The general solution of the differential equa | vedclass

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(B) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{2x^2 - xy - y^2}{x^2 - y^2}$.Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{

Vedclass · Accepted Answer

$\sqrt{2} \log \left|\frac{y^2 - 2x^2}{x^2}\right| + \log \left|\frac{y - \sqrt{2}x}{y + \sqrt{2}x}\right| + 2\sqrt{2} \log |x| = c$

Vedclass · Answer

(B) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{2x^2 - xy - y^2}{x^2 - y^2}$.Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting into the equation: $v + x\frac{dv}{dx} = \frac{2x^2 - x(vx) - (vx)^2}{x^2 - (vx)^2} = \frac{2 - v - v^2}{1 - v^2}$.$x\frac{dv}{dx} = \frac{2 - v - v^2}{1 - v^2} - v = \frac{2 - v - v^2 - v + v^3}{1 - v^2} = \frac{v^3 - v^2 - 2v + 2}{1 - v^2} = \frac{v^2(v - 1) - 2(v - 1)}{1 - v^2} = \frac{(v^2 - 2)(v - 1)}{-(v^2 - 1)}$.Separating variables: $\int \frac{v^2 - 1}{(v^2 - 2)(v - 1)} dv = -\int \frac{dx}{x}$.Using partial fractions on $\frac{v^2 - 1}{(v^2 - 2)(v - 1)} = \frac{(v-1)(v+1)}{(v^2-2)(v-1)} = \frac{v+1}{v^2-2} = \frac{v}{v^2-2} + \frac{1}{v^2-2}$.Integrating: $\frac{1}{2} \log |v^2 - 2| + \frac{1}{2\sqrt{2}} \log \left|\frac{v - \sqrt{2}}{v + \sqrt{2}}\right| = -\log |x| + C$.Multiplying by $2\sqrt{2}$: $\sqrt{2} \log |v^2 - 2| + \log \left|\frac{v - \sqrt{2}}{v + \sqrt{2}}\right| = -2\sqrt{2} \log |x| + C$.Substituting $v = \frac{y}{x}$: $\sqrt{2} \log \left|\frac{y^2 - 2x^2}{x^2}\right| + \log \left|\frac{y - \sqrt{2}x}{y + \sqrt{2}x}\right| + 2\sqrt{2} \log |x| = c$.

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

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