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.