Second-order IVP

1. Sep 18, 2009

jrsweet

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

Find y as a function of t if 36y''-132y'+121y=0, y(0)=5, y'(0)=4

3. The attempt at a solution

36y''-132y'+121y=0
36r^2-132r+121=0
(6r-11)^2
So, general solution

y(x) = C1*e^(11x/6)+C2*x*e^(11x/6)
y'(x)=(11/6)*C1*e^(11x/6)+C2*e^(11x/6)*((6x-11)-(36/121))

y(0)= C1=5
y'(0)= (11/6)*5+(-36/121)*C2=4
(-36/121)*C2=(-31/6)
C2=(3751/216)

So,
y(x)=5e^((11/6)t)+(3751/216)t*e^((11/6)t)

This is for an online homework and it's wrong, but I can't figure out where I went wrong. Can someone help me out?

2. Sep 18, 2009

Office_Shredder

Staff Emeritus
$$y'(x) = \frac{11}{6} C_1 e^{\frac{11x}{6}} + C_2 e^{\frac{11x}{6}} + C_2 x (\frac{11}{6} e^{ \frac{11x}{6}})$$

using the product rule on the second term I get something different than what you have... how did you get (6x-11) - 36/121?

3. Sep 18, 2009

Dick

The derivative y'(x) is wrong. I have no idea how you got what you did. To do it correctly just use the product rule on the C2*x*e^(11x/6) part.

4. Sep 18, 2009

jrsweet

Hmmm... I integrated it instead of derivating...haha. Well I feel stupid now...

Thanks for the help!!