Question

use the laplace transform and partial fractions to solve the differential equation with initial conditions:
y''+2y'+5y=45; y(0)=2, y'(0)=3

Answer 1

Do you recall definitions for the Laplace transform of a function's derivatives? Those will be particularly important for this problem.

Answer 2

yes, I believe I do

Answer 3

Alright, I can type them out here then because we will use them soon:
\( \displaystyle \mathcal{L} (y') = s\mathcal{L}(y) - y(0) \)
\( \displaystyle \mathcal{L}(y") = s^2 \mathcal{L}(y) - s y(0) - y'(0) \)

Answer 4

So, using our initial DE, we can start by taking the Laplace transform of both sides. Because of linearity, this condenses into taking the Laplace transform of the individual terms.
\(y"+ 2y' + 5y = 45 \)
\( \displaystyle \mathcal{L} (y" + 2y' + 5y) = \mathcal{L}(45) \)
\( \displaystyle \mathcal{L}(y") + 2 \mathcal{L}(y') + 5 \mathcal{L}(y) = \mathcal{L}(45) \)
Is that clear thus far?

Answer 5

yep

Answer 6

Great. With that, we want to apply those two identities we specified earlier.
\(\displaystyle \color{#aaaabb}{s^2 \mathcal{L}(y) - s y(0) - y'(0)} + 2 (\color{#aabbaa}{s \mathcal{L}(y) - y(0)}) + 5 \mathcal{L}(y) = \mathcal{L}(45) \)
Note how we now have a common factor of \(\mathcal{L}(y)\) floating around. We just have to use our initial conditions and some manipulating to get it on its own... :)

Answer 7

\(\displaystyle s^2 \mathcal{L}(y) - s y(0) - y'(0) + 2 (s \mathcal{L}(y) - y(0)) + 5 \mathcal{L}(y) = \mathcal{L}(45)\)
In case my colours are a little too bright.

Answer 8

Okay. So with the y and the y', the 2 and the 5 can be more or less ignored and multiplied out later? They don't have anything extra to do with them? and your colors are fine.

Answer 9

\( \displaystyle s^2 \mathcal{L}(y) - s y(0) - y'(0) + 2s \mathcal{L}(y) - 2y(0) + 5\mathcal{L}(y) = \mathcal{L}(45) \)
Basically, they're going to end up as coefficients of a polynomial in s when we factor the \(\mathcal{L}(y)\).

Like this:
\(\displaystyle (s^2 + 2s + 5 ) \mathcal{L}(y) - s y(0) - y'(0) - 2 y(0) = \mathcal{L}(45)\)

Answer 10

so this may be part of where I'm getting confused. What exactly are we solving for? Are we solving for y in terms of s?

Answer 11

It will be \(\mathcal{L}(y)\) in terms of s. Basically, what we will be doing with this is taking the inverse Laplace transform of both sides once it is alone. Using Partial Fraction Decomposition, we can break up the right-side to figure out the basic forms of simple Laplace transforms like 1/s and a/(s^2 + a^2) and such.

Answer 12

\( \displaystyle (s^2 + 2s + 5) \mathcal{L}(y) - s \times 2 - 3 - 2 \times 2 = 45 \mathcal{L}(1) \)
Just inserting the initial values into the equation so we can simplify.
\( \displaystyle (s^2 + 2s + 5) \mathcal{L}(y) - 2s - 7 = 45 \mathcal{L}(1) \)
We use: \( \displaystyle\mathcal{L}(1) = \frac{1}{s}\)
\(\displaystyle (s^2 + 2s + 5) \mathcal{L}(y) - 2s - 7 = \frac{45}{s} \)

Answer 13

Then move everything but the laplace over, right?

Answer 14

Yep. Once we do that, we will have that alone and the right side should be an interesting looking fraction. :)

Answer 15

then take the inverse and we get y, right? Will that be the answer then?

Answer 16

Yes, although certainly the right-hand side will need some breaking down and digesting to take the inverse cleanly. What should happen, basically, is we take the PFD to split it into terms of addition and take the inverse of each part. We should be able to manipulate each part into a form which has a known inverse.

Answer 17

\( \displaystyle (s^2 + 2s + 5) \mathcal{L}(y) - 2s - 7 = \frac{45}{s} \)
\( \displaystyle (s^2 + 2s + 5) \mathcal{L}(y) = \frac{45}{s} + 2s + 7 \)
\( \displaystyle (s^2 + 2s + 5) \mathcal{L}(y) = \frac{45 + 2s^2 + 7s}{s} \)
\( \displaystyle \mathcal{L}(y) = \frac{2s^2 + 7s + 45}{s(s^2 + 2s + 5)} \)
Like that...

Answer 18

okay. That's what I got on that part. do you have any recommendations on how to break it down?

