If \(y = y(x)\) is the solution of the differential equation \(x\displaystyle\frac{dy}{dx} = y + xe^{-\frac{y}{x}}\), \(x \gt 0\), with \(y(1) = 0\), then \(y(e)\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(e\log 2\)
Dividing by x:
\( \displaystyle \frac{dy}{dx} = \frac{y}{x} + e^{-\frac{y}{x}} \)
This is a homogeneous equation. With \(y = vx\), \(\frac{dy}{dx} = v + x\frac{dv}{dx}\):
\( \displaystyle v + x\frac{dv}{dx} = v + e^{-v} \)
\( \displaystyle e^{v}\,dv = \frac{dx}{x} \)
\( e^{v} = \log x + C \)
\( e^{\frac{y}{x}} = \log x + C \)
Using \(y(1) = 0\): \(e^0 = 0 + C\), so \(C = 1\). At \(x = e\):
\( e^{\frac{y}{e}} = 1 + 1 = 2 \)
\( \displaystyle \frac{y}{e} = \log 2 \)
\( y = e\log 2 \)
Why the other options are incorrect
Option (1) \(\log 2\) — Incorrect. This is the value of \(\frac{y}{x} = v\) at \(x = e\), not of y; it must be multiplied by \(x = e\).
Option (2) \(2e\) — Incorrect. This takes \(\frac{y}{e} = 2\), but 2 is the value of \(e^{\frac{y}{e}}\), not of \(\frac{y}{e}\).
Option (3) \(e\) — Incorrect. \(y = e\) gives \(e^{\frac{y}{e}} = e\), but the solution needs \(e^{\frac{y}{e}} = 1 + \log e = 2\).