The solution of the differential equation x y | vedclass

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(D) Given equation is $x \frac{dy}{dx} = 2x e^{-y/x} + y$.Dividing by $x$, we get $\frac{dy}{dx} = 2e^{-y/x} + \frac{y}{x}$.This is a homogeneous diff

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$e^{y / x} = 2 \log |C x|$

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(D) Given equation is $x \frac{dy}{dx} = 2x e^{-y/x} + y$.Dividing by $x$, we get $\frac{dy}{dx} = 2e^{-y/x} + \frac{y}{x}$.This is a homogeneous differential equation. Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation:$v + x \frac{dv}{dx} = 2e^{-v} + v$.Subtracting $v$ from both sides, we get $x \frac{dv}{dx} = 2e^{-v}$.Separating the variables: $e^v dv = \frac{2}{x} dx$.Integrating both sides: $\int e^v dv = \int \frac{2}{x} dx$.$e^v = 2 \log |x| + C_1$.Since $2 \log |x| + C_1 = \log |x^2| + \log |C| = \log |C x^2|$, or more simply $2 \log |x| + \log |C| = \log |C x^2|$ is not matching, let us write $C_1 = \log |C|$.Then $e^v = 2 \log |x| + \log |C| = \log |x^2| + \log |C| = \log |C x^2|$.Wait, checking the options, $e^{y/x} = 2 \log |Cx| = 2(\log |C| + \log |x|) = 2 \log |x| + 2 \log |C|$.If we set the constant $C_1 = 2 \log |C|$, then $e^{y/x} = 2 \log |x| + 2 \log |C| = 2 \log |Cx|$.Thus, the correct option is $D$.

The solution of the differential equation $x y^{\prime} = 2 x e^{-y / x} + y$ is

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