Question

Find the solution of y"-y=-20delta(t-3); y(0)=1, y'(0)=0.

Answer 1

homogeneous part solve as usual, nonhomogeneous part using laplace inverse, formula#17 said that it is = e^(-3s)

Answer 2

What about the -20?

Answer 3

time it then, that's it.

Answer 4

it is a constant, right?

Answer 5

It should be, so it's -20e^-3s?

Answer 6

yes,

Answer 7

So I got Y(s)=(-20e^(-3s)+s)/(s^2-1).

Answer 8

I don't know. @SithsAndGiggles check her work, please.

Answer 9

Sorry for the delay.

\[y''-y=20~\delta(t-3)\]
Applying the transform to both sides yields
\[\left(s^2Y(s)-sy(0)-y'(0)\right)-Y(s)=20e^{-3s}\\
s^2Y(s)-s-0-Y(s)=20e^{-3s}\\
\left(s^2-1\right)Y(s)=20e^{-3s}+s\\
Y(s)=\frac{20e^{-3s}}{s^2-1}+\frac{s}{s^2-1}\]
So you're right so far.

Answer 10

The first term is of the form \(e^{-as}Y(s)\), so its inverse transform would be \(u(t-c)f(t-c)\).

The second term has its own formula. Hyperbolic trig function, if I recall correctly.

Answer 11

But how did (-20e^(-3s)+s)/(s^2-1) became (20e^(-3s))/(s^2-1)+s/(s^2-1)?

Answer 12

the first term is u_c(t) f(t-c) ,not u (t-c)

Answer 13

I mean it is u _3 (t)

Answer 14

How did -20 became 20?

Answer 15

@Loser66, I'm saying \(u_c(t)=u(t-c)\).

@Idealist, you're right, that should be a negative. Didn't see it at first.
\[Y(s)=\frac{-20e^{-3s}}{s^2-1}+\frac{s}{s^2-1}\]

Answer 16

So the inverse of -20e^(-3s)/(s^2-1) is -20u3(t)sinh(t-3), right?

Answer 17

Yep

Answer 18

Great!