Question

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

$y=e^x+1$ and y''-y'=0.

Answer 1

Given, $y=e^x+1$
On differentiating both sides of this equation w.r.t. x, we get
$y^{\prime }=\frac{d}{dx}(e^x+1)=e^x....\left(i\right)$
Again differentiating both sides w.r.t. x, we get
$y^{\prime \prime }=\frac{d}{dx}\left(e^x\right)=e^x\Rightarrow y^{\prime \prime }=e^x$
On substituting the values of y' and y'' in the given differential equation,
$LHS=y^{\prime \prime }-y^{\prime }=e^x-e^x=0=RHS$
Hence, $y=e^x+1$ is a solution of the given differential equation.