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Question Number 123009 by aurpeyz last updated on 21/Nov/20

Commented by aurpeyz last updated on 21/Nov/20

$$plshelpwithquestion1$$

Answered by Dwaipayan Shikari last updated on 21/Nov/20

$${(x+y)}^2\frac{dy}{dx}=a^{2}x+y=v\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$$$$v^2(\frac{dv}{dx}-1)=a^2$$$$v^2\frac{dv}{dx}=v^2+a^2\Rightarrow \int 1-\frac{a^2}{v^2+a^2}dv=\int dx$$$$\Rightarrow v-{atan}^{-1}\frac{v}{a}=x$$$$\Rightarrow x+y-{atan}^{-1}\frac{x+y}{a}=x$$$$\Rightarrow \frac{y}{a}={tan}^{-1}\frac{x+y}{a}\Rightarrow atan\frac{y}{a}-y=x$$

Commented by aurpeyz last updated on 22/Nov/20

$$thatslikehomogenousequation?$$

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