Question

Solve the differential equation \(ye^{\frac xy}dx=(xe^{\frac xy}+y^2)dy, \ \ (y≠0)\)

Answer 1

\(ye^{\frac xy}dx=(xe^{\frac xy}+y^2)dy, \ \ (y≠0)\)

\(⇒ye^{\frac xy} \frac {dx}{dy}=xe^{\frac xy}+y^2\)

\(⇒e^{\frac xy}[y.\frac {dx}{dy}-x]=y^2\)

\(⇒e^{\frac xy}.\frac {[y.\frac {dx}{dy}-x]}{y^2}=1\)       ...(1)

\(Let\  e^{\frac xy}=z.\)

Differentiating it with respect to y,we get:

\(\frac {d}{dy}(e^{\frac xy})=\frac {dz}{dy}\)

\(⇒e^{\frac xy}.\frac {d}{dy}(\frac  xy)=\frac {dz}{dy}\)

\(⇒e^{\frac xy}.[\frac {y.\frac {dx}{dy}-x}{y^2}]=\frac {dz}{dy}\)       ...(2)

From equation (1) and equation (2), we get:

\(\frac {dz}{dy}=1\)

\(⇒dz=dy\)

Integrating both sides, we get:

\(z=y+C\)

\(⇒e^{\frac xy}=y+C\)