The solution of the differential equation x f | vedclass

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Question

(A) Given the differential equation $x \frac{dy}{dx} = y(\log y - \log x + 1)$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} (\log(\frac{y}{x})

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$y = x e^{cx}$

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(A) Given the differential equation $x \frac{dy}{dx} = y(\log y - \log x + 1)$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} (\log(\frac{y}{x}) + 1)$.This is a homogeneous differential equation.Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.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 $\int \frac{du}{u} = \log x + C$.$\log(\log v) = \log x + \log c = \log(cx)$.Taking the exponential of both sides: $\log v = cx$.Since $v = \frac{y}{x}$,we have $\log(\frac{y}{x}) = cx$.Therefore,$\frac{y}{x} = e^{cx}$,which implies $y = x e^{cx}$.

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

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