# Find the solution given the initial value problem

• Apr 23rd 2007, 05:31 PM
pakman
Find the solution given the initial value problem
Also determine its behavior as t increases.

6y'' - 5y' + y = 0 y(0) = 4 y'(0) = 0

I found the general solution to be y = C1*e^(t/2) + C2*e^(t/3)

C1 + C2 = 4
(1/2)*C1 + (1/3)*C2 = 0

I substituted in C1 and solved that way getting the 4 for C1 and 3 for C2. Obviously I did something wrong here, any ideas? Thanks!
• Apr 23rd 2007, 05:44 PM
Jhevon
Quote:

Originally Posted by pakman
Also determine its behavior as t increases.

6y'' - 5y' + y = 0 y(0) = 4 y'(0) = 0

I found the general solution to be y = C1*e^(t/2) + C2*e^(t/3)

C1 + C2 = 4
(1/2)*C1 + (1/3)*C2 = 0

I substituted in C1 and solved that way getting the 4 for C1 and 3 for C2. Obviously I did something wrong here, any ideas? Thanks!

Let y = e^mt
Then the characteristic equation is
6m^2 - 5m + 1 = 0
=> (3m - 1)(2m - 1) = 0
=> m = 1/3 or m = ½

So y = Ae^(t/3) + Be^(t/2)
=> y’ = (1/3)Ae^(t/3) + (1/2)Be^(t/2)

y(0) = 4
=> y(0) = 4 = Ae^0 + Be^0 = A + B
So A + B = 4 …………………(1)

y’(0) = 0
=> y’(0) = 0 = (1/3)Ae^0 + (1/2)Be^0 = (1/2)A + (1/2)B
So (1/3)A + (1/2)B = 0 ………...(2)

So we have the system:
A + B = 4 ………………..…(1)
(1/3)A + (1/2)B = 0 ………...(2)

A + B = 4 ………………..…(1)
A + (3/2)B = 0 ……………...(3) = (2) * 3

=> (1/2)B = -4
=> B = -8

But A + B = 4
=> A – 8 = 4
=> A = 12

So y = 12e^(t/3) – 8e^(t/2)
• Apr 23rd 2007, 06:01 PM
pakman
Thanks! If I used the quadratic to find the characteristic, how would I know which one is A or B?
• Apr 23rd 2007, 06:08 PM
Jhevon
Quote:

Originally Posted by pakman
Thanks! If I used the quadratic to find the characteristic, how would I know which one is A or B?

we decided at the beginnig. i used A to be the coefficient of e^(t/3) and B to be the coefficient of e^(t/2), so you choose where you want which constant from the beginning, they are just arbitrary constants