Solve the differential equation $x\,dy+( 2xe^{y/x}-y)\,dx=0$ if $y=0$ when $x=1$.

Solution Follow the steps for solving a homogeneous first-order differential equation:

Step 1 Both $x$ and $2xe^{y/x}-y$ are homogeneous of degree 1.

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Step 2 Let $y=xv$. Then $dy=x\,dv+v\,dx$. Substitute these into the differential equation. $$\begin{eqnarray*} x\,dy+( 2xe^{y/x}-y)\,dx &=&0 \\ x( x\,dv+v\,dx) +( 2xe^{v}-xv)\,dx &=&0 \\ x^{2}dv+xv\,dx+2xe^{v}dx-xv\,dx &=&0 \\ x^{2}dv+2xe^{v}dx &=&0 \\ x\,dv+2e^{v}dx &=&0 \end{eqnarray*}$$

Step 3 To separate the variables, we divide by $xe^{v}.$ Then for $x\neq 0$, $$\begin{equation*} \dfrac{2dx}{x}+\dfrac{dv}{e^{v}}=0 \end{equation*}$$

Step 4 Integrate. $$\begin{eqnarray*} \int \dfrac{2dx}{x}+\int \dfrac{dv}{e^{v}} &=&0 \\ 2\int \dfrac{dx}{x}+\int e^{-v}dv &=&0 \\ 2\ln \left\vert x\right\vert -e^{-v} &=&C \\ 2\ln \left\vert x\right\vert -e^{-y/x} &=&C\qquad \color{#0066A7}{{v}} \color{#0066A7}{{=\dfrac{y}{x}}} \end{eqnarray*}$$

This is the general solution to the differential equation. To find the particular solution, we substitute $x=1$ and $y=0$ to find $C$. $$\begin{eqnarray*} 2~\ln 1-e^{0} &=&C \\ C &=&-1 \end{eqnarray*}$$

The particular solution is $$2~\ln \left\vert x\right\vert -e^{-y/x}=-1$$