Originally Posted by
taurus yes for A and B
so i did:
- for y(0) = 1 i get
(-1/20) + A + B = 1 => A + B = (1/20) Mr F says: A + B = 1 + (1/20) = 21/20.
for y'(0) =2 i get
-3Ae^(-3x) -3Be^(-3x) + x^2 -3e^(-3x)
=> -3A - 3B = 5
Mr F says: I have no idea why you have 3's instead of 4's in the exponentials, and I don't know where the trig bits have gone. I doubt -3 A - 3 B = 5 is correct. I suggest you go back to your expression for y and differentiate it carefully. Then substitute x = 0 and get a seond equation. I've probably slipped up but I get 2 = (-1/20)(-1) - (1/40)(1) + A - B etc. ..
then i got a = 0 and b = 1
?