Question

use Laplace transforms to solve the initial value problem:

y''+x+2y=0
x''+2x+4y=0
x(0)=y(0)=1
x'(0)=y'(0)=0

Answer 1

hello again and thanks again

Answer 2

would you have to start by setting the two equations equal to eachother or solve for one then substitute into the other?

Answer 3

Hmm, I haven't seen one like this before. :P
I am thinking we should first take the Laplace transform of each equation and get it down to the Laplace transforms of x and y (taking care of those derivatives again). We would get something like a two-variable system of linear equations, I believe.

Answer 4

\( y" + x + 2y = 0 \)
\( \mathcal{L} (y" + x + 2y) = 0 \)
\( \mathcal{L}(y") + \mathcal{L}(x) + 2 \mathcal{L}(y) = 0 \)
\( s^2 \mathcal{L}(y) - s y(0) - y'(0) + \mathcal{L}(x) + 2 \mathcal{L}(y) = 0 \)
And similarly for the other equation as well.

Answer 5

\( x" + 2x + 4y = 0 \)
\( \mathcal{L} (x" + 2x + 4y) = 0 \)
\( \mathcal{L}(x") + 2 \mathcal{L}(x) + 4 \mathcal{L}(y) = 0 \)
\( s^2 \mathcal{L}(x) - s x(0) - x'(0) + 2 \mathcal{L}(x) + 4 \mathcal{L}(y) = 0 \)

Answer 6

#1:
\( s^2 \mathcal{L}(y) - s \times 1 - 0 + \mathcal{L}(x) + 2 \mathcal{L}(y) = 0 \)
\( (s^2 + 1) \mathcal{L}(y) + \mathcal{L}(x) - s = 0 \)
#2:
\( s^2 \mathcal{L}(x) - s \times 1 - 0 + 2 \mathcal{L}(x) + 4 \mathcal{L}(y) = 0 \)
\( 4 \mathcal{L}(y) + (s^2 + 2) \mathcal{L}(x) - s = 0 \)

If we just casually treat s as a constant, we just have \(\mathcal{L}(x) \) and \( \mathcal{L}(y) \) as the 'variables' to solve for.

Answer 7

in the first one, isn't it s^2+2?

Answer 8

Oh, yes, sorry.
\( (s^2 + 2) \mathcal{L}(y) + \mathcal{L}(x) - s = 0 \)
\( 4 \mathcal{L}(y) + (s^2 + 2) \mathcal{L}(x) - s = 0 \)

Thank you. I'm still thinking about this method. :P

Answer 9

Multiply by (s^2 + 2) so that we can eliminate the \(\mathcal{L}(x) \)
\( (s^2 + 2) \left( (s^2 + 2) \mathcal{L}(y) + \mathcal{L}(x) - s \right) = 0 \)
\( (s^2 + 2)^2 \mathcal{L}(y) + (s^2 + 2) \mathcal{L}(x) - s (s^2 + 2) = 0 \)


\( (s^2 + 2)^2 \mathcal{L}(y) + (s^2 + 2) \mathcal{L}(x) - s (s^2 + 2) = 0 \)
\( 4 \mathcal{L}(y) + (s^2 + 2) \mathcal{L}(x) - s = 0 \) Subtract the equations to eliminate

\( ((s^2 + 2)^2 - 4) \mathcal{L}(y) - s(s^2 + 2) + s = 0 \)
\( ((s^2 + 2)^2 - 4) \mathcal{L}(y) - s(s^2 + 2 - 1) = 0 \)
\( ((s^2 + 2)^2 - 4) \mathcal{L}(y) - s(s^2 + 1) = 0 \)
\( ((s^2 + 2)^2 - 4) \mathcal{L}(y) = s(s^2 + 1) \)

\( \displaystyle \mathcal{L}(y) = \frac{s(s^2 + 1)}{(s^2 + 2)^2 - 4} \)

Very interesting...

Answer 10

So, that definitely looks like we can break it down a bit because of the denominator being a difference of squares and thus factoring a bit. We can also use that to find \(\mathcal{L}(x) \) if we substitute into one of the equations. Does that make sense?

Answer 11

I think so. It looks like an interesting equation to try to break down. do you know of anything we can use off the top of your head? I'm thinking of a couple but seem to not be able to completely reduce it.

Answer 12

Well, since we'll be using inverse Laplace transforms on it, the technique we used before should be helpful -- Partial fractions. First, though, we need to get the actual factors in the denominator.



\( \displaystyle \mathcal{L}(y) = \frac{s(s^2 + 1)}{(s^2 + 2)^2 - 2^2} \)
\( \displaystyle \mathcal{L}(y) = \frac{s(s^2 + 1)}{(s^2 + 2 + 2)(s^2 + 2 - 2)} \)
\( \displaystyle \mathcal{L}(y) = \frac{s(s^2 + 1)}{(s^2 + 4)s^2} \)
Note that we can cancel a bit now.

