# Science and engineering math: non-homogeneous differential equation

1. Mar 18, 2012

### chatterbug219

1. The problem statement, all variables and given/known data

Solve the differential equation
y(iv)(t) - 16y(t) = 30sint
subject to y(0) = 0, y'(0) = 2, y"(∏) = 0, y'"(∏) = -18

2. Relevant equations

There is a Laplace transform table attached if needed :)

3. The attempt at a solution

I tried making it homogeneous and then taking the Laplace. But I don't think that was right because wouldn't the Laplace of y(t) just be the definition of a Laplace ∫ y(t)e-stdt? So that's where I got confused

#### Attached Files:

• ###### Laplace%20Transform%20Info.pdf
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2. Mar 19, 2012

### tiny-tim

hi chatterbug219!

just solve x4 - 16 = 0

3. Mar 19, 2012

### chatterbug219

(x2-4)(x2+4)=0
x=2

But how do I use that to solve the differential?

4. Mar 19, 2012

### tiny-tim

there are four roots

haven't you covered how to solve polynomial differential equations?

5. Mar 19, 2012

### Ray Vickson

If you take the Laplace Transform of the DE, you need y''(0) = a and y'''(0) = b, which you are not given. However, you can solve for the transform Y(s) as a function of the unknown parameters a and b. Then you can get the solution y(t) in terms of a and b; then you can impose the boundary conditions at t = π to finally determine a and b.

RGV

6. Mar 19, 2012

### chatterbug219

So I did that and I got
Y= 30[1/(s2+1)(s4-16)] + (2s2+as+b)/(s4-16)
Do I do partial fractions for this now to get a and b? Then take the inverse Laplace?

7. Mar 20, 2012

### Ray Vickson

Try it and see!

RGV

8. Mar 20, 2012

### chatterbug219

I've been trying to do the partial fractions for both...so far all I have is
1/(s2+1)(s4-16)
where I get
[As+B/(s2+1)] + [(Cs+D)/(s4-16)]
which simplifies to
[As(s4-16)]+[B(s4-16)]+[Cs(s2+1)]+[D(s2+1)]
and D=1 & C=-2/5...but I can't seem to get A or B to cancel so I can solve for them

the second one is (2s2+as+b)/(s4-16)
but I don't know how to do partial fractions for that

Last edited: Mar 20, 2012
9. Mar 20, 2012

### Ray Vickson

Write $(s^4 - 16) = (s^2+4)(s^2-4) = (s-2)(s+2)(s^2+4)$ and go on from there. Alternatively, if you want to avoid quadratic denominators altogether, you could take complex roots and write $(s^4 - 16) = (s-2)(s+2)(s-2i)(s+2i),$ but that is not really necessary, since the inverse Laplace of $1/(s^2 + w^2)$ is tabulated.

RGV

10. Mar 20, 2012

### chatterbug219

So I got the partial fractions for the first one, but not the second one becaue the a and b are throwing me off...I don't know what to do with them in order to get the partial fractions of
[2s2+as+b]/[(s-2)(s+2)(s2+4)]

11. Mar 20, 2012

### Ray Vickson

Figure out $$\frac{1}{(s-2)(s+2)(s^2+4)} = \frac{A}{s-2} + \frac{B}{s+2} + \frac{Cs + D}{s^2+4},$$ then multiply by the numerator.

RGV

12. Mar 20, 2012

### chatterbug219

Okay...
-1/8(s2+4) - 1/32(s+2) + 1/32(s-2)

So do I multiply 2s2+as+b by -1/8(s2+4) - 1/32(s+2) + 1/32(s-2)? Is that you were saying?
But how do I take the inverse Laplace of 2s2+as+b? That's where I'm really confused

Last edited: Mar 21, 2012
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