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)