I was reading some of my notes and I was not sure how the following works:
$\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$
Solve the above with the condition $y(0)=0$
$\Rightarrow y(x) = A(1-e^{-x})$ with $A$ an arbitrary constant.
I was just wondering how do you solve the above to get $y(x) = A(1-e^{-x})$? Because when I tried it, I got: $y=A + Be^{-x}$, and using the condition $y(0)=0$, I got $A+B =0 $.