# Verify that the given function(explicit or implicit)is a solution of the corresponding differential equation $y=e^x+1\qquad:\;y''-y'=0$

Toolbox:
• Differentiation of $e^x = e^x$
Step 1:
Given $y = e^x + 1$
Differentiating on both sides we get,
$y' = e^x + 0.$--------(1)
Step 2:
Differentiating again we get,
$y'' = e^x.$------(2)
Now subtracting equ (2) from equ (1)
$y'' - y' =e^2 -e^2 = 0$
Hence it is solution for the given function.