Question

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

Answer 1

x+y=tan-1y
Differentiating both sides of this equation with respect to x,we get:
$\frac{d}{dx}(x+y)=\frac{d}{dx}(tan^{-1}y)$
$⇒1+y'=[\frac{1}{1+y^2}]y'$
$⇒y'[\frac{1}{1+y2-1}]=1$
$⇒y'[\frac{1-(1+y^2)}{1+y^2}]=1$
$⇒y'[-\frac{y}{/1+y2}]=1$
$⇒y'=-(\frac{1+y^2)}{y^2}$
Substituting the value of y'in the given differential equation,we get:
L.H.S.=y2y'+y2+1=y2[-$(\frac{1+y2)}{y2}]+y^2+1$
$=-1-y^2+y^2+1$
=0
=R.H.S.
Henve,the given function is the solution of the corresponding differential equation.