The solution of (x+y)^{2} frac{dy}{dx} = a^{2 | vedclass

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Question

(A) Given the differential equation: $(x+y)^{2} \frac{dy}{dx} = a^{2}$.Let $v = x+y$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{d

Vedclass · Accepted Answer

$\frac{x+y}{a} = 	an \frac{y+C}{a}$,where $C$ is an arbitrary constant

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(A) Given the differential equation: $(x+y)^{2} \frac{dy}{dx} = a^{2}$.Let $v = x+y$. Then,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: $v^{2} (\frac{dv}{dx} - 1) = a^{2}$.Rearranging the terms: $v^{2} \frac{dv}{dx} = v^{2} + a^{2}$,so $\frac{dv}{dx} = \frac{v^{2} + a^{2}}{v^{2}}$.Separating the variables: $\frac{v^{2}}{v^{2} + a^{2}} dv = dx$.Integrating both sides: $\int \frac{v^{2}}{v^{2} + a^{2}} dv = \int dx$.This can be written as: $\int (1 - \frac{a^{2}}{v^{2} + a^{2}}) dv = x + C'$.Integrating gives: $v - a 	an^{-1}(\frac{v}{a}) = x + C'$.Substituting $v = x+y$: $(x+y) - a 	an^{-1}(\frac{x+y}{a}) = x + C'$.Simplifying: $y - a 	an^{-1}(\frac{x+y}{a}) = C'$.Rearranging: $\frac{y-C'}{a} = 	an^{-1}(\frac{x+y}{a})$.Taking the tangent on both sides: $	an(\frac{y-C'}{a}) = \frac{x+y}{a}$.Letting $-C' = C$,we get $\frac{x+y}{a} = 	an(\frac{y+C}{a})$.

The solution of $(x+y)^{2} \frac{dy}{dx} = a^{2}$ (where $a$ is a constant) is:

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