Answer
See below
Work Step by Step
Given: $y''+4y'-12y=F(x)$
Two linearly independent solutions to the associated homogeneous differential equation are:
$y_1(x)=e^{-6x}\\
y_2(x)=e^{2x}$
with Wronskian $W_{y_1,y_2}(x)=(e^{-6x})(e^{2x})-(-6e^{-6x})(e^{2x})=8e^{-4x}$
Obtain: $K(x,t)=\frac{1}{8}(e^{-6t}e^{6x}-e^{2t}e^{-2x})$
Consequently: $y_p(x)=\frac{1}{8} \int^x_{x_0} (\frac{1}{8}(e^{6x-6t}-e^{2t-2x}) F(t)dt$
The general solution to the given differential equation can therefore be expressed as
$y(x)=c_1e^{-6x}+c_2e^{2x}+\frac{1}{8} \int^x_{x_0} (\frac{1}{8}(e^{6x-6t}-e^{2t-2x}) F(t)dt$
