Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$x+y={\tan }^{-1}y:y^2{y}^{\prime }+y^2+1=0$

Answer 1

Let $x+y={\tan }^{-1}y$
Differentiating both sides w.r.t. $x$ we get,
$\frac{d}{dx}[x+y]=\frac{d}{dx}\left[{\tan }^{-1}y\right]$
$\frac{dx}{dx}+\frac{dy}{dx}=\frac{1}{1+y^2}.\frac{dy}{dx}$
$1=\frac{1}{1+y^2}.\frac{dy}{dx}-\frac{dy}{dx}$
$1=\frac{dy}{dx}[\frac{1}{1+y^2}-1]$
$1=\frac{dy}{dx}\left[\frac{1-1-y^2}{1+y^2}\right]$
$1=\frac{dy}{dx}\left[\frac{-y^2}{1+y^2}\right]$
$\frac{dy}{dx}=-\frac{1+x^2}{y^2}$
$\frac{dy}{dx}=y^{\prime }$
$y^{\prime }=-\frac{1+y^2}{y^2}$
Taking LHS
Putting
$y^{\prime }=-\frac{1+y^2}{y^2}$
$y^2{y}^{\prime }+y^2+1$
$=y^2[-\frac{(1-y^2)}{y^2}-1]+y^2+1$
$y^2{y}^{\prime }+y^2+1=-(1+y^2)+y^2+1$
$y^2{y}^{\prime }+y^2+1=-1-y^2+y^2+1$
$y^2{y}^{\prime }+y^2+1==0$
Thus $LHS=RHS$
Hence verified.

Final answer:
Hence, the function $x+y={\tan }^{-1}y$ is a solution of the differential equation $y^2{y}^{\prime }+y^2+1=0$.