Question

Laplace of differential equations

Answer 1

\[L[y"+16y=10\cos4x] y(o)=3 y'(0)=4\]

Answer 2

Is there a specific step you were uncertain of?

Answer 3

Can you walk me through it, because i'm not sure where i am going wrong

Answer 4

Sure. Well, our first step is applying linearity. The Laplace transform of a sum is equal to the sum of the individual Laplace transforms, and constants can permeate the operator:
L[y"] + 16 L[y] = 10 L[ cos(4t) ]
That part is straightforward. So we need to transform the second derivative of y and cos(4t). You know how to do these two?

Answer 5

yep, i'll write them out

Answer 6

\[S ^{2}Y(s)-sy(0)-y'(0)+16Y(s)=10(\frac{ s }{ s ^{2}+16})\]

Answer 7

Yup, that looks good so far. Then we can substitute our initial values y(0) = 3 and y'(0) = 4.
We need to solve for that Y(s) inevitably.

Answer 8

s^2 Y(s) - s * 3 - 4 + 16 Y(s) = 10 *(s / (s^2 + 16)
Get those Y(s) on one side alone:
s^2 Y(s) + 16 Y(s) - 3s - 4 = 10s / (s^2 + 16)
Y(s) * (s^2 + 16) = 10s/ (s^2 + 16) + 3s + 4
and then divide off the s^2 + 16
Y(s) = 10s / (s^2 + 16)^2 + (3s + 4)/(s^2 + 16)
That all looks good so far?

Answer 9

yep

Answer 10

At this point, we're looking at taking the inverse Laplace transform now.
The right-hand side looks interesting with a squared term in the denominator. Do you happen to have a table of Laplace transforms that includes a 2a s / (s^2 + a^2)^2 ?

Answer 11

yes transforms to tsin(st)

Answer 12

Well t sin (at), but yes.
So we need to do just a little manipulation to make sure we have the right a-value of 4 in the numerator (from 4^2 = 16). Constants are unimportant; we can just multiply and divide by something we want and toss the undesired parts to hang on the side.
\( \displaystyle \dfrac{10s}{(s^2 + 16)^2} = 10 \dfrac{8/8 \times s}{(s^2 + 16)^2} = \frac{10}{8}\frac{8s}{(s^2 + 16)^2} \)

Answer 13

So now that has inverse Laplace transform of 10/8 t sin (4t). That makes sense?

Answer 14

yep, it does, this is where i was getting lost

Answer 15

Alright. So I'll rewrite where we are at:
\( \displaystyle y(t) = \dfrac{5}{4} t \sin (4t) + \mathcal{L}^{-1} \left[ \frac{3s + 4}{s^2 + 16} \right] \)
Now we need to split up the second inverse Laplace transform into terms of sine and cosine. You have an idea of how we do this?

Answer 16

separate the second inverse laplace into two fractions

Answer 17

Yup.
\( \displaystyle \dfrac{3s}{s^2 + 16} + \dfrac{4}{s^2 + 16} \)
That's a start.

Answer 18

that willl give us 3sin(4t) +cos(4t)

Answer 19

Hm, I think the left one is cosine and the right one is sine. Otherwise that is the right structure.

Answer 20

yep, juss realised that

Answer 21

thanks a lot wouldn't have know to use 2a s / (s^2 + a^2)^2=tsin(at)

Answer 22

No problem! The square in the denominator tells us it isn't quite right for decomposing (you get the same thing out I believe). So it can be good to refer to the table if you have it handy, or keep those ones in mind. :)

Answer 23

** Well, the square in the denominator and no other outstanding factors

Answer 24

thanks

Answer 25

Glad to help! :)