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The given differential equation is: \( xe^{\frac{y}{x}} - y + x\frac{dy}{dx} = 0 \)
Rearrange the equation to isolate \(\frac{dy}{dx}\):
\( x\frac{dy}{dx} = y - xe^{\frac{y}{x}} \)
\( \frac{dy}{dx} = \frac{y}{x} - e^{\frac{y}{x}} \)
This is a homogeneous differential equation. Let \( y = vx \), so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Substitute \( y = vx \) into the equation:
\( v + x\frac{dv}{dx} = v - e^v \)
Simplify the equation:
\( x\frac{dv}{dx} = -e^v \)
Separate the variables:
\( e^{-v} dv = -\frac{dx}{x} \)
Integrate both sides:
\( \int e^{-v} dv = \int -\frac{dx}{x} \)
\( -e^{-v} = -\ln|x| + C \)
Multiply by -1:
\( e^{-v} = \ln|x| - C \)
Substitute back \( v = \frac{y}{x} \):
\( e^{-\frac{y}{x}} = \ln|x| - C \)
We can rewrite the constant as \( \ln|C_1| \) for simplicity:
\( e^{-\frac{y}{x}} = \ln|x| + \ln|C_1'| \)
\( e^{-\frac{y}{x}} = \ln|C_1 x| \)
Taking exponential on both sides is not necessary, the solution is:
\( e^{-\frac{y}{x}} = \ln|x| + C \)
Correct Answer: \( e^{-\frac{y}{x}} = \ln|x| + C \)