Question

If \( y = y(x) \) is the solution of the differential equation \[ x \frac{dy}{dx} = y + xe^{-\frac{y}{x}}, \quad y(1) = \log e \] , then \( y(e) = \)

• A. \( \log \left( \frac{1}{e} + 1 \right) \)
• B. \( e \log(1 + e) \)
• C. \( e \log \left( \frac{1}{e} + 1 \right) \)
• D. \( e \log \left( 1 - \frac{1}{e} \right) \)

Hint:

Try substitution like \( z = \frac{y}{x} \) in equations involving both \( y \) and \( \frac{y}{x} \) for easier separation of variables.

Answer 1

The Correct Option is B

We are given: \[ x \frac{dy}{dx} = y + x e^{- \frac{y}{x}} \]
Let us use substitution. Let \( z = \frac{y}{x} \Rightarrow y = zx \). Then: \[ \frac{dy}{dx} = z + x \frac{dz}{dx} \]
Substitute into the original equation: \[ x(z + x \frac{dz}{dx}) = zx + x e^{-z} \Rightarrow xz + x^2 \frac{dz}{dx} = zx + x e^{-z} \]
Subtracting \( xz \) from both sides: \[ x^2 \frac{dz}{dx} = x e^{-z} \Rightarrow x \frac{dz}{dx} = e^{-z} \]
Separate the variables: \[ e^z dz = \frac{dx}{x} \Rightarrow \int e^z dz = \int \frac{dx}{x} \Rightarrow e^z = \ln x + C \]
Substitute back \( z = \frac{y}{x} \), so: \[ e^{\frac{y}{x}} = \ln x + C \Rightarrow \frac{y}{x} = \ln(\ln x + C) \Rightarrow y = x \ln(\ln x + C) \]
Use the condition \( y(1) = \log e = 1 \): \[ 1 = 1 \cdot \ln(\ln 1 + C) \Rightarrow \ln(\ln 1 + C) = 1 \Rightarrow \ln(0 + C) = 1 \Rightarrow C = e \]
Now, find \( y(e) \): \[ y(e) = e \cdot \ln(\ln e + e) = e \cdot \ln(1 + e) \]