Exercise 10.7-Additional Problems - Chapter 10 Ordinary Differential Equations Class 12 Samacheer Kalvi Solutions

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Additional Problems
Question 1.
Solve: $\frac{d y}{d x}+y \cot x=2 \cos x$
Solution:
The given equation is of the form $\frac{d y}{d x}+P y=Q$. This is linear in $y$. Here $\mathrm{P}=\cot x$ and $\mathrm{Q}=2 \cos x$
$
\text { I.F }=e^{\int P d x}=e^{\int \cot x d x}=e^{\log \sin x}=\sin x
$
$\therefore$ The required solution is
$
\begin{aligned}
y(\text { I.F. }) & =\int \mathrm{Q}(\text { I.F. }) d x+c \\
y \sin x & =\int \sin 2 x d x+c \Rightarrow y(\sin x)=\int 2 \cos x \sin x d x+c \\
\Rightarrow y(\sin x) & =-\frac{\cos 2 x}{2}+c \\
\Rightarrow 2 y(\sin x)+\cos 2 x & =c
\end{aligned}
$

Question 2.
Solve: $\left(1-x^2\right) \frac{d y}{d x}+2 x y=x \sqrt{1-x^2}$
Solution:
The given equation is $\frac{d y}{d x}+\left(\frac{2 x}{1-x^2}\right) y=\frac{x}{\sqrt{\left(1-x^2\right)}}$. This is linear in $y$.
Here $\int \mathrm{P} d x=\int \frac{2 x}{1-x^2} d x=-\log \left(1-x^2\right)$
$
\text { I.F }=e^{\int P d x}=\frac{1}{1-x^2}
$
The required solution is
$
\begin{aligned}
& y \cdot \frac{1}{1-x^2}=\int \frac{x}{\sqrt{\left(1-x^2\right)}} \times \frac{1}{1-x^2} d x . \text { Put } 1-x^2=t \Rightarrow-2 x d x=d t \\
& \therefore \frac{y}{1-x^2}=\frac{-1}{2} \int t^{-3 / 2} d t+c \\
& \Rightarrow \frac{y}{1-x^2}=t^{-1 / 2}+c \\
& \Rightarrow \frac{y}{1-x^2}=\frac{1}{\sqrt{1-x^2}}+c
\end{aligned}
$

Question 3.
Solve: $(x+1) \frac{d y}{d x}-y=e^x(x+1)^2$
Solution:
The given equation can be written as $\frac{d y}{d x}-\frac{y}{x+1}=e^x(x+1)$
This is linear in $y$. Here $\int \mathrm{P} d x=-\int \frac{1}{x+1} d x=-\log (x+1)$
So I.F. $=e^{\int P d x}=e^{-\log (x+1)}=\frac{1}{x+1}$
The required solution is
$
\begin{aligned}
& y . \frac{1}{x+1}=\int e^x(x+1) \frac{1}{x+1} d x+c=\int e^x d x+c \\
& \text { i.e., } \frac{y}{x+1}=e^x+c
\end{aligned}
$
Question 4.
Solve: $\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y$

Solution:
The given equation can be written as $\frac{d x}{d y}+\frac{x}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2}$.
This is linear in $x$. There we have
$
\begin{array}{r}
\int \mathrm{P} d y=\int \frac{1}{1+y^2} d y=\tan ^{-1} y \\
\mathrm{I} \cdot \mathrm{F}=e^{\int \mathrm{P} d y}=e^{\tan ^{-1} y}
\end{array}
$
The required solution is
$
\begin{aligned}
& x e^{\tan ^{-1} y}=\int e^{\tan ^{-1} y}\left(\frac{\tan ^{-1} y}{1+y^2}\right) d y+c \\
\Rightarrow \quad x e^{\tan ^{-1} y} & =\int e^t \cdot t d t+c \\
\Rightarrow \quad x e^{\tan ^{-1} y} & =t e^t-e^t+c \\
\Rightarrow \quad x e^{\tan ^{-1} y} & =e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+c
\end{aligned}
$
Question 5.
Solve: $d x+x d y=e^{-y} \sec ^2 y d y$
Solution:
$
\begin{aligned}
d x & =e^{-y} \sec ^2 y d y-x d y \\
& =\left(e^{-y} \sec ^2 y-x\right) d y \\
\text { (or) } \frac{d x}{d y} & =e^{-y} \sec ^2 y-x ; \text { (i.e.,) } \frac{d x}{d y}=\frac{\sec ^2 y}{e^y}-x \\
\text { (i.e.,) } \frac{d x}{d y}+x & =\frac{\sec ^2 y}{e^y}
\end{aligned}
$
Comparing this equation with $\frac{d x}{d y}+\mathbf{P} x=\mathrm{Q}$, we get,
$
\begin{gathered}
\mathrm{P}=1, \mathrm{Q}=\frac{\sec ^2 y}{e^y} \\
\int \mathrm{P} d y=\int 1 d y=y
\end{gathered}
$

$\begin{aligned}
\text { Integrating factor } & =\mathrm{I} . \mathrm{F}=e^{\int \mathrm{P} d y}=e^y \\
\text { Solution is } x \text { (I.F.) } & =\int \mathrm{Q} \text { (I.F.) } d y+c \\
\text { (i.e.,) } x e^y & =\int \frac{\sec ^2 y}{e^y} e^y d y+c \\
x e^y & =\int \sec ^2 y d y+c \\
x e^y & =\tan y+c
\end{aligned}$