v}\right) d v=2 \int \frac{1}{x} d x
\end{aligned}
$
$\begin{aligned}
& \Rightarrow \int\left(\tan v-\frac{1}{v}\right) d v=2 \int \frac{1}{x} d x \\
& \Rightarrow \log |\sec v|-\log |v|=2 \log |x|+\log |c| \\
& \Rightarrow \log \left|\frac{\sec v}{v}\right|=\log |x|^2+\log |c| \\
& \Rightarrow \log \left|\frac{\text { secv }}{v}\right|=\log |c| x^2 \\
& \Rightarrow \frac{\sec v}{v}= \pm|c| x^2 \\
& \Rightarrow \sec v= \pm|c| x^2 v \\
& \text { Putting } v=\frac{y}{x} \\
& \Rightarrow \sec \frac{y}{x}=\operatorname{Cx}^2 \frac{y}{x} \quad \text { where } \mathrm{C}= \pm c \\
& \Rightarrow \sec \frac{y}{x}=\mathrm{C} x y \\
& \Rightarrow \frac{1}{\cos \frac{y}{x}}=\mathrm{C} x y \\
& \Rightarrow C x y \cdot \cos \frac{y}{x}=1 \\
& \Rightarrow x y \cdot \cos \frac{y}{x}=\frac{1}{\mathrm{C}} \\
&
\end{aligned}$
Ex 9.4 Question 8.
$\text {} x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$
Answer.
Given: Differential equation $x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$
$
\begin{aligned}
& \Rightarrow x \frac{d y}{d x}=y-x \sin \left(\frac{y}{x}\right) \\
& \Rightarrow \frac{d y}{d x}=\frac{y}{x}-\sin \left(\frac{y}{x}\right)=f\left(\frac{y}{x}\right)
\end{aligned}
$
Therefore, the given differential equation is homogeneous.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in eq. (i), we get
$
\begin{aligned}
& v+x \frac{d v}{d x}=v-\sin v \\
& \Rightarrow x \frac{d v}{d x}=-\sin v \\
& \Rightarrow x d v=-\sin v d x \\
& \Rightarrow \frac{d v}{\sin v}=\frac{-d x}{x} \\
& \Rightarrow \operatorname{cosec} v d v=\frac{-d x}{x}
\end{aligned}
$
Integrating both sides,
$\begin{aligned}
& \Rightarrow \int \operatorname{cosec} v d v=-\int \frac{1}{x} d x \\
& \Rightarrow \log |\operatorname{cosec} v-\cot v|=-\log |x|+\log |c| \\
& \Rightarrow \log |\operatorname{cosec} v-\cot v|=\log \left|\frac{c}{x}\right| \\
& \Rightarrow \operatorname{cosec} v-\cot v= \pm \frac{c}{x} \\
& \Rightarrow \operatorname{cosec} \frac{y}{x}-\cot \frac{y}{x}= \pm \frac{c}{x}\left[\text { putting } v=\frac{y}{x}\right] \\
& \Rightarrow \frac{1}{\sin \frac{y}{x}}-\frac{\sin \frac{y}{x}}{x}=\frac{\mathrm{C}}{x} \\
& \Rightarrow \frac{1-\cos \frac{y}{x}}{\sin \frac{y}{x}}=\frac{\mathrm{C}}{x} \text { where } \pm c=\mathrm{C} \\
& \Rightarrow x\left(1-\cos \frac{y}{x}\right)=\mathrm{C} \sin \frac{y}{x}
\end{aligned}$
Ex 9.4 Question 9.
$y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$
Answer.
