Question

Solve the differential equation:
$(1+e^{2x})dy+(1+y^2)e^{x}dx=0$.

Answer 1

$(1+e^{2x})dy+(1+y^2)e^{x}dx=0$

separating $y$ and $x$

$\Rightarrow \frac{dy}{-(1+y^2)}=\frac{e^x}{1+e^{2x}}dx$

Integrating both side

$-\int \frac{dy}{1+y^2}=\int \frac{e^{x}dx}{1+e^{2x}}$

Let $p=e^x$

$\Rightarrow \frac{dp}{dx}=e^x\Rightarrow dx=\frac{dp}{e^x}$

$\therefore -\int \frac{dy}{1+y^2}=\int e^x\frac{dp}{e^2(1+e^{2x})}=\int \frac{dp}{1+e^{2x}}=\int \frac{dp}{1+p^2}$

$(\therefore p=e^x,p^2=e^{2x})$

$\Rightarrow -\int \frac{dy}{1+y^2}=\int \frac{dp}{1+p^2}$

$\Rightarrow -{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(p\right)+c$

$\Rightarrow -{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(e^x\right)+c$

$\Rightarrow {\tan }^{-1}\left(y\right)=-\left[{\tan }^{-1}\right(e^x)+c$