Solve 2nd Order IVP: y(t) with y(0)=5, y'(0)=4

[jrsweet]

Homework Statement

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

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?

 

[Office_Shredder]

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

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

 

[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.

 

[jrsweet]

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

Thanks for the help!