Given: Differential equation $y d x+x\left(\log \frac{y}{x}\right) d y-2 x d y=0$
$
\Rightarrow y d x=2 x d y-x\left(\log \frac{y}{x}\right) d y
$
$
\begin{aligned}
& \Rightarrow y d x=x\left(2-\log \frac{y}{x}\right) d y \\
& \Rightarrow \frac{d y}{d x}=\frac{\frac{y}{x}}{2-\log \frac{y}{x}}=f\left(\frac{y}{x}\right)
\end{aligned}
$
Therefore, the given differential equation is homogeneous.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in eq. (i), we get
$
\begin{aligned}
& \Rightarrow v+x \frac{d v}{d x}=\frac{v}{2-\log v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{v}{2-\log v}-v \\
& =\frac{v-2 v+v \log v}{2-\log v} \\
& \Rightarrow x \frac{d v}{d x}=\frac{-v+v \log v}{2-\log v}
\end{aligned}
$
$\begin{aligned}
& \Rightarrow x \frac{d v}{d x}=\frac{v(\log v-1)}{2-\log v} \\
& \Rightarrow x(2-\log v) d v=v(\log v-1) d x
\end{aligned}$
$
\begin{aligned}
& \Rightarrow \frac{2-\log v}{v(\operatorname{og} v-1)}=\frac{d x}{x} \\
& \Rightarrow \frac{1-(\log v-1)}{v(\log v-1)} d v=\frac{d x}{x} \\
& \Rightarrow\left[\frac{1}{v(\log v-1)}-\frac{1}{v}\right] d v=\frac{d x}{x}
\end{aligned}
$
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int\left[\frac{1}{v(\log v-1)}-\frac{1}{v}\right] d v=\int \frac{1}{x} d x \\
& \Rightarrow \log |\log v-1|-\log |v|=\log |x|+\log |c| \\
& \Rightarrow \log \left|\frac{\mid \log v-1}{v}\right|=\log |c x| \\
& \Rightarrow\left|\frac{\log v-1}{v}\right|=|c x| \\
& \Rightarrow \frac{\log v-1}{v}= \pm c x=\mathrm{C} x \text { where } \mathrm{c}= \pm c \\
& \Rightarrow \log v-1=\mathrm{C} x v \\
& \Rightarrow \log \frac{y}{x}-1=\mathrm{C} x \frac{y}{x}\left[\text { Putting } v=\frac{y}{x}\right] \\
& \Rightarrow \log \frac{y}{x}-1=\mathrm{C} y
\end{aligned}
$
Ex 9.4 Question 10.
$\text {}\left(1+e^{x / y}\right) d x+e^{x / y}\left(1-\frac{x}{y}\right) d y=0$
Answer.
Given: Differential equation $\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$
$
\begin{aligned}
& \Rightarrow\left(1+e^{\frac{x}{y}}\right) \frac{d x}{d y}+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)=0[\text { Dividing by } d y] \\
& \Rightarrow\left(1+e^{\frac{x}{y}}\right) \frac{d x}{d y}=-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) \\
& \Rightarrow \frac{d x}{d y}=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{\left(1+e^{\frac{x}{y}}\right)}=f\left(\frac{x}{y}\right) \ldots \ldots . . \text { (i) }
\end{aligned}
$
Therefore, it is a homogeneous.
