The solution of the differential equation x^{ | vedclass

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(A) Given the differential equation $x^{2} \frac{dy}{dx} = y^{2} + xy$.Dividing by $x^{2}$,we get $\frac{dy}{dx} = \frac{y^{2} + xy}{x^{2}} = \left(\f

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$\frac{x}{y} + \log |x| = c$

Vedclass · Answer

(A) Given the differential equation $x^{2} \frac{dy}{dx} = y^{2} + xy$.Dividing by $x^{2}$,we get $\frac{dy}{dx} = \frac{y^{2} + xy}{x^{2}} = \left(\frac{y}{x}\right)^{2} + \frac{y}{x}$.This is a homogeneous differential equation. Let $y = ux$,then $\frac{dy}{dx} = u + x \frac{du}{dx}$.Substituting these into the equation: $u + x \frac{du}{dx} = u^{2} + u$.Subtracting $u$ from both sides: $x \frac{du}{dx} = u^{2}$.Separating the variables: $\frac{du}{u^{2}} = \frac{dx}{x}$.Integrating both sides: $\int u^{-2} du = \int \frac{1}{x} dx$.$-u^{-1} = \log |x| + c$.Since $u = \frac{y}{x}$,we have $-\frac{x}{y} = \log |x| + c$.Rearranging the terms: $\frac{x}{y} + \log |x| = -c$,which can be written as $\frac{x}{y} + \log |x| = C$ (where $C = -c$).

The solution of the differential equation $x^{2} \frac{dy}{dx} = y^{2} + xy$ is

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