Answer
False
Work Step by Step
The second derivative of $Ae^x$ is $Ae^x$. Plugging this into the differential equation, we get:
$Ae^x - Ae^x = 0$
This shows that no matter what value of A we chose, plugging $Ae^x$ into the differential equation will always yield 0, rather than $e^x$. So the claim is false.
