Solve the differential equation \(x\,dy+( 2xe^{y/x}-y)\,dx=0\) if \(y=0 \) when \(x=1\).
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 \[ 2~\ln \left\vert x\right\vert -e^{-y/x}=-1 \]