#### Answer

To verify the solution we find $y'$, $y''$ and then put $y$, $y'$ and $y''$ into the differential equation. Since we see that the left and the right side are equal we verified the solution.

#### Work Step by Step

Let us find $y'$ and $y''$.
$$y' = \left(\frac{2}{5}(e^{-4x}+e^x)\right)' = \frac{2}{5}(e^{-4x}+e^x)' = \frac{2}{5}((e^{-4x})'+(e^x)').$$
From table of derivatives $$(e^x)'=e^x.$$
Now to find $(e^{-4x})'$ use the chain rule:
$$(e^{-4x})'=e^{-4x}(-4x)'=-4e^{-4x}.$$
Putting this into the expression for $y'$ we have
$$y'=\frac{2}{5}e^x -\frac{8}{5}e^{-4x}.$$
Now lets find the second derivative $y''$:
$$y''=(y')' = \left(\frac{2}{5}e^x -\frac{8}{5}e^{-4x}\right)' = \frac{2}{5}(e^x)'-\frac{8}{5}(e^{-4x})'.$$
We already found $(e^x)' = e^x$ and $(e^{-4x})'=-4e^{-4x}$ so putting this into previous expression gives us
$$y'' =\frac{2}{5}e^x -(-4) \frac{8}{5}e^{-4x} =\frac{2}{5}e^x + \frac{32}{5}e^{-4x}.$$
Puttiong $y$, $y'$ and $y''$ into the differential equation we obtain:
The Left side:
$$y''+4y' = \frac{2}{5}e^x + \frac{32}{5}e^{-4x} + 4(\frac{2}{5}e^x -\frac{8}{5}e^{-4x}) = \frac{2}{5}e^x + \frac{32}{5}e^{-4x} + \frac{8}{5}e^x - \frac{32}{5}e^{-4x} = \frac{10}{5}e^x=2e^x.$$
The Right side is just $$2e^x.$$
We see that the left and the right sides are equal and this verifies that given $y$ is a solution to the given differential equation.