Question

For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

$x+y=ta{n}^{-1}yand{y}^2{y}^{\prime }+y^2+1=0$

Answer 1

Given, $x+y=ta{n}^{-1}y....\left(i\right)$
On differentiating both sides w.r.t. x, we get
$\frac{d}{dx}(x+y)=\frac{d}{dx}ta{n}^{-1}y\Rightarrow 1+y^{\prime }=\frac{1}{1+y^2}\left(y^{\prime }\right)(\because \frac{dy}{dx}=y^{\prime })\Rightarrow (1+y^{\prime })(1+y^2)=y^{\prime }\Rightarrow 1+y^2+y^{\prime }+y^2{y}^{\prime }=y^{\prime }\Rightarrow 1+y^2+y^2{y}^{\prime }=0\Rightarrow y^{\prime }=\frac{-(1+y^2)}{y^2}$
But we have to verify, $y^2{y}^{\prime }+y^2+1=0...\left(ii\right)$
On substituting the value of y'in Eq.(ii), we get
$LHS=y^2\left\{\frac{-(1+y^2)}{y^2}\right\}+y^2+1=-(1+y^2)+(1+y^2)=0=RHS$
Hence, $x+y=ta{n}^{-1}y$ is a solution of the given differential equation.