Answer
$y=2e^x+Ce^{-x}$
Work Step by Step
Use an integration factor of:
$$\mu (x)=e^{\int P(x)dx}$$
where $P(x) = 1$
$$\mu (x)=e^{\int 1dx}=e^{\int 1dx}=e^{x}$$
Multiply the entire equation by this factor.
$$e^x(\frac{dy}{dx}+y=4e^x)$$
$$e^x\frac{dy}{dx}+e^xy=4e^{2x}$$
Integrate each side. Note how the left side becomes
$$\int e^x\frac{dy}{dx}+e^xy=\int 4e^{2x}dx$$
$$e^xy=2e^{2x}+C$$
Solve for y.
$$y=2e^x+Ce^{-x}$$
