Verify that the function y = e^x + 1(explicit or implicit) is a s… - Vidyadip

Question

Verify that the function $y = e^x + 1($explicit or implicit$)$ is a solution of differential equation $y" - y' = 0$

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Answer

It is given that $y = e^x + 1$
Now, differentiating both sides w.r.t. $x,$ we get,
$\frac{d y}{d x}=y^\prime=\frac{d}{d x}\left(e^{x}\right)=e^x ...(i)$
Now, Again, differentiating both sides w.r.t. $x,$ we get,
$\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^{x}\right)$
$\Rightarrow y'' = e^x=y' ...[$Using $(i)]$
$\Rightarrow y'' = y'$
$\Rightarrow y'' - y'=0$

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