Question

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

Answer 1

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

$⇒\frac {dy}{1+y^2}+\frac {e^xdx}{1+e^{2x}}=0$

Integrating both sides,we get:

$tan^{-1}y+∫\frac {e^xdx}{1+e^{2x}}=C$ ...(1)

$Let\ e^x=t⇒e^{2x}=t^2.$

$⇒\frac {d}{dx}(e^x)=\frac {dt}{dx}$

$⇒e^x=\frac {dt}{dx}$

$⇒e^xdx=dt$

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

$tan^{-1}y+∫\frac {dt}{1+t^2}=C$

$⇒tan^{-1}y+tan^{-1}t=C$

$⇒tan^{-1}y+tan^{-1} (e^x)=C$ ...(2)

Now, y=1 at x=0.

Therefore, equation (2) becomes:

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

$⇒\frac \pi4+\frac \pi4=C$

$⇒C=\frac \pi2$

Substituting $\frac \pi2$ in equation (2), becomes:

$tan^{-1}y+tan^{-1}(e^x)=\frac \pi2$

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