Verify that the function y = ex + 1(explicit or implicit) is a so… - 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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