Question

solve the given initial - value problem. the de is homogeneous.
(x + ye^{y/x})dx−xe^{y/x}dy = 0, y(1)=0

Explanation:

Step1: Rewrite the differential equation

Given \((x + ye^{y/x})dx-xe^{y/x}dy = 0\), we can rewrite it as \(\frac{dy}{dx}=\frac{x + ye^{y/x}}{xe^{y/x}}=\frac{1}{e^{y/x}}+\frac{y}{x}\). Let \(v=\frac{y}{x}\), then \(y = vx\) and \(\frac{dy}{dx}=v + x\frac{dv}{dx}\).

Step2: Substitute \(y = vx\) into the differential - equation

Substituting gives \(v + x\frac{dv}{dx}=\frac{1}{e^{v}}+v\). Then \(x\frac{dv}{dx}=\frac{1}{e^{v}}\).

Step3: Separate the variables

We get \(e^{v}dv=\frac{1}{x}dx\).

Step4: Integrate both sides

Integrating \(\int e^{v}dv=\int\frac{1}{x}dx\). We know that \(\int e^{v}dv=e^{v}+C_1\) and \(\int\frac{1}{x}dx=\ln|x|+C_2\). So \(e^{v}=\ln|x| + C\).

Step5: Substitute back \(v=\frac{y}{x}\)

We have \(e^{y/x}=\ln|x| + C\).

Step6: Use the initial - condition \(y(1) = 0\)

When \(x = 1\) and \(y = 0\), \(e^{0}=\ln(1)+C\). Since \(e^{0}=1\) and \(\ln(1)=0\), then \(C = 1\).

Step7: Write the final solution

The solution of the initial - value problem is \(e^{y/x}=\ln|x| + 1\).

Answer:

\(e^{y/x}=\ln|x| + 1\)