The solution of x frac{d y}{d x} = y(log y - | vedclass

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(A) Given differential equation is $x \frac{d y}{d x} = y(\log(\frac{y}{x}) + 1)$.Dividing by $x$,we get $\frac{d y}{d x} = \frac{y}{x}(\log(\frac{y}{

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$y = x e^{c x}$

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(A) Given differential equation is $x \frac{d y}{d x} = y(\log(\frac{y}{x}) + 1)$.Dividing by $x$,we get $\frac{d y}{d x} = \frac{y}{x}(\log(\frac{y}{x}) + 1)$.This is a homogeneous differential equation. Let $y = v x$,then $\frac{d y}{d x} = v + x \frac{d v}{d x}$.Substituting these into the equation:$v + x \frac{d v}{d x} = v(\log v + 1)$$v + x \frac{d v}{d x} = v \log v + v$$x \frac{d v}{d x} = v \log v$Separating variables: $\frac{d v}{v \log v} = \frac{d x}{x}$.Integrating both sides: $\int \frac{1}{v \log v} d v = \int \frac{1}{x} d x$.Let $\log v = t$,then $\frac{1}{v} d v = d t$.$\int \frac{1}{t} d t = \int \frac{1}{x} d x \implies \log t = \log x + \log c$.$\log(\log v) = \log(c x) \implies \log v = c x$.Since $v = \frac{y}{x}$,we have $\log(\frac{y}{x}) = c x \implies \frac{y}{x} = e^{c x} \implies y = x e^{c x}$.

The solution of $x \frac{d y}{d x} = y(\log y - \log x + 1)$ is

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