Find the particular solution of the differential equation : <img alt="\left ( 1+e^{2x} \right )dy+\left ( 1+y^{2} \right )e^{x}dx= 0," src="https://entrancecorner.codecogs.com/gif.latex?%5Cleft%20%28%201+e%5E%7B2x%7D%20%5Cright%20%29dy&

Find the particular solution of the differential equation :
$\left ( 1+e^{2x} \right )dy+\left ( 1+y^{2} \right )e^{x}dx= 0,$ given that $y\left ( 0 \right )= 1\cdot$

Answers (1)

$\left ( 1+e^{2x} \right )dy+\left ( 1+y^{2} \right )e^{x}dx= 0$
$\Rightarrow \int \frac{dy}{1+y^{2}}+\int \frac{e^{x}dx}{1+e^{2x}}= 0$
put $e^{x}= t\; \; \Rightarrow e^{x}dx= dt$ in the second integral
$\therefore \tan^{-1}y+\int \frac{dt}{1+t^{2}}= 0$
$\Rightarrow \tan^{-1}y+\tan^{-1}t= \tan^{-1}c$
$\Rightarrow \tan^{-1}y+\tan^{-1}e^{x}= \tan^{-1}c$
given that $y\left ( 0 \right )= 1,\tan^{-1}1+\tan^{-1}e^{0}= \tan^{-1}c$
$c= \tan \left ( \frac{\pi }{4}+\frac{\pi }{4} \right )= \tan \frac{\pi }{2}= \frac{1}{0}$
$\therefore$ The solution is $\tan^{-1}y+\tan^{-1}e= \tan^{-1}\frac{1}{0}$
ie $\tan^{-1}\frac{y+e^{x}}{1-ye^{x}}= \tan^{-1}\frac{1}{0}$
$\Rightarrow \frac{y+e^{x}}{1-ye^{x}}= \frac{1}{0}$ $\therefore 1-ye^{x}= 0$
$\therefore y=e^{-x}$ is the required solutions.