How to find $y$ from $y' = e^{2x}-e^x y$? The problem asks me to find $y(x)$ from the equation
$$y' = e^{2x}-e^x y$$
The $y'$ is $dy/dx$ right, so wouldn't the correct step be to integrate right away? If not, should I change some terms before integrating? I'm fairly new to this, and am unaware of rules so please be clear in explanation, thank you very much.

You can use the method of integrating factors. First write
$$y' + e^x y = e^{2x}.$$
An integrating factor for the equation is $\exp(\int e^x\, dx) = e^{e^x}$. So we multiply both sides by $e^{e^x}$.
$$e^{e^x}y' + e^x e^{e^x}y = e^{2x}e^{e^x}.$$
The left hand side is $(e^{e^x} y)'$. So 
$$(e^{e^x}y)' = e^{2x}e^{e^x}.$$
By integration,
$$e^{e^x} y = \int e^{2x} e^{e^x}\, dx + c,$$
where $c$ is a constant. Let $u = e^x$. Then $du = e^x\, dx$, thus
$$\int e^{2x}e^{e^x}\, dt = \int e^x e^{e^x} e^x\, dx = \int ue^u\, du = (u - 1)e^u + c' = (e^x - 1)e^{e^x} + c',$$
where $c'$ is constant. Hence
$$e^{e^x}y = (e^x - 1)e^{e^x} + C,$$
where $C$ is a constant. So 
$$y = e^x - 1 + Ce^{-e^x}.$$
