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 \(y\) and \(\frac{dy}{dx}\)

Substituting into the differential - equation, we get \(v + x\frac{dv}{dx}=\frac{1}{e^{v}}+v\).

Step3: Simplify the equation

Subtracting \(v\) from both sides gives \(x\frac{dv}{dx}=\frac{1}{e^{v}}\), which can be rewritten as \(e^{v}dv=\frac{dx}{x}\).

Step4: Integrate both sides

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

Step5: Back - substitute \(v\)

Since \(v = \frac{y}{x}\), we have \(e^{y/x}=\ln|x|+C\).

Step6: Use the initial condition

Given \(y(1) = 0\), substitute \(x = 1\) and \(y = 0\) into \(e^{y/x}=\ln|x|+C\). We get \(e^{0}=\ln(1)+C\), and since \(e^{0}=1\) and \(\ln(1) = 0\), then \(C = 1\).

Answer:

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