$
\begin{aligned}
& \text { Now putting } \frac{x}{y}=v \\
& \Rightarrow x=v y \\
& \Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y}
\end{aligned}
$
Putting these values of $\frac{x}{y}$ and $\frac{d x}{d y}$ in eq. (i), we have
$
v+y \frac{d v}{d y}=\frac{e^v(v-1)}{1+e^v}
$
$\Rightarrow y \frac{d v}{d y}=\frac{e^v(v-1)}{1+e^v}-v$
$
\begin{aligned}
& \Rightarrow y \frac{d v}{d y}=\frac{v e^v-e^v-v-v e^v}{1+e^v}=\frac{-e^v-v}{1+e^v} \\
& \Rightarrow y\left(1+e^v\right) d v=-\left(e^v+v\right) d y \\
& \Rightarrow \frac{1+e^v}{v+e^v} d v=-\frac{d y}{y} \text { [Separating variables] }
\end{aligned}
$
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{1+e^v}{v+e^v} d v=-\int \frac{1}{y} d y \\
& \Rightarrow \log \left|v+e^v\right|=-\log |y|+\log |c|
\end{aligned}
$
Now putting $v=\frac{x}{y}$,
$
\begin{aligned}
& \Rightarrow \log \left|\frac{x}{y}+e^{\frac{x}{y}}\right|=-\log |y|+\log |c| \\
& \Rightarrow \log \left|\frac{x}{y}+e^{\frac{x}{y}}\right|=\log \left|\frac{c}{y}\right| \\
& \Rightarrow\left|\frac{x}{y}+e^{\frac{x}{y}}\right|=\left|\frac{c}{y}\right| \\
& \Rightarrow \frac{x}{y}+e^{\frac{x}{y}}= \pm \frac{x}{y}
\end{aligned}
$
$
\Rightarrow x+y e^{\frac{x}{y}}=\mathrm{C} \text { where } \mathrm{C}= \pm c
$
For each of the differential equations in Questions from 11 to 15, find the particular
solution satisfying the given condition
Ex 9.4 Question 11.
$(x+y) d y+(x-y) d x=0 ; y=1$ when $x=1$
Answer.
Given: Differential equation $(x+y) d y+(x-y) d x=0, y=1$ when $x=1$
$
\begin{aligned}
& \Rightarrow(\mathrm{x}+\mathrm{y}) \mathrm{dy}+(\mathrm{x}-\mathrm{y}) \mathrm{dx}=0 \\
& \Rightarrow \frac{d y}{d x}=\frac{-(x-y)}{(x+y)} \\
& \Rightarrow \frac{d y}{d x}=\frac{y-x}{y+x} \\
& \Rightarrow \frac{d y}{d x}=\frac{x\left(\frac{y}{x}-1\right)}{x\left(\frac{y}{x}+1\right)} \\
& \Rightarrow \frac{d y}{d x}=\frac{\left(\frac{y}{x}-1\right)}{\left(\frac{y}{x}+1\right)}=f\left(\frac{y}{x}\right) .
\end{aligned}
$
Therefore the given differential equation is homogeneous because each coefficient of $d x$ and $d y$ is same i.e., degree 1.
$
\text { Putting } \frac{y}{x}=v
$
$
\begin{aligned}
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d v}{d x}$ in eq. (ii), we have
$
\begin{aligned}
& v+x \frac{d v}{d x}=\frac{v-1}{v+1} \\
& \Rightarrow x \frac{d v}{d x}=\frac{v-1}{v+1}-v \\
& \Rightarrow x \frac{d v}{d x}=\frac{v-1-v^2-v}{v+1}=\frac{-v^2-1}{v+1} \\
& \Rightarrow x \frac{d v}{d x}=-\frac{v^2+1}{v+1} \\
& \Rightarrow x(v+1) d v=-\left(v^2+1\right) d x \\
& \Rightarrow \frac{v+1}{v^2+1} d v=-\frac{d x}{x} \text { [Separating variables] }
\end{aligned}
$
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \frac{v}{v^2+1} d v+\int \frac{1}{v^2+1} d v=-\int \frac{1}{x} d x \\
& \Rightarrow \frac{1}{2} \int \frac{2 v}{v^2+1} d v+\tan ^{-1} v=-\log x+c \\
& \Rightarrow \frac{1}{2} \log \left(v^2+1\right)+\tan ^{-1} v=-\log x+c
\end{aligned}
$
Now putting $v=\frac{y}{x}$,
$
