Question

Use the Laplace transform to solve y"-2y'+2y=cos(t); y(0)=1, y'(0)=0.

Answer 1

\[y''-2y'+2y=\cos t\]
Applying the transform on both sides, you have
\[\left(s^2Y(s)-sy(0)-y'(0)\right)-2\left(sY(s)-y(0)\right)+2Y(s)=\frac{s}{s^2+1}\\
Y(s)\left(s^2-2s+2\right)-s-0+2(1)=\frac{s}{s^2+1}\\
Y(s)\left(s^2-2s+2\right)=\frac{s}{s^2+1}+s-2\\
Y(s)\left(s^2-2s+2\right)=\frac{s+(s-2)(s^2+1)}{s^2+1}\\
Y(s)=\frac{s+s^3-2s^2+s-2}{\left(s^2+1\right)\left(s^2-2s+2\right)}\\
Y(s)=\frac{s^3-2s^2+2s-2}{\left(s^2+1\right)\left(s^2-2s+2\right)}\]
Is this what you have so far? Just checking to see if I made any errors...

Anyway, partial fraction decomposition should be the next step.

Answer 2

How did you get the laplace transform for cos(t)? Is that also on the laplace transform table?

Answer 3

It should be, yes. \[\mathcal{L}\left\{\cos at\right\}=\frac{s}{s^2+a^2}\]
You could derive it from the definition of the transform.

Answer 4

Also, do you suppose to factor out the numerator for Y(s)?

Answer 5

Yes, that was the step between line one and two.

Answer 6

What's the fastest way to factor out the long polynomial?

Answer 7

Are you asking whether you can factor the numerator?

Answer 8

Could you factor by grouping?

Answer 9

No, I don't think so. You don't have to, PDF will work regardless.

Answer 10

partial fraction decomp @Loser66

Answer 11

@Idealist, do you know how to do that?

Answer 12

Not really.

Answer 13

But isn't it (s^3-2s^2+s-2)/((s^2+1)(s^2-2s+2))? What you got is (s^3-2s^2+2s-2)/((s^2+1)(s^2-2s+2)).

Answer 14

\[\left(s^2Y(s)-sy(0)-y'(0)\right)-2\left(sY(s)-y(0)\right)+2Y(s)=\frac{s}{s^2+1}\\
s^2Y(s)-s-2sY(s)+2+2Y(s)=\frac{s}{s^2+1}\\
\left(s^2-2s+2\right)Y(s)=\frac{s}{s^2+1}+s-2\\
\left(s^2-2s+2\right)Y(s)=\frac{s}{s^2+1}+\frac{(s-2)(s^2+1)}{s^2+1}\\
\left(s^2-2s+2\right)Y(s)=\frac{s}{s^2+1}+\frac{s^3-2s^2+s-2}{s^2+1}\\
\left(s^2-2s+2\right)Y(s)=\frac{\color{red}s+s^3-2s^2+\color{red}s-2}{s^2+1}
\]

Answer 15

Never mind. I made a mistake, so sorry. Please continue.

Answer 16

That's okay!
\[Y(s)=\frac{s^3-2s^2+2s-2}{\left(s^2+1\right)\left(s^2-2s+2\right)}=\frac{As+B}{s^2+1}+\frac{Cs+D}{s^2-2s+2}\]
\[s^3-2s^2+2s-2=(As+B)(s^2-2s+2)+(Cs+D)(s^2+1)\\
s^3-2s^2+2s-2=As^3+Bs^2-2As^2-2Bs+2As+2B+Cs^3+Ds^2+Cs+D\\
s^3-2s^2+2s-2=(A+C)s^3+(-2A+B+D)s^2+(2A-2B+C)s+(2B+D)\]
Matching up coefficients yields the system
\[\begin{cases}A+C=1\\-2A+B+D=-2\\2A-2B+C=2\\2B+D=-2 \end{cases}\]

Answer 17

Solving that by hand looks tricky, so I'd suggest the use of a calculator if you can:
http://www.wolframalpha.com/input/?i=a%2Bc%3D1%2C+-2a%2Bb%2Bd%3D-2%2C+2a-2b%2Bc%3D2%2C+2b%2Bd%3D-2

Answer 18

On the test, were you allowed to solve that using a calculator?

Answer 19

On a non-calculator test, you're likely to get a much easier problem where the coefficients are much easier to solve. If you're intent on solving this by hand, I'd suggest using a matrix method, where you would solve the matrix equation
\[\begin{pmatrix}1&0&1&0\\-2&1&0&1\\2&-2&1&0\\0&2&0&1\end{pmatrix}\begin{pmatrix}A\\B\\C\\D\end{pmatrix}=\begin{pmatrix}1\\-2\\2\\-2\end{pmatrix}\]

Answer 20

Okay, so once you get A, B, C, D, how do you write the Laplace?

Answer 21

So you have
\[Y(s)=\frac{1}{5}\left(\frac{s-2}{s^2+1}\right)+\frac{1}{5}\left(\frac{4s-6}{s^2-2s+2}\right)\]
Take the inverse Laplace transform of both sides and you have
\[y(t)=\frac{1}{5}\mathcal{L}^{-1}\left\{\frac{s-2}{s^2+1}\right\}+\frac{1}{5}\mathcal{L}^{-1}\left\{\frac{4s-6}{s^2-2s+2}\right\}\]
Following so far?

I think it would be easier to approach this one term at a time.

Answer 22

Okay, thanks.

Answer 23

HAHAHA!

Answer 24

Do you think you can take care of the inverse transforming? It shouldn't be too difficult if you break it up into multiple fractions.

Answer 25

Yes. I get the idea.

Answer 26

@Loser66, Of course there are plenty of tougher things than this. The only tough part about this problem is the algebra haha

Answer 27

I know, right? I used to be good at algebra...

Answer 28

I have question, what math subject comes after partial differential equations for physics and electrical engineering majors?

Answer 29

I don't know about that, I'm just a math major. I get to have my pick of subjects.

Answer 30

I see.