Exercise 10.6 - Chapter 10 Ordinary Differential Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $10.6$
Solve the following differential equations:
Question 1.
$
\left[x+y \cos \left(\frac{y}{x}\right)\right] d x=x \cos \left(\frac{y}{x}\right) d y
$
Solution:
$
\begin{aligned}
{\left[x+y \cos \left(\frac{y}{x}\right)\right] d x } & =x \cos \left(\frac{y}{x}\right) d y \\
\frac{d y}{d x} & =\frac{x+y \cos \left(\frac{y}{x}\right)}{x \cos \left(\frac{y}{x}\right)}
\end{aligned}
$
This is a Homogeneous differential equation
$
\begin{aligned}
\text { Put } y=v x \Rightarrow \frac{d y}{d x} & =v+x \frac{d v}{d x} \\
\text { (1) } \Rightarrow v+x \frac{d v}{d x} & =\frac{x+v x \cos v}{x \cos v} \\
v+x \frac{d v}{d x} & =\frac{x(1+v \cos v)}{x \cos v} \\
x \frac{d v}{d x} & =\frac{1+v \cos v}{\cos v}-v \\
x \frac{d v}{d x} & =\frac{1+v \cos v-v \cos v}{\cos v}=\frac{1}{\cos v}
\end{aligned}
$
Seperating the variables
$
\begin{aligned}
\int \cos v \cdot d v & =\int \frac{d x}{x} \\
\sin v & =\log x+c \\
\sin \left(\frac{y}{x}\right) & =\log x+\log c \quad\left[\because v=\frac{y}{x}\right]
\end{aligned}
$

$
\sin \left(\frac{y}{x}\right)=\log |c x|
$
Question 2.
$
\left(x^3+y^3\right) d y-x^2 y d x=0
$
Solution:
$
\begin{aligned}
& \left(x^3+y^3\right) d y-x^2 y d x=0 \\
& \left(x^3+y^3\right) d y=x^3 y d x \\
& \frac{d y}{d x}=\frac{x^2 y}{x^3+y^3}
\end{aligned}
$
This is a Homogeneous differential equation.
$
\begin{aligned}
\text { Put } y=v x \Rightarrow \frac{d y}{d x} & =v+x \frac{d v}{d x} \\
(1) \Rightarrow v+x \frac{d v}{d x} & =\frac{x^2(v x)}{x^3+v^3 x^3} \\
v+x \frac{d v}{d x} & =\frac{x^3 v}{x^3\left(1+v^3\right)} \\
x \frac{d v}{d x} & =\frac{v}{1+v^3}-v \\
x \frac{d v}{d x} & =\frac{v-v-v^4}{1+v^3}=\frac{-v^4}{1+v^3}
\end{aligned}
$

Separating the variables
$
\begin{aligned}
& \int \frac{\left(1+v^3\right)}{v^4} d v=-\int \frac{d x}{x} \\
& \int\left(\frac{1}{v^4}+\frac{v^3}{v^4}\right) d v=-\int \frac{d x}{x} \\
& \int\left(v^{-4}+\frac{1}{v}\right) d v=-\int \frac{d x}{x} \\
& \frac{v^{-3}}{-3}+\log v=-\log x+\log c \\
& \frac{-1}{3 v^3}=\log c-\log x-\log v \\
& \frac{-1}{3 v^3}=\log \left(\frac{c}{v x}\right) \\
& \text { Put } v^{\prime}=\frac{y}{x} \\
& \frac{1}{3\left(\frac{y^3}{x^3}\right)}=\log \left(\frac{c}{\frac{y}{x} x}\right)^{-1} \\
& \frac{x^3}{3 y^3}=\log \left(\frac{y}{c}\right) \Rightarrow \frac{y}{c}=e^{\frac{x^3}{3 y^3}} \Rightarrow \therefore y=c e^{\frac{x^3}{3 y^3}} \\
&
\end{aligned}
$
Question 3.
$
y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y\right) d y
$
Solution:

