Question

Find the particular solution of the 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

We have

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

$\Rightarrow \frac{dx}{dy}=\frac{2x{e}^{\frac{x}{y}}-y}{2y{e}^{\frac{x}{y}}}$ ...............(1)

Here, the given differential equation is a homogeneous differential equation. As the right hand side of equation (1) is expressed as function of $\frac{x}{y}.$

Therefore, we put $x=vy$

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

substituting $x=vy$ in the given differential equation we will get,

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

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

$\Rightarrow 2e^{v}dv=\frac{-1}{y}dy,y\ne 0$

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

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

$\Rightarrow 2e^{\frac{x}{y}}+\log |y|=C$ ...........(2)

It is given that $x=0$ when $y=1$,

So putting $x=0,y=1$ in equation (2), we get

$2e^0+\log 1=C$

$\Rightarrow C=2$

Putting $C=2$ in equation (2), we get the required particular solution

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