\Rightarrow \frac{1}{2} \log \left(\left(\frac{y}{x}\right)^2+1\right)+\tan ^{-1} \frac{y}{x}=-\log x+c
$
$\Rightarrow \frac{1}{2} \log \left(\frac{y^2+x^2}{x^2}\right)+\tan ^{-1} \frac{y}{x}=-\log x+c$
$
\begin{aligned}
& \Rightarrow \frac{1}{2} \log \left(y^2+x^2\right)-\frac{1}{2} \log x^2+\tan ^{-1} \frac{y}{x}=-\log x+c \\
& \Rightarrow \frac{1}{2} \log \left(y^2+x^2\right)-\frac{1}{2} \times 2 \log x+\tan ^{-1} \frac{y}{x}=-\log x+c \\
& \Rightarrow \frac{1}{2} \log \left(y^2+x^2\right)+\tan ^{-1} \frac{y}{x}=c \ldots \ldots \ldots \text { (iii) }
\end{aligned}
$
Now again given $y=1$ when $x=1$, therefore putting these values in eq. (iii),
$
\begin{aligned}
& \frac{1}{2} \log (1+1)+\tan ^{-1} 1=c \\
& \Rightarrow c=\frac{1}{2} \log 2+\frac{\pi}{4}
\end{aligned}
$
Putting this value of $c$ in eq. (iii), we get
$
\begin{aligned}
& \Rightarrow \frac{1}{2} \log \left(y^2+x^2\right)+\tan ^{-1} \frac{y}{x}=\frac{1}{2} \log 2+\frac{\pi}{4} \\
& \Rightarrow \log \left(y^2+x^2\right)+2 \tan ^{-1} \frac{y}{x}=\log 2+\frac{\pi}{4}
\end{aligned}
$
Ex 9.4 Question 12.
$x^2 d y+\left(x y+y^2\right) d x=0 ; y=1$ when $x=1$
Answer.
Given: Differential equation $x^2 d y+\left(x y+y^2\right) d x=0$
$
\begin{aligned}
& \Rightarrow x^2 d y=-\left(x y+y^2\right) d x \\
& \Rightarrow \frac{d y}{d x}=-\frac{y(x+y)}{x^2}=\frac{x y\left(1+\frac{y}{x}\right)}{x^2} \\
& \Rightarrow \frac{d y}{d x}=-\frac{y}{x}\left(1+\frac{y}{x}\right)=f\left(\frac{y}{x}\right) \ldots \ldots \ldots .(\text { i) }
\end{aligned}
$
Therefore the given differential equation is homogeneous.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d v}{d x}$ in eq. (i), we have
$
\begin{aligned}
& \Rightarrow v+x \frac{d v}{d x}=-v(1+v)=-v-v^2 \\
& \Rightarrow x \frac{d y}{d x}=-v^2-2 v \\
& \Rightarrow x \frac{d y}{d x}=-v(v+2) \\
& \Rightarrow \frac{d v}{v(v+2)}=-\frac{d x}{x}
\end{aligned}
$
Integrating both sides, $\int \frac{1}{v(v+2)} d v=-\int \frac{1}{x} d x$
$
\begin{aligned}
& \Rightarrow \frac{1}{2} \int \frac{2}{v(v+2)} d v=-\log |x|+\log |c| \\
& \Rightarrow \frac{1}{2} \int \frac{(v+2)-v}{v(v+2)} d v=-\log |x|+\log |c| \\
& \Rightarrow \int\left(\frac{1}{v}-\frac{1}{v+2}\right) d v=-2 \log |x|+\log |c|
\end{aligned}
$
$\Rightarrow \log \left|\frac{v}{v+2}\right|=\log \left|x^{-2}\right|+\log |c|$
$
\begin{aligned}
& \Rightarrow \log \left|\frac{v}{v+2}\right|=\log \left|c x^{-2}\right| \\
& \Rightarrow \frac{v}{v+2}= \pm c x^{-2}
\end{aligned}
$
Putting $v=\frac{y}{x}$
$
\begin{aligned}
\Rightarrow & \frac{\frac{y}{x}}{\frac{y}{x}+2}= \pm c x^{-2} \\
\Rightarrow & \frac{y}{y+2 x}= \pm c x^{-2} \\
\Rightarrow & x^2 y=\mathrm{C}(y+2 x) \text { where } \mathrm{c}= \pm c
\end{aligned}
$
Now putting $x=1$ and $y=1$ in eq. (ii), we get $1=3 C \Rightarrow \mathrm{C}=\frac{1}{3}$
Putting value of $\mathrm{C}$ in eq. (ii),
$
\begin{aligned}
& x^2 y=\frac{1}{3}(y+2 x) \\
& \Rightarrow 3 x^2 y=y+2 x
\end{aligned}
$
Ex 9.4 Question 13.