$
\begin{aligned}
y e^{\frac{x}{y}} d x & =\left(x e^{\frac{x}{y}}+y\right) d y \\
\frac{d x}{d y} & =\frac{x e^{x / y}+y}{y e^{x / y}}
\end{aligned}
$
This is a Homogeneous differential equation
Put $x=v y \Rightarrow \frac{d x}{d y}=v+y \cdot \frac{d v}{d y}$
$
\text { (1) } \begin{aligned}
\Rightarrow v+y \cdot \frac{d v}{d y} & =\frac{v y e^v+y}{y e^v} \\
v+y \cdot \frac{d v}{d y} & =\frac{y\left(v e^v+1\right)}{y e^v} \\
y \frac{d v}{d y} & =\frac{v e^v+1}{e^v}-v \\
y \frac{d v}{d y} & =\frac{v e^v+1-v e^v}{e^v} \\
y \frac{d v}{d y} & =\frac{1}{e^v}
\end{aligned}
$
Seperating the variables
$
\begin{aligned}
\int e^v d v & =\int \frac{d y}{y} \\
e^v & =\log y+\log c \\
e^v & =\log |c y| \\
\text { i.e., } e^{x y y} & =\log |c y| \quad\left[\because v=\frac{x}{y}\right]
\end{aligned}
$

Question 4.
$2 x y d x+\left(x^2+2 y^2\right) d y=0$
Solution:
$2 x y d x+\left(x^2+2 y^2\right) d y=0$
$\left(x^2+2 y^2\right) d y=-2 x y d x$
$
\frac{d x}{d y}=\frac{x^2+2 y^2}{-2 x y}
$
This is a Homogeneous differential equation
Put $x=v y \Rightarrow \frac{d x}{d y}=v+y \cdot \frac{d v}{d y}$
$
\text { (1) } \begin{aligned}
\Rightarrow v+y \cdot \frac{d v}{d y} & =\frac{v^2 y^2+2 y^2}{-2 v y^2}=\frac{y^2\left(v^2+2\right)}{-2 v y^2} \\
y \frac{d v}{d y} & =\frac{v^2+2}{-2 v}-v=\frac{v^2+2+2 v^2}{-2 v} \\
y \frac{d v}{d y} & =\frac{3 v^2+2}{-2 v}
\end{aligned}
$
Separating the variables
$
\begin{aligned}
\int \frac{2 v}{3 v^2+2} d v & =-\int \frac{d y}{y} & & \frac{2 v}{3 v^2+2} d v \\
\frac{1}{3} \int \frac{d t}{t} & =-\int \frac{d y}{y} & & \text { Put } t=3 v^2+2 \\
\frac{1}{3} \log t & =-\log y+\log c & & d t=6 v \cdot d v \\
\log t^{1 / 3} & =\log \left(\frac{c}{y}\right) & &
\end{aligned}
$

$
\begin{aligned}
\log t^{1 / 3} & =\frac{c}{y} \quad \Rightarrow t=\frac{c^3}{y^3} \\
\left(3 v^2+2\right) & =\frac{c^3}{y^3} \\
y^3\left(\frac{3 x^2}{y^2}+2\right) & =c^3 \quad\left[\because v=\frac{x}{y}\right] \\
y^3\left(\frac{3 x^2+2 y^2}{y^2}\right) & =c^3 \\
3 x^2 y+2 y^3 & =c
\end{aligned}
$
Question 5.
$
\left(y^2-2 x y\right) d x=\left(x^2-2 x y\right) d y
$

