Question

Find the particular solution of differential equation : $2y{e}^{\frac{x}{y}}dx+(y-2x{e}^{\frac{x}{y}})dy=0$ given that x = 0 when y = 1.

Answer 1

Here $2y{e}^{\frac{x}{y}}dx+(y-2x{e}^{\frac{x}{y}})dy=0$ ~~~ $\Rightarrow \frac{dx}{dy}=\frac{2x{e}^{\frac{x}{y}}-y}{2y{e}^{\frac{x}{y}}}$

Put x=vy $\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$ $\therefore v+y\frac{dv}{dy}=\frac{2vy{e}^v-y}{2y{e}^v}=\frac{2v{e}^v-1}{2e^v}\Rightarrow y\frac{dv}{dy}=\frac{2v{e}^v-1}{2e^v}$

$\Rightarrow y\frac{dv}{dy}=\frac{-1}{2e^v}\Rightarrow 2\int e^{v}dv=-\int \frac{dy}{y}\Rightarrow 2e^v=-log|y|+C\Rightarrow 2e^{\frac{x}{y}}=-log|y|+C.$

As y = 1 when x = 0 so, $2e^{\frac{0}{1}}=-log|1|+C\Rightarrow C=2.$

Hence the required solution is, $2e^{\frac{x}{y}}=2-log|y|.$