Answer
See below
Work Step by Step
Given: $y''-y=F(x)$
Two linearly independent solutions to the associated homogeneous differential equation are:
$y_1(x)=e^x\\
y_2(x)=e^{-x}$
with Wronskian $W_{y_1,y_2}(x)=(e^x)(-e^{-x})-(e^x)(e^{-x})=-1$
Obtain: $K(x,t)=-\frac{1}{2}(e^te^{-x}-e^{-t}e^x)$
Consequently: $y_p(x)=\frac{1}{2} \int^x_{x_0} (e^{x-t}-e^{t-x} F(t)dt$
The general solution to the given differential equation can therefore be expressed as
$y(x)=c_1e^x+c_2e^{-x}+\frac{1}{2} \int^x_{x_0} (e^{x-t}-e^{t-x} F(t)dt$
