Find the particular solution of the differential equation $x e^{y / x}-y+x \frac{d y}{d x}=0$, given that $y(1)=0$.

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Find the particular solution of the differential equation $x e^{y / x}-y+x \frac{d y}{d x}=0$ , given that $y(1)=0$ .

Find the particular solution of the differential equation $x e^{y / x}-y+x \frac{d y}{d x}=0$ , given that $y(1)=0$ .

Solution:

$\begin{array}{l} \Rightarrow x e^{\frac{y}{x}}-y+x \frac{d y}{d x}=0 \\ \Rightarrow x \frac{d y}{d x}=y-x e^{\frac{y}{x}} \\ \Rightarrow \frac{d y}{d x}=\left(\frac{y}{x}\right)-e^{\frac{y}{x}} \\ \Rightarrow \frac{d y}{d x}=f\left(\frac{y}{x}\right) \end{array}$
$\Rightarrow$ the given differential eq. is a homogenous eq..
Solution of the given differential eq. is:
Putting $y=v x$
$\begin{array}{l} \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ v+x \frac{d v}{d x}=\left(\frac{v x}{x}\right)-e^{\frac{v x}{x}} \\ \Rightarrow x \frac{d v}{d x}=-e^{v} \\ \Rightarrow \frac{d v}{e^{v}}=\frac{-d x}{x} \end{array}$
On integrating both sides we obtain:
$\begin{array}{l} \Rightarrow \int \frac{d v}{\mathrm{e}^{v}}=-\int \frac{\mathrm{dx}}{\mathrm{x}}+c \\ \Rightarrow-\mathrm{e}^{-\mathrm{v}}=-\ln |\mathrm{x}|+\mathrm{c} \end{array}$
By resubstituting the value of $\mathrm{y}=\mathrm{v} \mathrm{x}$ we obtain
$\begin{array}{l} \Rightarrow-e^{-\left(\frac{y}{x}\right)}=-\ln |\mathrm{x}|+c \\ \text { Now, } y(1)=0 \\ \Rightarrow-e^{-(0)}=-\ln |1|+c \\ \Rightarrow c=-1 \\ \Rightarrow \log |x|+e^{-y / x}=1 \end{array}$