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

$2y{e}^{\frac{x}{y}}dx+(y-2x{e}^{\frac{x}{y}})dy=0$
$\Rightarrow 2e^{\frac{x}{y}}=-(1-\frac{2x}{y}{e}^{\frac{x}{y}})\frac{dy}{dx}$
Put $\frac{x}{y}=t$
$x=yt$
$1=\frac{dy}{dx}t+y\frac{dt}{dy}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{t}(1-y\frac{dt}{dx})$
$\Rightarrow 2t{e}^t=-(1-2t{e}^t)\frac{1}{t}(1-\frac{ydt}{dx})$
$\Rightarrow 2t{e}^t=(2t{e}^t-1)(1-y\frac{dt}{dx})=2t{e}^t-1-y2t{e}^t\frac{dt}{dx}+\frac{ydt}{dx}$
$\Rightarrow 1=y(1-2t{e}^t)\frac{dt}{dx}=\frac{x}{t}(1-2t{e}^t)\frac{dt}{dx}$
$\int \frac{dx}{x}=\int (\frac{1}{t}-2e^t)dt$
$\Rightarrow lnx=lnt-2e^t+C$
$\Rightarrow lnx=ln\frac{x}{y}-2e^{\frac{x}{y}}+C$
$\Rightarrow lny=-2e^{\frac{x}{y}}+C$
At $x=0,y=1$
$\Rightarrow 0=-2e^0+C\Rightarrow C=2$
$lny=2-2e^{\frac{x}{y}}$