# Homogeneous equation (third order)

1. Oct 17, 2009

### ihumayun

1. The problem statement, all variables and given/known data

Find y as a function of x if

y'''−11y''+28y'=0 y(0)=1 y'(0)=7 y''(0)=2

I have one attempt left on this question. Could someone verify my answer for me?

2. Relevant equations

3. The attempt at a solution
(use t as lamda)
t^3-11t^2+28t=0
t(t-4)(t-7)=0
t= 0, 4, 7

y = C1 e^(4x) + C2 e^(7x) + C3

1 = C1 (1) + C2 (1) + C3 ...(1)

y' = 4 C1 e^(4x) + 7 C2 e^(7x)

7 = 4 C1 + 7 C2 ... (2)

y'' = 16 C1 e^(4x) + 49 C2 e^(7x)

2 = 16 C1 + 49 C2 ...(3)

Using (2) and (3) to solve for C1 and C2:

28 = 16 C1 + 28 C2 --> (2)*2
2 = 16 C1 + 49 C2 ---> (3)
------------------------------
26 = -21 C2
C2 = (26/-21)

2 = 16 C1 + 49 (-26/21) ... (3)
C1 = 47/12

1 = -26/21 + 47/12 + C3

C3 = 1+ 26/21 - 47/12
C3 = 121/84

y = (47/12)e^(4x) - (26/21) e^(7x) +121/84

2. Oct 17, 2009

### Staff: Mentor

You've done all the hard work. Checking your solution is easy.
See if your function satisfies the differential equation. Check that for your function, y''' - 11y'' + 28y' = 0 is a true statement.
See if your function satisfies the initial conditions.

3. Oct 17, 2009

### ihumayun

I've submitted the answer and it is incorrect. Can anyone tell me why?

4. Oct 18, 2009

### Staff: Mentor

Did you check your solution as I suggested in my previous post? I haven't worked the problem, so can't vouch for your solution.