Answer
See below
Work Step by Step
Given: $y''+5y'+4y=F(x)$
Two linearly independent solutions to the associated homogeneous differential equation are:
$y_1(x)=e^{-x}\\
y_2(x)=e^{-4x}$
with Wronskian $W_{y_1,y_2}(x)=(e^{-x})(-4e^{-4x})-(e^{-4x})(-e^{-x})=-3e^{-5x}$
Obtain: $K(x,t)=-\frac{1}{3}(e^{-t}e^{-4x}-e^{-4t}e^{-x})$
Consequently: $y_p(x)=\frac{1}{3} \int^x_{x_0} (e^{4x-4t}-e^{x-t} F(t)dt$
The general solution to the given differential equation can therefore be expressed as
$y(x)=c_1e^{-x}+c_2e^{-4x}+\frac{1}{3} \int^x_{x_0} (e^{4x-4t}-e^{x-t} F(t)dt$
