Question

Use the Laplace transform to solve the given initial value problem:

y''-8y'-65y=0

y(0) = 9
y'(0) = -9

Answer 1

So far I was able to get that
\[\text{Y(s)}=\frac{9s+63}{(s-12)(s+5)}\]

After PFD I got\[\text{Y(s)}=\frac{10}{s-13}- \frac{1}{s+5}\]

Where do I go after here? I feel like it's so easy and I'm just being stupid.

Answer 2

That's supposed to be a 13, not 12! X)

Answer 3

You're ready to take the inverse transform

Answer 4

Just write it down. You're almost done. \(10e^{-13t} - \)...

Answer 5

Wait, is it just \[10e^{13t}-e^{-5t}\]

Answer 6

Hey double check once. You should get
\[\text{Y(s)}=\frac{9s\color{red}{-81}}{(s-12)(s+5)}\]

Answer 7

\[\large\rm s^2Y-sy_o-y_o'-8(sY-y_o)-65Y=0\]\[\large\rm s^2Y-9s-(-9)-8(sY-9)-65Y=0\]\[\large\rm Y(s^2-8s-65)-9s+9+72=0\]Ya I'm still getting a different coefficient for the numerator.

Answer 8

...assuming you did the first part correctly and that's what led you to this glorious moment if insight. :-)

Answer 9

Not coefficient, I mean constant >.< yah like ganesh said

Answer 10

Oops.. :( But I get the rest now anyway! Thanks all! and haha @tkhunny X)