Quote:

2) (d2y/dx2)-3(dy/dx)+2y=e^5x

Look for particular of form,

http://www.mathhelpforum.com/math-he...8cf2fb9a-1.gif
i have used the particular solution of $\displaystyle y=Cxe^{5x} $

so

$\displaystyle y' = Ce^{5x} + 5Cxe^{5x} $

and $\displaystyle y'' = 5Ce^{5x} + 5Ce^{5x} +25 Cxe^{5x} $

which is $\displaystyle Ce^{5x}(25x+10) $

then $\displaystyle Ce^{5x}(25x+10) $ -3 $\displaystyle Ce^{5x} + 5Cxe^{5x} $ +2Cxe^{5x} = e^{5x}

giving 10c - 3c = 1

and c=1/7

so the general solution is $\displaystyle y= Ae^x + Be^{2x} + 1/7 xe^{5x} $

is this right?