Science and engineering math: non-homogeneous differential equation

[chatterbug219]

Homework Statement

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

Homework Equations

There is a Laplace transform table attached if needed :)

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

 

[tiny-tim]

hi chatterbug219!

just solve x⁴ - 16 = 0

 

[chatterbug219]

(x²-4)(x²+4)=0
x=2

But how do I use that to solve the differential?

 

[tiny-tim]

(x²-4)(x²+4)=0
x=2

there are four roots

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

There is a Laplace transform table attached if needed :)

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

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

 

[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?

 

[Ray Vickson]

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?

Try it and see!

RGV

 

[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

 

[Ray Vickson]

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

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

RGV

 

[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)]

 

[Ray Vickson]

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)]

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

RGV

 

[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