Yes, that is the correct solution to the "associated homogeneous equation". I presume you know that the general solution to the entire equation is that general solution to the associated homogeneous equation plus a **single solution** to the entire equation.

One method for finding that one solution, called "undetermined coefficients" is to look for a solution of the form \(\displaystyle Pe^{2s}\) (we need to "determine" the coefficient P).

If \(\displaystyle y= Pe^{2s}\) then \(\displaystyle y'= 2Pe^{2s}\) and \(\displaystyle y''= 4Pe^{2s}\) so that \(\displaystyle y''- 4y'+ 5y= 4Pe^{2s}- 4Pe^{2s}+ 5Pe^{2s}= 5Pe^{2s}= e^{2s}\) so that P= 1/5.

The general solution to the differential equation is \(\displaystyle y(s)= e^{2s}(A cos(s)+ B sin(s))+ (1/5)e^{2s}\).