- #1

badman

- 57

- 0

y''+y'-6y=10e^2x-18e^3x-6x-11.

complementary solution:c1e^2x+c2e^-3x

s1={e^2x}

s2={e^3x}

s3=(x,1}

ok since e^2x exists in the complimentary solution, it is therefore a solution, so i multiply it by x to get s'1{xe^2x}, so now i have that and my s2 and s3.

so i end up with this:

yp=Axe^2x+Be^3x+Cx+D

heres where i get confused do i use the product rule for Axe^2x, if so do i end up with these two derivatives: y'p=Axe^2x+2Ae^2x+...etc, y''p=Axe^2x+4Ae^2x+...etc. i really get confused with this product rule thing.