The general solution of the differential equa | vedclass

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(B) Given the differential equation $(x-y)^2 \frac{dy}{dx} = a^2$. Let $v = x - y$. Then $\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{

Vedclass · Accepted Answer

$x - y = a 	an \left( \frac{y+c}{a} ight)$

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(B) Given the differential equation $(x-y)^2 \frac{dy}{dx} = a^2$. Let $v = x - y$. Then $\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$. Substituting these into the equation: $v^2 (1 - \frac{dv}{dx}) = a^2$. $1 - \frac{dv}{dx} = \frac{a^2}{v^2} \implies \frac{dv}{dx} = 1 - \frac{a^2}{v^2} = \frac{v^2 - a^2}{v^2}$. Separating variables: $\frac{v^2}{v^2 - a^2} dv = dx$. Integrating both sides: $\int \frac{v^2 - a^2 + a^2}{v^2 - a^2} dv = \int dx$. $\int (1 + \frac{a^2}{v^2 - a^2}) dv = x + c$. $v + a^2 \cdot \frac{1}{2a} \log \left| \frac{v-a}{v+a} \right| = x + c$. $v + \frac{a}{2} \log \left| \frac{v-a}{v+a} \right| = x + c$. Substituting $v = x - y$: $(x-y) + \frac{a}{2} \log \left| \frac{x-y-a}{x-y+a} \right| = x + c$. $-y + \frac{a}{2} \log \left| \frac{x-y-a}{x-y+a} \right| = c$. $y = \frac{a}{2} \log \left| \frac{x-y-a}{x-y+a} \right| + c'$.

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

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