Answer 19

If I recall correctly, it should be a standard partial fraction decomp: We have an irreducible quadratic and a linear term, so it should decompose into something like this: A/s + (Bs + C)/(s^2 + 2s + 5).

Answer 20

With A, B, and C all being constants, correct?

Answer 21

yes. :)

Answer 22

so would it be something like \[\frac{(2s+7)}{s^2+2s+5} +\frac{45}{s}\]

Answer 23

Nope, that does not look correct.
The process of finding the coefficients is pretty much equating the old form and the new form:
\(\displaystyle \frac{2s^2 + 7s + 45}{s(s^2 + 2s + 5)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 2s + 5} \)
and then multiplying out the denominators so you only have polynomials. You can then equate the coefficients of the two polynomials and create a system of linear equations with A, B, and C.

Answer 24

Okay. I remeber doing that in one of my calculus classes a while back, but I honestly can't remember how to do it... :( would you be able to jump my memory?

Answer 25

\(\displaystyle \frac{2s^2 + 7s + 45}{s(s^2 + 2s + 5)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 2s + 5} \)
\(\displaystyle \color{#889988}{s(s^2 + 2s + 5)} \times \frac{2s^2 + 7s + 45}{s(s^2 + 2s + 5)} = \color{#889988}{s(s^2 + 2s + 5)} \times \left(\frac{A}{s} + \frac{Bs + C}{s^2 + 2s + 5} \right)\)

Answer 26

\( 2s^2 + 7s + 45 = A (s^2 + 2s + 5) + (Bs + C)s \)
Like that... and then we can distribute a bit and combine like terms.

Answer 27

\(2s^2 + 7s + 45 = As^2 + 2As + 5A + Bs^2 + Cs \)
\( 2s^2 + 7s + 45 = (A+B)s^2 + (2A+C)s + 5A \)

We have: A+B = 2, 2A+C = 7, 5A = 45

Answer 28

The 5A = 45 is easy to solve now, and then you can substitute it into the other two equations to find the other coefficients. This one is quite forgiving, sometimes they are not as forgiving. :P

Answer 29

making A=9, B=-7, and C=-9, correct? Then we can plug that back in to the (laplace)y= equation to get y, right?

Answer 30

2*9 + C = 7; C = 7-18 = -11
That is the only adjustment, otherwise yes. :)

I just check here to confirm we didn't make any mistakes: http://www.wolframalpha.com/input/?i=partial+fraction+decomposition+%282s%5E2+%2B+7s+%2B+45%29%2F%28s%28s%5E2+%2B+2s+%2B+5%29%29
Wolfram agrees, we are good. :P

So yes, we then plug back our coefficients into our expanded form of the right-side.
\( \displaystyle \mathcal{L}(y) = \frac{9}{s} + \frac{-7s - 11}{s^2 + 2s + 5} \)

Answer 31

whoops. sadly simple mistake on my part. then the inverse laplace of 9/s would be 9, but does the other term need to be simplified some more or is there a known inverse for it?

Answer 32

I just threw that into wolfram alpha quickly, and it gave me a huge mess...

Answer 33

In that term's current state, we need to do some detective work to figure it out. Basically, all of the known Laplace transforms with fractions usually take the shape of (something with s and/or a)/(s^2 + a^2) and then there are shifted forms. What we first need to do is complete the square in the denominator.

Sorry, it was lagging so I had to take my typing to Notepad. :x

Answer 34

\( \displaystyle \frac{-7s - 11}{s^2 + 2s + 5} \)
Factor out the -1 just for simplicity.
\( \displaystyle - \frac{7s + 11}{s^2 + 2s + 5} \)
Now, we complete the square in \(s^2 + 2s + 5\). The square would be \(s^2 + 2s + 1 = (s + 1)^2\). If you are not familiar / not recalling how that works, we just take the linear term, divide by 2 and then square the result for what we add at the end. \(x^2 + 2a x + \left(\frac{2a}{2}\right)^2 = x^2 + 2a x + a^2 = (x + a)^2 \)

Answer 35

but that then isn't in s^2+a^2. how do we get it in to that orm from what we have?

Answer 36

So, now we have:
\( s^2 + 2s + 5 = \color{#557755}{s^2 + 2s + 1} + 5 \color{#557755}{- 1} = (s + 1)^2 + 4\)
\( \displaystyle - \frac{7s + 11}{(s + 1)^2 + 4} \)

It is 'close', we jus have that (s+1) instead of the s^2.

Answer 37

ooooh... I forgot about the four that got factored out... so then we can use the laplace for cos(at) = s/(s^2+a^2)?

Answer 38

Yes, we will use that (and the one for sine, most likely also) along with the shifting theorem.to take the inverse.

Answer 39

What we can do with the numerator, is take it apart into something like 7s + 7 + 4. We would have a 7(s + 1) for the numerator for a cosine transform (with the shift), and then the extra 4 makes a sine transform with another shift.