Solution:
$
\begin{aligned}
& \left(\mathrm{y}^2-2 \mathrm{xy}\right) \mathrm{dx}=\left(\mathrm{x}^2-2 \mathrm{xy}\right) \mathrm{dy} \\
& \frac{d y}{d x}=\frac{y^2-2 x y}{x^2-2 x y}
\end{aligned}
$
This is a Homogeneous differential equation
$
\begin{aligned}
& \text { Put } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& \text { (1) } \Rightarrow v+x \cdot \frac{d v}{d x}=\frac{v^2 x^2-2 x^2 v}{x^2-2 x^2 v}=\frac{x^2\left(v^2-2 v\right)}{x^2(1-2 v)} \\
& x \cdot \frac{d v}{d x}=\frac{v^2-2 v}{1-2 v}-v \\
& x \cdot \frac{d v}{d x}=\frac{v^2-2 v-v+2 v^2}{1-2 v}=\frac{3 v^2-3 v}{1-2 v}
\end{aligned}
$
Seperating the variables
$
\begin{aligned}
\int \frac{1-2 v}{3 v^2-3 v} d v & =\int \frac{d x}{x} \\
\frac{1}{3} \int \frac{1-2 v}{v^2-v} d v & =\int \frac{d x}{x} \\
-\frac{1}{3} \int \frac{d t}{t} & =\int \frac{d x}{x}
\end{aligned}
$

Consider $\int \frac{1-2 v}{v^2-v} d v$
Put $t=v^2-v$
$
\begin{aligned}
d t & =(2 v-1) d v \\
d t & =-(1-2 v) d v \\
-d t & =(1-2 v) d v
\end{aligned}
$

$\begin{aligned}
-\frac{1}{3} \log t & =\log x+\log c \\
-\frac{1}{3} \log \left(v^2-v\right) & =\log c x \\
\log \left(v^2-v\right)^{-1 / 3} & =\log c x \\
\left(v^2-v\right) & =(c x)^{1 / 3} \\
\left(\frac{y^2}{x^2}-\frac{y}{x}\right) & =c^{\frac{1}{3}} x^{\frac{1}{3}} \\
x^3\left(\frac{y^2-x y}{x^2}\right) & =c \\
x y^2-x^2 y & =c
\end{aligned}$

Question 6.
$
x \frac{d y}{d x}=y-x \cos ^2\left(\frac{y}{x}\right)
$
Solution:
$
x \frac{d y}{d x}=y-x \cos ^2\left(\frac{y}{x}\right)
$
Dividing throughout by ' $x$ '
$
\frac{d y}{d x}=\frac{y}{x}-\cos ^2\left(\frac{y}{x}\right)
$
This is a Homogeneous differential equation.
$
\begin{aligned}
\text { Put } y=v x \Rightarrow \frac{d y}{d x} & =v+x \frac{d v}{d x} \\
(1) \Rightarrow v+x \cdot \frac{d v}{d x} & =v-\cos ^2 v \\
x \cdot \frac{d v}{d x} & =v-\cos ^2 v-v \\
x \cdot \frac{d v}{d x} & =-\cos ^2 v
\end{aligned}
$
Seperating the variables
$
\begin{aligned}
\int \sec ^2 v d v & =-\int \frac{d x}{x} \\
\tan v & =-\log x+\log c \\
\tan \left(\frac{y}{x}\right) & =\log \left(\frac{c}{x}\right) \quad\left[\because v=\frac{y}{x}\right] \\
\frac{c}{x} & =e^{\tan \left(\frac{y}{x}\right)} \Rightarrow \therefore c=x e^{\tan \left(\frac{y}{x}\right)}
\end{aligned}
$

Question 7.
$\left(1+3 e^{\frac{y}{x}}\right) d y+3 e^{\frac{y}{x}}\left(1-\frac{y}{x}\right) d x=0$, given that $y=0$ when $x=1$
Solution:
$
\begin{aligned}
\left(1+3 e^{\frac{y}{x}}\right) d y+3 e^{\frac{y}{x}}\left(1-\frac{y}{x}\right) d x & =0 \\
\left(1+3 e^{\frac{y}{x}}\right) d y \cdot & =-3 e^{\frac{y}{x}}\left(1-\frac{y}{x}\right) d x \\
\frac{d y}{d x} & =\frac{-3 e^{y / x}\left(1-\frac{y}{x}\right)}{1+3 e^{y / x}}
\end{aligned}
$
This is a Homogeneous differential equation
$
\begin{aligned}
& \text { Put } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& \begin{aligned}
(1) \Rightarrow v+x \cdot \frac{d v}{d x} & =\frac{-3 e^v(1-v)}{1+3 e^v} \\
x \frac{d v}{d x} & =\frac{-3 e^v+3 v e^v}{1+3 e^v}-v \\
x \frac{d v}{d x} & =\frac{-3 e^v+3 v e^v-v-3 v e^v}{1+3 e^v}
\end{aligned}
\end{aligned}
$

