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The solution of the differential equation $x^2 \frac{d y}{d x}=x^2+x y+y^2$ is
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The correct answer is:
$\tan ^{-1}\left(\frac{y}{x}\right)=\log x+c$
It is homogeneous equation which can be written in the form $\frac{d y}{d x}=\frac{x^2+x y+y^2}{x^2}$
Now put $\boldsymbol{y}=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
Therefore, $\quad v+x \frac{d v}{d x}=\frac{x^2+v x^2+v^2 x^2}{x^2}=1+v+v^2$
$\begin{aligned} & \Rightarrow x \frac{d v}{d x}=1+v^2 \\ & \Rightarrow \frac{d v}{1+v^2}=\frac{d x}{x}\end{aligned}$
Now integrating both sides, we get $\tan ^{-1} v=\log x+c$
$\Rightarrow \quad \tan ^{-1}\left(\frac{y}{x}\right)=\log x+c$ {$\left.y=v x \Rightarrow v=y / x\right\}$
Now put $\boldsymbol{y}=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
Therefore, $\quad v+x \frac{d v}{d x}=\frac{x^2+v x^2+v^2 x^2}{x^2}=1+v+v^2$
$\begin{aligned} & \Rightarrow x \frac{d v}{d x}=1+v^2 \\ & \Rightarrow \frac{d v}{1+v^2}=\frac{d x}{x}\end{aligned}$
Now integrating both sides, we get $\tan ^{-1} v=\log x+c$
$\Rightarrow \quad \tan ^{-1}\left(\frac{y}{x}\right)=\log x+c$ {$\left.y=v x \Rightarrow v=y / x\right\}$
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