General solution of (x+y)^{2} frac{d y}{d x}= | vedclass

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(D) Given equation: $(x+y)^{2} \frac{d y}{d x}=a^{2}, a 
eq 0$Let $x+y=t$. Then $1+\frac{d y}{d x}=\frac{d t}{d x}$,which implies $\frac{d y}{d x}=\f

Vedclass · Accepted Answer

$	an \frac{y+C}{a}=\frac{x+y}{a}$

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(D) Given equation: $(x+y)^{2} \frac{d y}{d x}=a^{2}, a 
eq 0$Let $x+y=t$. Then $1+\frac{d y}{d x}=\frac{d t}{d x}$,which implies $\frac{d y}{d x}=\frac{d t}{d x}-1$.Substituting this into the given equation:$t^{2}(\frac{d t}{d x}-1)=a^{2}$$t^{2} \frac{d t}{d x} = a^{2}+t^{2}$Separating the variables:$\frac{t^{2}}{a^{2}+t^{2}} d t = d x$Integrating both sides:$\int \frac{t^{2}+a^{2}-a^{2}}{t^{2}+a^{2}} d t = \int d x$$\int (1 - \frac{a^{2}}{t^{2}+a^{2}}) d t = x + C'$$t - a^{2} \cdot \frac{1}{a} 	an^{-1}(\frac{t}{a}) = x + C'$$t - a 	an^{-1}(\frac{t}{a}) = x + C'$Substituting $t = x+y$ back:$(x+y) - a 	an^{-1}(\frac{x+y}{a}) = x + C'$$y - C' = a 	an^{-1}(\frac{x+y}{a})$$\frac{y-C'}{a} = 	an^{-1}(\frac{x+y}{a})$$	an(\frac{y-C'}{a}) = \frac{x+y}{a}$Let $C = -C'$. Then the general solution is $	an(\frac{y+C}{a}) = \frac{x+y}{a}$.

General solution of $(x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0$ is ($C$ is an arbitrary constant)

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