# Find the particular solution of the differential equation

Question:

Find the particular solution of the differential equation

$\left(1+e^{2 x}\right) d y+\left(1+y^{2}\right) e^{x} d x=0$, given that $y=1$ when $x=0$

Solution:

$\left(1+e^{2 x}\right) d y+\left(1+y^{2}\right) e^{x} d x=0$

$\Rightarrow \frac{d y}{1+y^{2}}+\frac{e^{x} d x}{1+e^{2 x}}=0$

Integrating both sides, we get:

$\tan ^{-1} y+\int \frac{e^{x} d x}{1+e^{2 x}}=\mathrm{C}$              ...(1)

Let $e^{x}=t \Rightarrow e^{2 x}=t^{2} .$

$\Rightarrow \frac{d}{d x}\left(e^{x}\right)=\frac{d t}{d x}$

$\Rightarrow e^{x}=\frac{d t}{d x}$

$\Rightarrow e^{x} d x=d t$

Substituting these values in equation (1), we get:

$\tan ^{-1} y+\int \frac{d t}{1+t^{2}}=\mathrm{C}$

$\Rightarrow \tan ^{-1} y+\tan ^{-1} t=\mathrm{C}$

$\Rightarrow \tan ^{-1} y+\tan ^{-1}\left(e^{x}\right)=\mathrm{C}$                ...(2)

Now, y = 1 at x = 0.

Therefore, equation (2) becomes:

$\tan ^{-1} 1+\tan ^{-1} 1=C$

$\Rightarrow \frac{\pi}{4}+\frac{\pi}{4}=\mathrm{C}$

$\Rightarrow \mathrm{C}=\frac{\pi}{2}$

Substituting $\mathrm{C}=\frac{\pi}{2}$ in equation ( 2 ), we get:

$\tan ^{-1} y+\tan ^{-1}\left(e^{x}\right)=\frac{\pi}{2}$

This is the required particular solution of the given differential equation.