laplace - initial value problem with discontinuous inputs?

Question

can someone walk me through this problem? I am having trouble with imaginary numbers in it...

Find laplace transform:

y'' + 9y = r(t), r(t) = 8sin(t) if 0 < t < pi and r(t) = 0 if t > pi, y(0) = 0, y'(0) = 4

Thanks so much!

Answer 1

Note: I have slightly amended my answer based on what the second answerer said. I did overlook something. As it turns out, the final answer does not change.



After taking the transforms:

s^2*Y - 4 + 9Y = 8/(s^2 + 1) for the first phase



For the first phase

Y = (4s^2 + 12)/[(s^2 + 1)(s^2 + 9)]

= (As+B)/(s^2 + 1) + (Cs + D)/(s^2 + 9)

or (As+B)(s^2 + 9) + (Cs + D)(s^2 + 1) = 4s^2 + 12



To find A,B,C,D, let me show you the easiest way

Let s^2 = -1, ie s = i

(Ai + B)*(9-1) + 0 = 12-4

Ai + B = 1 + 0*i

A = 0, B = 1

Now let s^2 = -9, ie s - 3i

(3C*i + D)*(-9+1) = -4*9 + 12

3C*i + D = 3 + 0*i

D = 3, C = 0



Y = 1/(s^2 + 1) + 3/(s^2 + 9)

y(t) = sin(t) + sin(3t)





For the second phase, we now start with t = π

If you wish, you could make the sub τ = t - π

y" + 9y = 0

y(π) = 0 from the phase 1 solution

y'(π) = -4 from the phase 1 solution



The transform for the second phase becomes:

s^2*Y + 4 + 9Y = 0



Y = -4/(s^2 + 9)

y(τ) = -4/3 * sin(3τ)

y(t) = -4/3 * sin(3t - 3π)

= +4/3 * sin(3t)



Final solution:

for 0 < t < π, y(t) = sin(t) + sin(3t)

for t > π, y(t) = 4/3 * sin(3t)



This solution actually makes sense because the system has a natural frequency of 3, so during the second phase only the sin(3t) appears. During the first phase there is a forcing frequency of 1, so both frequencies appear.

Answer 2

that is unusual that the question asks you to apply the Laplace rework to sparkling up an preliminary fee situation after which provides no preliminary situations. although, that's achievable to apply the Laplace rework to acquire the final answer fairly than a definite answer (this is what you seek for in an preliminary fee situation), in spite of the shown fact that different strategies are favourite while fixing for a ordinary answer. purely cope with y(0) and y'(0) as arbitrary constants. Divide the two sides of the "laplace bit" via (s^2+3s+2) to isolate Y(s). Then use partial fractions and tables (or the Mellin inversion necessary would desire to you know a thank you to apply it) to locate the inverse rework of Y(s) as customary. on the grounds that Laplace transformation is a linear operation and y(0) and y'(0) are constants, the inverse rework of y(0)*(s+3)/(s^2+3s+2) is comparable to y(0) situations the inverse rework of (s+3)/(s^2+3s+2), and a similar good judgment holds for y'(0)/(s^2+3s+2). Your answer would be in terms of y(0) and y'(0) a similar way that a ordinary answer of a 2d order linear differential equation is oftentimes solved for in terms of c1 and c2.

Answer 3

Be careful.



With piecewise forcing terms you have to be very careful. In this case things workout OK but that's not always going to be true.



What would the solution be if r(t) = 8sin(t) for 01?