Question

Solve the following initial value problem:
\[x e^{y/x} - y + x\frac{dy}{dx} = 0, y\left( e \right) = 0\]

Answer 1

\[x e^\frac{y}{x} - y + x\frac{dy}{dx} = 0 y\left( e \right) = 0\]
This is also a homogenous equation,

Put y = vx
\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[x e^v - vx + x\left( v + x\frac{dv}{dx} \right) = 0\]
\[x e^v - vx + xv + x^2 \frac{dv}{dx} = 0\]
\[x e^v + x^2 \frac{dv}{dx} = 0\]
\[ e^v = - x\frac{dv}{dx}\]
\[\frac{dx}{x} = - \frac{1}{e^v}dv\]
On integration both sides we get,
\[\int\frac{dx}{x} = - \int\frac{1}{e^v}dv\]
\[ \log_e x = - \int e^{- v} dv\]
\[ \Rightarrow \log_e x = e^{- \frac{y}{x}} + c ............\left( \because y = vx \right)\]
\[\text{ As given }y\left( e \right) = 0\]
\[ \log_e e = e^{- \frac{0}{e}} + c\]
\[1 = 1 + c\]
\[ \Rightarrow c = 0\]
\[ \therefore \log_e x = e^{- \frac{y}{x}}\]