y'' - 2y' +5y = e^X cos(2x)
im just a bit lost on this one...i know the sub will be Ae^x cos 4x + B e^x sin 4x ....
no, we will guess a particular solution of the form:
$\displaystyle y_p = Axe^x \cos 2x + Bx e^x \sin 2x$
since the homogeneous solution is of the form $\displaystyle A e^x \cos 2x + B e^x \sin 2x$ we can't choose that as the particular solution, you'll end up with zero, so we add a factor of x