The general solution of the differential equa | vedclass

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(D) Given the differential equation: $y y^{\prime} = x \left[ \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi^{\prime}\left(\frac{y^2}{

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$\phi\left(\frac{y^2}{x^2}\right) = c x^2$

Vedclass · Answer

(D) Given the differential equation: $y y^{\prime} = x \left[ \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi^{\prime}\left(\frac{y^2}{x^2}\right)} \right]$.Divide both sides by $x^2$: $\frac{y}{x} \frac{dy}{dx} = \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi^{\prime}\left(\frac{y^2}{x^2}\right)}$.Let $v = \frac{y^2}{x^2}$. Then $v x^2 = y^2$. Differentiating with respect to $x$: $2v x + x^2 \frac{dv}{dx} = 2y \frac{dy}{dx}$.Thus,$y \frac{dy}{dx} = v x + \frac{x^2}{2} \frac{dv}{dx}$.Substituting into the equation: $\frac{1}{x} (v x + \frac{x^2}{2} \frac{dv}{dx}) = v + \frac{\phi(v)}{\phi^{\prime}(v)}$.$v + \frac{x}{2} \frac{dv}{dx} = v + \frac{\phi(v)}{\phi^{\prime}(v)}$.$\frac{x}{2} \frac{dv}{dx} = \frac{\phi(v)}{\phi^{\prime}(v)}$.Separating variables: $\frac{\phi^{\prime}(v)}{\phi(v)} dv = \frac{2}{x} dx$.Integrating both sides: $\ln|\phi(v)| = 2 \ln|x| + \ln|c| = \ln|c x^2|$.Therefore,$\phi(v) = c x^2$.Substituting $v = \frac{y^2}{x^2}$,we get $\phi\left(\frac{y^2}{x^2}\right) = c x^2$.

The general solution of the differential equation $y y^{\prime} = x \left[ \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi^{\prime}\left(\frac{y^2}{x^2}\right)} \right]$,where $\phi$ is an arbitrary function,is

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