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(D) Given differential equation is $(x + y)^2 \frac{dy}{dx} = a^2$. Let $x + y = v$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \

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None of these

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(D) Given differential equation is $(x + y)^2 \frac{dy}{dx} = a^2$. Let $x + y = v$. Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$. Substituting these into the original equation: $v^2 (\frac{dv}{dx} - 1) = a^2$ $\frac{dv}{dx} - 1 = \frac{a^2}{v^2}$ $\frac{dv}{dx} = \frac{a^2}{v^2} + 1 = \frac{a^2 + v^2}{v^2}$ Separating the variables: $\frac{v^2}{a^2 + v^2} dv = dx$ Integrating both sides: $\int \frac{v^2 + a^2 - a^2}{a^2 + v^2} dv = \int dx$ $\int (1 - \frac{a^2}{a^2 + v^2}) dv = x + c$ $v - a 	an^{-1}(\frac{v}{a}) = x + c$ Substituting $v = x + y$ back: $(x + y) - a 	an^{-1}(\frac{x + y}{a}) = x + c$ $y - a 	an^{-1}(\frac{x + y}{a}) = c$ Thus,the correct option is $(d)$.

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

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