The solution of the equation frac{dy}{dx} = f | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} \left( \log \frac{y}{x} + 1 \right)$.This is a homogeneous differential equation. Le

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$\log \left( \frac{y}{x} \right) = cx$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} \left( \log \frac{y}{x} + 1 \right)$.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} = v (\log v + 1)$$v + x \frac{dv}{dx} = v \log v + v$$x \frac{dv}{dx} = v \log v$Separating the variables:$\frac{dv}{v \log v} = \frac{dx}{x}$Integrating both sides:$\int \frac{dv}{v \log v} = \int \frac{dx}{x}$Let $u = \log v$,then $du = \frac{1}{v} dv$. The integral becomes:$\int \frac{du}{u} = \int \frac{dx}{x}$$\log |u| = \log |x| + \log |c|$$\log |\log v| = \log |xc|$$\log v = xc$Substituting $v = \frac{y}{x}$ back:$\log \left( \frac{y}{x} \right) = cx$.

The solution of the equation $\frac{dy}{dx} = \frac{y}{x} \left( \log \frac{y}{x} + 1 \right)$ is

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