$\left[x \sin ^2\left(\frac{y}{x}\right)-y\right] d x+x d y=0 ; y=\frac{\pi}{4}$ when $x=1$
Answer.
Given: Differential equation $\left(x \sin ^2 \frac{y}{x}-y\right) d x+x d y=0 ; y=\frac{\pi}{4}, x=1$
$
\Rightarrow x d y=-\left(x \sin ^2 \frac{y}{x}-y\right) d x
$
$
\Rightarrow \frac{d y}{d x}=-\sin ^2 \frac{y}{x}+\frac{y}{x}=f\left(\frac{y}{x}\right)
$
Therefore, the given differential equation is homogeneous.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d v}{d x}$ in eq. (i), we have
$
\begin{aligned}
& \Rightarrow v+x \frac{d v}{d x}=-\sin ^2 v+v \\
& \Rightarrow x \frac{d v}{d x}=-\sin ^2 v \\
& \Rightarrow \frac{d v}{\sin ^2 v}=-\frac{d x}{x} \text { [Separating variables] }
\end{aligned}
$
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \operatorname{cosec}^2 v d v=-\int \frac{1}{x} d x \\
& \Rightarrow-\cot v=-\log |x|+c \\
& \Rightarrow \cot v=\log |x|-c
\end{aligned}
$
$
\Rightarrow \cot \frac{y}{x}=\log |x|+c \quad\left[\text { Putting } \frac{y}{x}=v\right]
$
Now putting $y=\frac{\pi}{4}, x=1$ in eq. (ii), $\cot \frac{\pi}{4}=\log 1-c$
$
\Rightarrow c=-1
$
Putting the value of $c$ in eq. (ii),
$
\begin{aligned}
& \cot \frac{y}{x}=\log |x|+1 \\
& \Rightarrow \cot \frac{y}{x}=\log |x|+\log e \\
& \Rightarrow \cot \frac{y}{x}=\log x e
\end{aligned}
$
Ex 9.4 Question 14.
$\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0 ; y=0$ when $x=1$
Answer.
Given: Differential equation $\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec} \frac{y}{x}=0 ; y=0, x=1$
$
\Rightarrow \frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec} \frac{y}{x}=f\left(\frac{y}{x}\right)
$
Therefore, the given differential equation is homogeneous.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d v}{d x}$ in eq. (i), we have
$
\Rightarrow v+x \frac{d v}{d x}=v-\operatorname{cosec} v
$
$
\begin{aligned}
& \Rightarrow x \frac{d v}{d x}=\frac{-1}{\sin v} \\
& \Rightarrow \sin v d v=-\frac{d x}{x} \text { [Separating variables] }
\end{aligned}
$
Integrating both sides,
$
\begin{aligned}
& \Rightarrow \int \sin v d v=-\int \frac{1}{x} d x \\
& \Rightarrow-\cos v=-\log |x|+c \\
& \Rightarrow \cos v=\log |x|-c \\
& \Rightarrow \cos \frac{y}{x}=\log |x|+c \quad\left[\text { Putting } \frac{y}{x}=v\right]
\end{aligned}
$
Now putting $y=0, x=1$ in eq. (ii), $\cos 0=\log 1-c$
$
\Rightarrow c=-1
$
Putting the value of $c$ in eq. (ii),
$
\begin{aligned}
& \cos \frac{y}{x}=\log |x|+1 \\
& \Rightarrow \cos \frac{y}{x}=\log |x|+\log e \\
& \Rightarrow \cos \frac{y}{x}=\log x e
\end{aligned}
$
Ex 9.4 Question 15.
