Question

Hey. Anybody wants to check this out.
Solve the initial value problem.
y'' + 4y' + 13y = 2e^(-t)

Answer 1

Condition is y(0) = 0 and y'(0) = 1

Answer 2

fun such fun.
write out the auxillary equation, while i pull up my trustie calc book..

Answer 3

What I did is:
[s^2 F(s) - s y(0) - y'(0) + 4 [ s F(s) - y(0)] +13 F(s) = 2/(s+1)
(s^2 + 4s + 13) F(s) = 3+s/(s+1)

Answer 4

F(s) = (3+s)/(s+1)(s^2+4s+13), then right?

Answer 5

hold on, refreshing my memory on this stuff, haven't done these in a while..

Answer 6

nvm, I can't remember what to do with the constant on the other side.. hold on, i'll call some higher-ups

Answer 7

term* not constant

Answer 8

Sure sure. Just wanna know, where i am getting wrong.

Answer 9

y = e^(alpha*x) (c_1 cos Beta*x + c_2 sin Beta*x)
that's the formula that i remember when dealing with a complex root.
where alpha = -b/(2a) and beta = sqrt( 4ac-b^2)/(2a)

Answer 10

all I got so far for the auxillary equation is:
r^2 + 4r + 13 = 2e^(-t)

what i'm not sure is that , does the 2e^(-t) become zero?..

Answer 11

Ooops. I think were having different approach. I'm using laplace transformations here. No, 2e^(-t) = 2/(s+1) --> thru laplace transformation

Answer 12

Oh, I'm using the formula for a second-order linear equation..
no wonder, i haven't learned that yet xD

Answer 13

who can help me tho? T.T

Answer 14

So you chose to solve this using Laplace Transform?

Answer 15

sec lemme read the previous posts XD

Answer 16

That's what I know, so far. But you still can teach me the other way.

Answer 17

Mmm ya the Method of Undetermined Coefficients would be kind of nice for this one.
You haven't learned about that yet? :o

Answer 18

Ahhhhh. Already know that. I think that will be lot easier, right?
But we are required to solve this using laplace transforms

Answer 19

Oh ok ok fair enough.
I'm a little rusty on taking inverse Laplace, but we'll see what we can do.

Answer 20

See my above post. :)

Answer 21

This step looks good so far.
(s^2 + 4s + 13) F(s) = 3+s/(s+1)

And this, ok ok good.
F(s) = (3+s)/[(s+1)(s^2+4s+13)]

So now we need to try and uhhhhhh...
Mmm.. complete the square on the s maybe?
And then Partial Fraction Decomposition afterwards.

Answer 22

That's somewhat confuse me.
Then it should be F(s) = (3+s)/[(s+1)((s+2)^2+9)], right?

Answer 23

Mmmmmmmm ya that looks right.
I'm trying to think if that's the approach we want.

Answer 24

Ya I think we're on the right track.

Remember how to do Partial Fraction Decomp?

Answer 25

I've done that part, but I think it's getting messy. Well let's try again.
(3+s)/[(s+1)(s^2+4s+13)] = A/(s+1) + (Cs+D)/(s^2+4s+13)

Answer 26

You didn't use a B, which is kinda funny lol

But setup looks good so far.

Answer 27

Umm I would always try to stick in easy numbers before expanding it all out like that D:

Remember how one of our terms had an (s+1) ?
We can substitute s=-1 to easily solve for A.

Hmm you should be getting 1/5's I think.. Not 1/4... Ahhhhhhh

Answer 28

http://www.twiddla.com/1474743

Answer 29

Darn I labeled them A B and C -_-
Hopefully that's not confusing.

Answer 30

What's B, then?

Answer 31

B = -1/5 right?

Answer 32

Ya I think so.

Answer 33

You sure with that?

Answer 34

But there's no wrong with my solution either.

Answer 35

There probably is.. I don't have the attention span this late to check it though :'c

I was able to check the expansion on wolfram, they seem to be coming up with the 1/5's also.

Answer 36

`s: 1 = 4A +D`

I think your s's should have given you:

1 = 4A + C + D

Answer 37

3+s = A (s^2+4s+13) + Cs^2 + `C` + Ds +D



3+s = A (s^2+4s+13) + Cs^2 + `Cs` + Ds +D

Answer 38

Yeah The hell

Answer 39

3+s = A (s^2+4s+13) + Cs^2 + Cs + Ds +D
Comparing coefficients:
s^2: 0 = A+C A = -C
s: 1 = 4A +C + D
k: 3 = 13A + D

eqn 2 to 3, sub eqn 1 to 3
1 = -3C + D
3 = -13C +D

-2 = 10C
C = -(1/5)
A = 1/5
D = 2/5

Answer 40

Is that right now?

Answer 41

A Mathematica solution is attached.

Answer 42

It seems like you're on the right track.
I'm thinking the coefficient on sine might be off... Mmmmmm

Answer 43

Then it will now be:
f(t) = 1/5 [ 1/s+1 - s-2/[(s+2)^+9]
f(t) = 1/15 [3e^(-t) - e^(-2t) (3cos3t - 2sin3t)]
This is my final answer.

Answer 44

Sorry but i don't have papers here. I'm just doing mental math :/

Answer 45

Ok fair enough XD Job well done!

Answer 46

Do we have same answers?