Find the particular solution of the following differential equation, given that $y=1$, when $x=0…
Find the particular solution of the following differential equation, given that $y=1$, when $x=0,\left(1+x^2\right) \frac{d y}{d x}=e^{m\left(\tan ^{-1} x\right)}-y$
$x e^{\tan ^{-1}(x)}=\tan ^{-1}(x)+1$
$x e^{\tan ^{-1}(x)}=\tan ^{-1}(x)-1$
$y e^{\tan ^{-1}(x)}=\tan ^{-1}(x)+1$
$y e^{\tan ^{-1}(x)}=\tan ^{-1}(x)-1$
Solution
Given, differential equation
$\left(1+x^2\right) \frac{d y}{d x}=e^{m\left(\tan ^{-1} x\right)}-y, y(0)=1$
$\frac{d y}{d x}=\frac{e^{m \tan ^{-1} x}}{1+x^2}-\frac{y}{1+x^2}$
$\Rightarrow \quad \frac{d y}{d x}+y\left(\frac{1}{1+x^2}\right)=\frac{e^{m \tan ^{-1} x}}{1+x^2}$ ...(i)
On comparing with Bernoulli's equation
$\frac{d y}{d x}+P y=Q$
$\Rightarrow \quad P=\frac{1}{1+x^2}$,
$Q=\frac{e^{m \tan ^{-1} x}}{1+x^2}$
$\therefore \quad I F=e^{\int P d x}=e^{\int \frac{1}{1+x^2} d x}=e^{\tan ^{-1} x}$
Solution $y$ Eq. (i), we get
$y \cdot I F=\int Q \cdot I F d x+C$
$\Rightarrow \quad y \cdot e^{\tan ^{-1} x}=\int \frac{e^{m \tan ^{-1} x}}{1+x^2} \cdot e^{\tan ^{-1} x} d x+C$
Put $\tan ^{-1} x=t$
$\begin{aligned} & \Rightarrow \quad \frac{1}{1+x^2} d x=d t \\ & \Rightarrow \quad y \cdot e^{\tan ^{-1} x}=\int e^{m t} \cdot e^t d t+C \\ & \Rightarrow \quad y \cdot e^{\tan ^{-1} x}=\frac{e^{(m+1) t}}{m+1}+C\end{aligned}$
$=\frac{e^{(m+1) \tan ^{-1} x}}{m+1}+C$
$\because \quad y=1$, when, $x=0$
$1 \cdot e^0=\frac{e^{(m+1) \cdot 0}}{m+1}+C$
$\Rightarrow \quad 1=\frac{1}{m+1}+C$
$C=1-\frac{1}{m+1}=\frac{m}{m+1}$
$\therefore \quad y e^{\tan ^{-1} x}=\frac{e^{(m+1) \tan ^{-1} x}}{m+1}+\frac{m}{m+1}$