Answer
See below
Work Step by Step
Given: $y''+y'-2y=F(x)$
Two linearly independent solutions to the associated homogeneous differential equation are:
$y_1(x)=e^{-2x}\\
y_2(x)=e^{x}$
2
with Wronskian $W_{y_1,y_2}(x)=(e^{-2x})(e^{x})-(-2e^{-2x})(e^{x})=3e^{-x}$
Obtain: $K(x,t)=\frac{1}{3}(e^{-2t}e^{x}-e^{t}e^{-x})$
Consequently: $y_p(x)=\frac{1}{3} \int^x_{x_0} (e^{2x-2t}-e^{t-x} F(t)dt$
The general solution to the given differential equation can therefore be expressed as
$y(x)=c_1e^{-2x}+c_2e^{x}+\frac{1}{3} \int^x_{x_0} (e^{2x-2t}-e^{t-x} F(t)dt$
