The solution of the differential equation y f | vedclass

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(C) Given the differential equation: $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} \right]$.Dividing by $y$, we g

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

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(C) Given the differential equation: $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} \right]$.Dividing by $y$, we get $\frac{dy}{dx} = \frac{y}{x} + \frac{x \phi(y^2/x^2)}{y \phi'(y^2/x^2)}$.Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = v + \frac{x \phi(v^2)}{vx \phi'(v^2)} = v + \frac{\phi(v^2)}{v \phi'(v^2)}$.Subtracting $v$ from both sides: $x \frac{dv}{dx} = \frac{\phi(v^2)}{v \phi'(v^2)}$.Separating variables: $\frac{v \phi'(v^2)}{\phi(v^2)} dv = \frac{dx}{x}$.Integrating both sides: $\int \frac{v \phi'(v^2)}{\phi(v^2)} dv = \int \frac{dx}{x}$.Let $u = v^2$, then $du = 2v dv$, so $v dv = \frac{1}{2} du$.The integral becomes $\frac{1}{2} \int \frac{\phi'(u)}{\phi(u)} du = \ln|x| + C_1$.$\frac{1}{2} \ln|\phi(u)| = \ln|x| + C_1 \Rightarrow \ln|\phi(v^2)| = 2 \ln|x| + 2C_1 = \ln|x^2| + \ln|c|$.Thus, $\phi(v^2) = c x^2$.Substituting $v^2 = y^2/x^2$, we get $\phi(y^2/x^2) = c x^2$.

The solution of the differential equation $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} \right]$ is (where $c$ is a constant):

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