# Particular Solution of $y'' - 3y' - 4y = 3e^{2t}$

I took this example out of the textbook but I am unable to understand one part - after hours staring at it.

For just the particular solution, $$Y(t) = Ae^{2t}$$ $$Y'(t) = 2Ae^{2t}$$ $$Y''(t)=4Ae^{2t}$$

Here is the part where I don't quite understand: $$(4A-6A-4A)e^{2t} = 3e^{2t}$$

I know we are equating the RHS of the differential equation with the particular solution to solve for A, but I'm not sure how did the textbook get $(4A-6A-4A)$ from.

-
Those are the coefficients of $e^{2t}$ in the original differential equation. – Sammy Black Apr 1 '13 at 6:33
By direct substitution into the left-hand side of the DE. – André Nicolas Apr 1 '13 at 6:35