Question

y'' - 4y' + 4y = 3 δ ( t - 1) + δ ( t-2) by solving Laplace. help?

Answer 1

This looks like Satan's work.

Answer 2

lol

Answer 3

I'm not there yet but I heard it is nastayt

Answer 4

i understand the first part but the last one is kinda weird. What section is this on Laplace?

Answer 5

the initial values it is y(0) = 1 and y'(0) = 1

Answer 6

is it one of those subsitutue forms?

Answer 7

it is function delta for Dirac

Answer 8

@ybarrap

Answer 9

The Laplace of the delta functions on the right side of the equal sign is:
$$
\Large{
e^{-2 s}+3 e^{-s}
}
$$

Combining with Laplace on the left:

$$
\Large{

s^2 L_t[y(t)](s))+4 L_t[y(t)](s)\\
\qquad -4 (s (L_t[y(t)](s))-y(0))-s y(0)-y'(0)\\
=s^2Y(s)+4Y(s)-4sY(s)-y(0)-sy(0)-y'(0)\\
=Y(s)(s^2-4s+4)-y(0)(1-s)-y'(0)\\
=Y(s)(s^2-4s+4)-1(1-s)-1\\
=Y(s)(s^2-4s+4)-2+s=e^{-2 s}+3 e^{-s}\\
Y(s)=\cfrac{e^{-2 s}+3 e^{-s}+2-s}{s^2-4s+4}
}
$$
Find the inverse of this to get y(t).

Answer 10

@ybarrap you made mistake at 4L'(y) it is 4SL(y) - 4 , not just -1

Answer 11

exact

Answer 12

the numerator is s -3 ( the last term) not -s +2

Answer 13

however, it's not big deal, next step is most important

Answer 14

you're right, but this is the basic idea. Hopefully it helps :)

Answer 15

@ybarrap can you help me with my question after this?

Answer 16

sure

Answer 17

THANK GAWD!

Answer 18

@ybarrap not finish yet, Socrates' friend.

Answer 19

D:

Answer 20

nooo!!!!! ((cries**

Answer 21

Just saw your work there, quite involved @UsukDoll

Answer 22

but but...I just ughhhhhh I know I have to work backwards but man I'm stuck as the hexagon hills of the eternal realm darkness stuff

Answer 23

The inverse here will also take some work - @Loser66 is an expert at Laplace inverses

Answer 24

She can get the job done quicker than everyone on this site

Answer 25

please save my differential equations soul. x(

Answer 26

please. stop

Answer 27

@marielen123 - Does the work here make sense so far?

Answer 28

No

Answer 29

The Laplace transform part or reducing it after the transform?

Answer 30

reducing it after the transform,

Answer 31

y(s) = [3e^-s + e ^-2s + s - 3] / s^2 -4s + 4

Answer 32

do not think the answer table

Answer 33

how? without looking at the table, how can you find out the inverse of them?

Answer 34

you need the tables...

Answer 35

I could settle for partial fraction or looking table

Answer 36

but I do not see

Answer 37

sadly, I have to say: no way. even you can partial fraction the denominator, you don't have formula for the numerator one. I mean for \(\dfrac{3e^{-3s}}{s^2-4s+4}\) after doing partial, what do you have? you cannot because they don't have the same variables or coefficient

Answer 38

Take the 1st part, the inverse of
y(s) = [3e^-s ] / s^2 -4s + 4
is
$$
\Large{
3 e^{2 (t-1)} (t-1) u(t-1)
}
$$
u(t) is the unit step function. Note that s^2 - 4s + 4 factors to (s-2)^2 in the denominator. Then use rules for time shifting.

You can use tables or wolfram:
http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of++ [3e^-s+]+%2F+%28s^2+-4s+%2B+4+%29

Answer 39

When you partial expand this1st term you get:

$$

\Large {\\
(3 e^{-s} s)/(s-2)^2-(3 e^{-s})/(s-2)-(3 e^{-s})/(s-2)^2
}
$$

Answer 40

I did no understand

Answer 41

Do you know who to use partial fraction expansion? This is critical.

Answer 42

oh I remember that calculus ii stuff

Answer 43

when the bottom is factorable you have something like A B C on the numerator but if the equation on the denominator can't be factorable, you have something like Ax+B

Answer 44

Here is a quick refresher: http://en.wikipedia.org/wiki/Partial_fraction_decomposition

Answer 45

I know A/ (s-2)^2 + B / (s-2) ^2

Answer 46

nooooooooooooooooooooooo you need to split that up