\( \displaystyle \mathcal{L}(y) = \frac{s^2 + 1}{(s^2 + 4)s} \)

Answer 13

Page glitch. o.o

I see that if we distribute the denominator to each individual part, we can cancel in s^2/((s^2+4)s) to get an identity for inverse Laplace, and then use partial fractions on the remaining 1/((s^2+ 4)s). Can you see how that will work out?

Answer 14

\( \displaystyle \mathcal{L}(y) = \frac{s^2 + 1}{(s^2 + 4)s} \)
\( \displaystyle \mathcal{L}(y) = \frac{s^2}{(s^2 + 4) s} + \frac{1}{(s^2 + 4) s} \)
\( \displaystyle \mathcal{L}(y) = \frac{s^\cancel{2}}{(s^2 + 4) \cancel{s}} + \frac{1}{(s^2 + 4)s} \)
\( \displaystyle \mathcal{L}(y) = \frac{s}{s^2 + 4} + \frac{1}{(s^2 + 4)s} \)

Like that. :)

Answer 15

so the first one is cos(2t) but what's the second? I can't seem to find a similar one in the table that I have

Answer 16

We need to do a bit more to that one. Notice that we have two factors in the denominator, so now we have to use Partial fractions. Do you recall that method we used earlier on (2s^2 + 7s + 45)/(s(s^2 + 2s + 5)) ? We set it equal to A/s + (Bs + C)/(s^2 + 2s + 5). The same thing will be done here.

\( \displaystyle \frac{1}{(s^2 + 4)s} = \frac{A}{s} + \frac{Bs + C}{s^2 + 4} \)

Answer 17

sorry I'm still writing a few things down. I'll be caught up in a second :)

Answer 18

Alright. :)

Answer 19

From here, it is really the same process as last time. We just have to pretend the left-hand-side has 0s^2 + 0s + 1 when we equate the coefficients.

Answer 20

so A=1/4, B=-1/4, and C=0 is what I got

Answer 21

Yep, that is what I have as well. :)

So, we can then split it up like so:

\( \displaystyle \frac{1}{4} \frac{1}{s} + \frac{1/4\; s + 0}{s^2 + 4} \)
Oh look, we have another single-unit denominator. That looks like a couple of our transforms now. :)

Answer 22

and I forgot a negative there, it should have been there at the 1/4s. :P

Answer 23

\( \displaystyle \mathcal{L}(y) = \frac{s}{s^2 + 4} + \frac{1}{(s^2 + 4)s} \)
\( \displaystyle \mathcal{L}(y) = \frac{s}{s^2 + 4} + \frac{1}{4} \frac{1}{s} - \frac{1}{4} \frac{s}{s^2 + 4} \)
Looks good now. :)

Answer 24

so y=cos(2t)+1/4-1/4cos(2t), right? So how would we go about solving for x?

Answer 25

you may like to combine the cos(2t) and -1/4 cos(2t) in that part into 3/4 cos(2t).

Anyways, we need to take the Laplace transform of y and basically substitute it back into one of the two equations we had:

\( (s^2 + 2) \mathcal{L}(y) + \mathcal{L}(x) - s = 0 \)
\( 4 \mathcal{L}(y) + (s^2 + 2) \mathcal{L}(x) - s = 0 \)

We can use any form of the Laplace transform of y in the equation of choice. I am thinking we should use the second equation.

I feel like we should use
\( \displaystyle \mathcal{L}(y) = \frac{s^2 + 1}{(s^2 + 4)s} \)
because its more compact for substituting.


\( \displaystyle 4 \color{#aaaa55}{\frac{s^2 + 1}{(s^2 + 4)s}} + (s^2 + 2) \mathcal{L}(x) - s = 0 \)

Answer 26

wow... that looks like a mess

Answer 27

Actually, I am not sure if the expanded form might be easier or not... I could check, I suppose.

Answer 28

Actually, if we simplify the expanded form a bit, it should work out better. Sorry to backtrack... >.>
\( \displaystyle \mathcal{L}(y) = \frac{s}{s^2 + 4} + \frac{1}{4} \frac{1}{s} - \frac{1}{4} \frac{s}{s^2 + 4} \)
\( \displaystyle \mathcal{L}(y) = \frac{3}{4} \frac{s}{s^2 + 4} + \frac{1}{4} \frac{1}{s} \)

Because then we have:
\( \displaystyle 4 \left( \frac{3}{4} \frac{s}{s^2 + 4} + \frac{1}{4} \frac{1}{s} \right) + (s^2 + 2) \mathcal{L}(x) - s = 0 \)
\( \displaystyle 3 \frac{s}{s^2 + 4} + \frac{1}{s} + (s^2 + 2) \mathcal{L}(x) - s = 0 \)

I mean, it will work out either way but it would be nice to have the least issues while working. Either way, we got some work to do with dividing off the (s^2 + 2) on the Laplace transform.