Answer 40

how would we get the cosine transform from that? wouldn't that give us s+5 in the denominator?

Answer 41

\(\displaystyle \frac{7s + 11}{(s + 1)^2 + 4} \)
\(\displaystyle \frac{7s + 7 + 4}{(s+1)^2 + 4} \)
\(\displaystyle \frac{7s + 7}{(s+1)^2 + 4} + \frac{4}{(s+1)^2 + 4} \)

Can you see how that will work out? Sorry if I am somewhat rushing, although the lag on my end is getting pretty bad. It won't let me type the messages out anymore in the box. ;x

Answer 42

\(\displaystyle \frac{7(s + 1)}{(s+1)^2 + 4} + \frac{2(2)}{(s+1)^2 + 4} \)
This is where we are going with that.

Answer 43

so it would give us 7cos(2t)+2sin(2t), right?

Answer 44

Yes, although we have a small transformation with the (s+1). Are you familiar with how that works?

There is an identity for a Laplace transform of a function with a factor of e^(at) that makes this (s + 1) come about:

\( \displaystyle \mathcal{L}( e^{at} \cos (bt) ) = \frac{s - a}{(s - a)^2 + b^2} \)
\( \displaystyle \mathcal{L}( e^{at} \sin (bt) ) = \frac{a}{(s - a)^2 + b^2} \)

Answer 45

ur case is just where a = -1, as s + 1 = s - (-1).

Answer 46

Yay, inconvenient maintenance times. >.>
Well, now that it has finished up, I would like to recap everything that we have done for this problem.

We started with the initial DE and took the Laplace transform of both sides.
Using the identities for the Laplace transform of derivatives, we broke it down into the Laplace transform of only y.
We applied our initial values to clean up a bit.
Now, we just solved for the Laplace transform of y in terms of s.

\( \displaystyle \mathcal{L}(y) = \frac{2s^2 + 7s + 45}{s(s^2 + 2s + 5)} \)
We took the Partial Fraction decomposition, filled that in.
\( \displaystyle \mathcal{L}(y) = \frac{9}{s} + \frac{-7s - 11}{s^2 + 2s + 5} \)
Now, we put the \(s^2 + 2s + 5\) into a completed square form.
\( \displaystyle \mathcal{L}(y) = \frac{9}{s} - \frac{7s + 11}{(s + 1)^2 + 4} \)
We break the last part into \(7s + 7\)
\( \displaystyle \mathcal{L}(y) = \frac{9}{s} - \frac{7s + 7 + 4}{(s + 1)^2 + 4} \)
\( \displaystyle \mathcal{L}(y) = \frac{9}{s} - \left(\frac{7s + 7}{(s + 1)^2 + 4} + \frac{4}{(s + 1)^2 + 4} \right) \)
Now, we prepare to take the inverse Laplace transform.
\( \displaystyle \mathcal{L}(y) = 9 \times \frac{1}{s} - 7 \times \frac{s + 1}{(s + 1)^2 + 4} - 2 \frac{2}{(s + 1)^2 + 2^2} \)
Take the inverse Laplace transform. We use our three identified forms:
* \( \displaystyle \mathcal{L}(1) = \frac{1}{s} \)
* \( \displaystyle \mathcal{L}(e^{at} \sin bt) = \frac{b}{(s - a)^2 + b^2} \)
* \( \displaystyle \mathcal{L}(e^{at} \cos bt) = \frac{s - a}{(s - a)^2 + b^2} \)

\( \displaystyle \mathcal{L}^{-1} (\mathcal{L}(y)) = \mathcal{L}^{-1} \left(9 \times \frac{1}{s} - 7 \times \frac{s + 1}{(s + 1)^2 + 4} - 2 \frac{2}{(s + 1)^2 + 2^2}\right) \)
\( \displaystyle y = 9 \mathcal{L}^{-1} \left( \frac{1}{s} \right) - 7 \mathcal{L}^{-1} \left( \frac{s + 1}{(s + 1)^2 + 2^2} \right) - 2 \mathcal{L}^{-1} \left( \frac{2}{(s + 1)^2 + 2^2} \right) \)
\( \displaystyle y = 9 \times 1 - 7 e^{-t} \cos 2t - 2 e^{-t} \sin 2t \)

Thus, our final answer at last is: \(y = 9 - 7e^{-t} \cos 2t - 2e^{-t} \sin 2t\).

Answer 47

@AccessDenied okay. Thank you very much. This helped a lot.

Answer 48

You're welcome! :)
Glad to see you made it back after that. :P

Answer 49

Thanks again and I have a couple more if you aren't busy and want to help.

Answer 50

Sure. :)

Answer 51

okay thanks! I'll close this one and post it in one second