Question

Find the particular solution of the differential equation
$(1+e^{2x})dy+(1+y^2)e^{x}dx=0$, given that $y=1$ when $x=0$.

Answer 1

$(1+e^{2x})dy+(1+y^2)e^{x}dx=0\frac{dy}{(1+y^2)}=-\frac{e^{x}dx}{(1+e^{2x})}{\tan }^{-1}y=-\frac{e^{x}dx}{(1+e^{2x})}$
Let $t=e^{x}dt=e^{x}dx{\tan }^{-1}y=-\frac{dt}{(1+t^2)}{\tan }^{-1}y=-{\tan }^{-1}t+c{\tan }^{-1}y=-{\tan }^{-1}\left(e^x\right)+c$
When $x=0$ and $y=1$,
${\tan }^{-1}1=-{\tan }^{-1}\left(e^0\right)+c\frac{\pi }{4}=-\frac{\pi }{4}+cc=\frac{\pi }{2}$
${\tan }^{-1}y=-{\tan }^{-1}\left(e^x\right)+\frac{\pi }{2}$