Question

Expressing a solution to an initial value problem in terms of a convolution integral. . .

Answer 1

y"+4y'+4y = g(t), where y(0) = 3 and y'(0) = 4

Answer 2

Were you able to solve for X(s)=expression for this?

Answer 3

No I left it as a definite integral from zero to t

Answer 4

Lets say that L(f(x) represents our Lagrange operator....

Answer 5

Ok, yes.

Answer 6

Using these 2 facts we can express the left of the equation all in terms of L(y).

Answer 7

Right. So we a working with the Laplace operator then, correct? Isn't L(y') = s*L(y)-y(0)?

Answer 8

L(y'')=s^2*L(y)-s*y(0)-y'(0)
L(y')=s*L(y)-y(0)
oops

Answer 9

yes

Answer 10

Ok, so then this results in\[s^2F(s)-3s-4 + 4(sF(s) - 3) + 4F(s) = G(s)\] Correct?

Answer 11

correct. Now we solve for F(s)

Answer 12

Then, \[F(s) = \frac{ 3s+16+G(s) }{ (s+2 )^2 }\]

Answer 13

perfect. now just separate that so the G(s) is in it's own fraction.

Answer 14

So,\[F(s) = \frac{ 3s+16 }{ (s+2)^2 } + \frac{ G(s) }{ (s+2)^2 }\]

Answer 15

So we find the inverse L transform of \[L ^{-1}(\frac{ 3s+16 }{ (s+2)^{2} })\]

Answer 16

To me, I thought this should equal 3e^(-2t) +16te^(-2t). But it doesn't. Why is that?

Answer 17

Wait.
I think I see it.

Answer 18

We can first do partial fraction decomp

Answer 19

\[\frac{ 3 }{ s+2}+\frac{ 10 }{ (s+2)^{2}}\]

Answer 20

Hmmm yes we can. This is what I was thinking,\[\frac{ 3s+16 }{ (s+2)^2 } = \frac{ 3s+6 + 10 }{ (s+2)^2 } = \frac{ 3 }{ (s+2) } +\frac{ 10 }{ (s+2)^2 }\]

Answer 21

I have to get going I am really sorry.
http://tutorial.math.lamar.edu/Classes/DE/ConvolutionIntegrals.aspx
Look at example #2

Answer 22

Wonderful. I think this was what my problem was. Thank you for your help

Answer 23

I can come back later if someone else doesn't step in, or if example #2 isn't illustrative enough.

Answer 24

Okay, thank you.