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:

$z=y+C$

$⇒e^{\frac xy}=y+C$