Question

Show that the differential equation $2y{e}^{x/y}dx+(y-2x{e}^{x/y})dy=0$ is homogeneous. Find the particular solution of this differential equation, given that $x=0$ when $y=1$.

Answer 1

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

$\frac{dy}{dx}=\frac{2y{e}^{x/y}}{2x{e}^{x/y}-y}$

$F(\lambda x,\lambda y)=\frac{2\lambda y{e}^{\lambda x/\lambda y}}{2\lambda x{e}^{\lambda x/\lambda y}-\lambda y}=\frac{2y{e}^{x/y}}{2x{e}^{x/y}-y}=F(x,y)$

Hence, it is a homogenous function.

Putting $\frac{x}{y}=v$

$\frac{dx}{dy}=v+y\frac{dv}{dy}$

$\frac{dx}{dy}=\frac{2v{e}^v-1}{2e^v}$

$v+y\frac{dv}{dy}=\frac{2v{e}^v-1}{2e^v}$

$\frac{dy}{y}=-2e^{v}dv$

$\log y=-2e^v+C$

$\log y+2e^{x/y}=C$

$C=2$

$P.I.-\log y+2e^{x/y}-2=0$