solve the given initial - value problem. the de is homogeneous.\n(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)