The general solution of the differential equa | vedclass

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(B) Given the differential equation: $x \frac{dy}{dx} = y \ln \left( \frac{y}{x} \right)$.This is a homogeneous differential equation. Let $y = vx$,th

Vedclass · Accepted Answer

$y = xe^{1 + cx}$

Vedclass · Answer

(B) Given the differential equation: $x \frac{dy}{dx} = y \ln \left( \frac{y}{x} \right)$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $x(v + x \frac{dv}{dx}) = vx \ln(v)$.Dividing by $x$: $v + x \frac{dv}{dx} = v \ln(v)$.Rearranging: $x \frac{dv}{dx} = v \ln(v) - v = v(\ln(v) - 1)$.Separating variables: $\frac{dv}{v(\ln(v) - 1)} = \frac{dx}{x}$.Integrating both sides: $\int \frac{dv}{v(\ln(v) - 1)} = \int \frac{dx}{x}$.Let $u = \ln(v) - 1$,then $du = \frac{1}{v} dv$.So,$\int \frac{du}{u} = \ln|x| + C$.$\ln|\ln(v) - 1| = \ln|x| + C$.$\ln(v) - 1 = Cx$ (where $C$ is a constant).$\ln(\frac{y}{x}) = 1 + Cx$.$\frac{y}{x} = e^{1 + Cx}$.$y = xe^{1 + Cx}$.

The general solution of the differential equation $x \left( \frac{dy}{dx} \right) = y \ln \left( \frac{y}{x} \right)$ is:

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