Answer 29

\( \displaystyle (s^2 + 2) \mathcal{L}(x) = s - \frac{1}{s} - 3 \frac{s}{s^2 + 4} \)
\( \displaystyle \mathcal{L}(x) = \frac{s}{s^2 + 2} - \frac{1}{s(s^2 + 2)} - 3 \frac{s}{(s^2 + 4)(s^2 + 2)} \)

We pretty much have to use partial fractions on the latter two terms to break those up into useful forms now. The first one is virtually a repeat of the one we did previously, but the second one is just slightly more difficult because there are two irreducible quadratics in the denominator.

Answer 30

what would the very first one be? would we use a square root of 2 for that one?

Answer 31

Yeah. Since it is an s^2 + (sqrt(2))^2, we need that sqrt2.

Answer 32

@AccessDenied A = 1/2, B=-1/2, and C=0? So we have the first and the second. I'm drawing a blank on the third one. Would the top part be as+B and Cs+D?

Answer 33

Yes, the first coefficients are correct. Also, yes, we have (As + B)/... + (Cs+D)/...

Answer 34

I apologize in advance if something happens and I disconnect soon. I seem to be disconnecting from OS every couple minutes. o.O

Answer 35

\( \displaystyle \frac{s}{(s^2+4)(s^2+2)} = \frac{As + B}{s^2+4} + \frac{Cs + D}{s^2+2} \)
The set up

You will have a four-variable system of equations, but luckily again it does not seem to be too difficult to solve. It should turn out to be almost like two two-variable systems.

Answer 36

I got A=-1/2,B=0, C=1/2, and D=0.

Answer 37

Yep, that looks correct to me. :) Now we substitute:
\( \displaystyle \mathcal{L}(x) = \frac{s}{s^2 + 2} - \left( \frac{1}{2} \frac{1}{s} - \frac{1}{2} \frac{s}{s^2+2} \right) - 3 \left( \frac{1}{2} \frac{s}{s^2 + 2} - \frac{1}{2} \frac{s}{s^2 + 4} \right) \)
\( \displaystyle \mathcal{L}(x) = \frac{s}{s^2 + 2} - \frac{1}{2} \frac{1}{s} + \frac{1}{2} \frac{s}{s^2+2} - 3 \frac{1}{2} \frac{s}{s^2 + 2} + 3 \frac{1}{2} \frac{s}{s^2 + 4} \)

Answer 38

I get x=cos(sqrt2 t)-1/2+1/2 cos(sqrt2 t)+3/2 cos(2t)-3/2 cos(sqrt2 t)

Answer 39

Hmm, you might be able to simplify that up quite a bit.

Answer 40

wow... ya you're right. not sure why I didn't...

Answer 41

x=3/2 cos(2t) - 1/2, yes?

Answer 42

Yes! I actually did the simplifications before taking the inverse, which also works out...
\( \displaystyle \mathcal{L}(x) = \frac{s}{s^2 + 2} - \frac{1}{2} \frac{1}{s} + \color{#aa5555}{\frac{1}{2} \frac{s}{s^2+2} - \frac{3}{2} \frac{s}{s^2 + 2}} + 3 \frac{1}{2} \frac{s}{s^2 + 4} \)
1/2 - 3/2 = -1
\( \displaystyle \mathcal{L}(x) = \frac{s}{s^2 + 2} - \frac{1}{2} \frac{1}{s} - \frac{s}{s^2 + 2} + 3 \frac{1}{2} \frac{s}{s^2 + 4} \)
\( \displaystyle \mathcal{L}(x) = \cancel{\frac{s}{s^2 + 2}} - \frac{1}{2} \frac{1}{s} - \cancel{\frac{s}{s^2 + 2}} + 3 \frac{1}{2} \frac{s}{s^2 + 4} \)
Incredible cancelling!
\( \displaystyle \mathcal{L}(x) = -\frac{1}{2} \frac{1}{s} + 3 \frac{1}{2} \frac{s}{s^2 + 4} \)

Answer 43

ya definitely. Thank you very much and do you have some more time to help or do you have something else to do? If you have some more time, I have two more questions, if not, that's fine too.

Answer 44

Well, I might take a break for a while. Are they similar to these last two where you may try those on your own and we can check the work?

Answer 45

the first one has to do with the translation(or shifting)theorem and the second has to do with the differential of transforms theorem. I'm not too familiar with either of them.

Answer 46

Alright. Well, I am probably going to take around a 10 to 20 minute break and then I will check back again. I think at this time there may be more people around who are keen on DE, though, so you could post the question and see if somebody else will help you as well. :)

Answer 47

will do. Maybe I'll see you again. Thanks a lot for the help though :)

Answer 48

You're welcome! Thanks for the interesting problems here. :D

Answer 49

I've got more if I see you again!