# Thread: second order initial value problem

1. ## second order initial value problem

to solve this:
y = (-1/20)*e^(-x)*cos(4*x) - (1/40)*e^(-x)*sin(4*x) + A*e^x + B*e^(-x)
using
y(0) = 1, y'(0) = 2

i got:
y = (-1/20)*e^(-x)*cos(4*x) - (1/40)*e^(-x)*sin(4*x) + 0+ 1

which is not right. Whats wrong?

2. Originally Posted by taurus
to solve this:
y = (-1/20)*e^(-x)*cos(4*x) - (1/40)*e^(-x)*sin(4*x) + A*e^x + B*e^(-x)
using
y(0) = 1, y'(0) = 2

i got:
y = (-1/20)*e^(-x)*cos(4*x) - (1/40)*e^(-x)*sin(4*x) + 0+ 1

which is not right. Whats wrong?
Are you trying to solve for the constants A and B? It might help if you give the DE.

3. yes for A and B
so i did:

- for y(0) = 1 i get
(-1/20) + A + B = 1 => A + B = (1/20)

for y'(0) =2 i get
-3Ae^(-3x) -3Be^(-3x) + x^2 -3e^(-3x)
=> -3A - 3B = 5

then i got a = 0 and b = 1
?

4. 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
?
I would suggest that greater care in doing the calculations rather than explicit help is what's need here.