$
x \frac{d v}{d x}=\frac{-\left(3 e^v+v\right)}{1+3 e^v}
$
Separating the variables
$
\begin{aligned}
\int \frac{\left(1+3 e^v\right) d v}{3 e^v+v} & =-\int \frac{d x}{x} \\
\text { Put } t & =3 e^v+v \\
d t & =\left(3 e^v+1\right) d v \\
\int \frac{d t}{t} & =-\int \frac{d x}{x}
\end{aligned}
$

$
\begin{aligned}
\log t & =-\log x+\log c \\
\log t=\log \left(\frac{c}{x}\right) \Rightarrow t & =\frac{c}{x} \\
3 e^v+v & =\frac{c}{x} \\
x\left(3 e^{y / x}+\frac{y}{x}\right) & =c \quad\left[\because v=\frac{y}{x}\right]
\end{aligned}
$
Given $y=0$ when $x=1$
$
1\left(3 e^0+0\right)=c \Rightarrow c=3
$
$\therefore$ The solution is $x\left(3 e^{y / x}+\frac{y}{x}\right)=3$
$
\Rightarrow 3 x e^{y / x}+y=3
$

Question 8.
$\left(x^2+y^2\right) d y=x y d x$. It is given that $y(1)=1$ and $y\left(x_0\right)=e$. Find the value of $x_0$.
Solution:
$
\begin{aligned}
\left(x^2+y^2\right) d y & =x y d x \\
\frac{d x}{d y} & =\frac{x^2+y^2}{x y}
\end{aligned}
$
This is a Homogeneous differential equation
$
\begin{aligned}
\text { Put } x=v y \Rightarrow \frac{d x}{d y} & =v+y \cdot \frac{d v}{d y} \\
(1) \Rightarrow v+y \cdot \frac{d v}{d y} & =\frac{v^2 y^2+y^2}{v y^2}=\frac{y^2\left(v^2+1\right)}{v y^2} \\
y \frac{d v}{d y} & =\frac{v^2+1}{v}-v \\
y \frac{d v}{d y} & =\frac{v^2+1-v^2}{v}=\frac{1}{v}
\end{aligned}
$

Seperating the variables
$
\begin{aligned}
\int v d v & =\int \frac{d y}{y} \\
\frac{v^2}{2} & =\log y+\log c \\
\frac{x^2}{2 y^2} & =\log c y \quad\left[\because v=\frac{x}{y}\right] \\
c y & =e^{x^2 / 2 y^2}
\end{aligned}
$
Given $y(1)=1$
i.e., When $x=1, y=1$
From (2) $\Rightarrow c=e^{\frac{1}{2}}$
$
\therefore \quad \text { (2) } \Rightarrow e^{\frac{1}{2}} y=e^{x^2 / 2 y^2}
$
Now, given $y\left(x_{\mathrm{o}}\right)=e$
i.e., When $x=x_0, y=e$
$
\begin{aligned}
\therefore e^{\frac{1}{2}} \cdot e & =e^{x_0^2} / 2 e^2 \\
e^{3 / 2} & =e^{x_0^2} / 2 e^2 \\
\frac{3}{2} & =\frac{x_0^2}{2 e^2} \\
\therefore x_0^2 & =3 e^2 \\
x_0 & =\pm \sqrt{3} e
\end{aligned}
$