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$x+y=\tan ^{-1} y \quad: y^{2} y^{\prime}+y^{2}+1=0$


$x+y=\tan ^{-1} y$

Differentiating both sides of this equation with respect to x, we get:

$\frac{d}{d x}(x+y)=\frac{d}{d x}\left(\tan ^{-1} y\right)$

$\Rightarrow 1+y^{\prime}=\left[\frac{1}{1+y^{2}}\right] y^{\prime}$

$\Rightarrow y^{\prime}\left[\frac{1}{1+y^{2}}-1\right]=1$

$\Rightarrow y^{\prime}\left[\frac{1-\left(1+y^{2}\right)}{1+y^{2}}\right]=1$

$\Rightarrow y^{\prime}\left[\frac{-y^{2}}{1+y^{2}}\right]=1$

$\Rightarrow y^{\prime}=\frac{-\left(1+y^{2}\right)}{y^{2}}$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{2} y^{\prime}+y^{2}+1=y^{2}\left[\frac{-\left(1+y^{2}\right)}{y^{2}}\right]+y^{2}+1$



$=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

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