$2 x y+y^2-2 x^2 \frac{d y}{d x}=0 ; y=2$ when $x=1$
Answer.
Given: Differential equation $2 x y+y^2-2 x^2 \frac{d y}{d x}=0$
$
\begin{aligned}
& \Rightarrow-2 x^2 \frac{d y}{d x}=-2 x y-y^2 \\
& \Rightarrow \frac{d y}{d x}=\frac{-2 x y}{-2 x^2}-\frac{y^2}{-2 x^2} \\
& \Rightarrow \frac{d y}{d x}=\frac{y}{x}+\frac{1}{2}\left(\frac{y}{x}\right)^2=f\left(\frac{y}{x}\right)
\end{aligned}
$
Therefore the given differential equation is homogeneous because each coefficient of $d x$ and $d y$ is same.
$
\begin{aligned}
& \text { Putting } \frac{y}{x}=v \\
& \Rightarrow y=v x \\
& \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}
\end{aligned}
$
Putting these values of $\frac{y}{x}$ and $\frac{d v}{d x}$ in eq. (ii), we have
$
\begin{aligned}
& v+x \frac{d v}{d x}=v+\frac{1}{2} v^2 \\
& \Rightarrow x \frac{d v}{d x}=\frac{1}{2} v^2 \\
& \Rightarrow 2 x d v=v^2 d x
\end{aligned}
$
$\Rightarrow 2 \frac{d v}{v^2}=\frac{d x}{x}$ [Separating variables]
Integrating both sides,
$
2 \int v^{-2} d v=\int \frac{1}{x} d x
$
$
\begin{aligned}
& \Rightarrow 2 \frac{v^{-1}}{-1}=\log |x|+c \\
& \Rightarrow \frac{-2}{v}=\log |x|+c \\
& \Rightarrow \frac{-2}{\left(\frac{y}{x}\right)}=\log |x|+c \quad\left[\text { Putting } \frac{y}{x}=v\right] \\
& \text { Now putting } y=2, x=1 \text { in } \frac{-2}{\left(\frac{y}{x}\right)}=\log |x|+c, \frac{-2}{2}=\log 1+c \\
& \Rightarrow c=-1
\end{aligned}
$
Again putting $c=-1$, in $\frac{2 x}{y}=\log |x|+c$, we get
$
\begin{aligned}
& \Rightarrow \frac{-2}{\left(\frac{y}{x}\right)}=\log |x|+c \\
& \Rightarrow y(\log |x|-1)=-2 x \\
& \Rightarrow y=\frac{-2 x}{\log |x|-1} \\
& \Rightarrow y=\frac{2 x}{1-\log |x|}
\end{aligned}
$
Choose the correct answer:
Ex 9.4 Question 16.
A homogeneous differential equation of the form $\frac{d y}{d x}=h\left(\frac{x}{y}\right)$ can be solved by making the substitution:
(A) $y=v x$
(B) $v=y x$
(C) $x=v y$
(D) $x=v$
Answer.
We know that a homogeneous differential equation of the form $\frac{d x}{d y}=h\left(\frac{x}{y}\right)$ can be solved by the substitution $\frac{x}{y}=v$ i.e., $x=v y$.
Therefore, option (C) is correct.
Ex 9.4 Question 17.
Which of the following is a homogeneous differential equation:
(A) $(4 x+6 y+5) d y-3(3 y+2 x+4) d x=0$
(B) $(x y) d x-\left(x^3+y^3\right) d y=0$
(C) $\left(x^3+2 y^2\right) d x+2 x y d y=0$
(D) $y^2 d x+\left(x^2-x y-y^2\right) d y=0$
Answer.
(D) $y^2 d x+\left(x^2-x y-y^2\right) d y=0$
