# Use Laplace to solve this differential equation

## Homework Statement

Obtain the solution of the differential equation

x'' + w2nx = t

My use of L refers to the Laplace

## The Attempt at a Solution

L{x'' + w2nx = t}

I decided to do the Laplace of each part individually starting with x''

L{x''} = sL{x'} - x'(0)

then

L{x'} = sL{x} - x(0)

Putting this together gives

L{x''} = s[sL{x} - x(0)] - x'(0)

Also, the L{x} = x/s

L{x''} = sx - sx(0) - x'(0)

Then the Laplace of w2nx is:

L{w2nx} = w2nx∫inf0e-st dt

because I believe that w2nx is a constant, hence I can move it out of the integral

evaluating the integral then gives me

L{w2nx} = w2nx/s

Then for L{t} I can see from the Laplce tables that this is just 1/s2

Consequently, my Laplace transformation is

L{x'' + w2nx = t} = sx - sx(0) - x'(0) + wn2x/s + 1/s2

After this step I'm pretty sure I need to do the inverse Laplace transform. But I wanted to check if what I had at the moment looked right, and if so, what is the best way to do the inverse of this laplace transformation?

vela
Staff Emeritus
Homework Helper

## Homework Statement

Obtain the solution of the differential equation

x'' + w2nx = t

My use of L refers to the Laplace

## The Attempt at a Solution

L{x'' + w2nx = t}

I decided to do the Laplace of each part individually starting with x''

L{x''} = sL{x'} - x'(0)

then

L{x'} = sL{x} - x(0)

Putting this together gives

L{x''} = s[sL{x} - x(0)] - x'(0)

Also, the L{x} = x/s

L{x''} = sx - sx(0) - x'(0)

Then the Laplace of w2nx is:

L{w2nx} = w2nx∫inf0e-st dt

because I believe that w2nx is a constant, hence I can move it out of the integral

evaluating the integral then gives me

L{w2nx} = w2nx/s

Then for L{t} I can see from the Laplce tables that this is just 1/s2

Consequently, my Laplace transformation is

L{x'' + w2nx = t} = sx - sx(0) - x'(0) + wn2x/s + 1/s2

After this step I'm pretty sure I need to do the inverse Laplace transform. But I wanted to check if what I had at the moment looked right, and if so, what is the best way to do the inverse of this laplace transformation?
You're confusing the function x with the variable t. L[x] = L[x(t)] = X(s) while L[t] = 1/s. X(s), the Laplace transform of x(t), is what you're solving for.

I was thinking it looked a bit odd.

[sL{x''} - x'(0)] + wn2[X(s) - x(0)] = 1/s2

This simplifies to:

s2X(s) - sx(0) - x'(0) + wn2X(s) - wn2x(0) = 1/s2

vela
Staff Emeritus
Homework Helper
Yup, looks right.

So, now I would solve for X(s), but what would I do after this point? Would I need partial fractions?

vela
Staff Emeritus
Homework Helper
After you solve for X(s), you want to invert it to find x(t). Partial fractions will probably be useful to get X(s) into a form where you can invert it using a table.

Okay, I've solved it for X(s) and used the initial conditions x(0)=0 and x'(0)=0

This gives me:

x(s) = 1/(s2(s2+wn2))

Which I can break down into:

s/(s2+ wn2) * 1/s3

From the tables this, I think, gives me:

cos(wnt) * t2/2

How does this look?

vela
Staff Emeritus
Homework Helper
The Laplace transform is a linear operator. That means if you have two transform pairs x(t), X(s) and y(t), Y(s), then L[ax(t)+by(t)] = aL[x(t)]+bL[y(t)] = aX(s) + bY(s). What you can't say is that x(t)y(t) corresponds to X(s)Y(s).

You want to use partial fractions to split your X(s) up into a sum of terms you can invert using the tables.
$$X(s) = \frac{1}{s^2(s^2+\omega^2)} = \frac{A}{s^2} + \frac{B}{s} + \frac{C}{s^2+\omega^2} + \frac{Ds}{s^2+\omega^2}$$

So, as I understand it. I have s=0, 0, -w, w.

s2*1/(s2(s2+w2)) evaluated at s=0

So, then I'd have A = 1/w2

Ray Vickson
Homework Helper
Dearly Missed
Okay, I've solved it for X(s) and used the initial conditions x(0)=0 and x'(0)=0

This gives me:

x(s) = 1/(s2(s2+wn2))

Which I can break down into:

s/(s2+ wn2) * 1/s3

From the tables this, I think, gives me:

cos(wnt) * t2/2

How does this look?

It looks wrong. The L.T. of a convolution is the product of the individual L.Ts, so the inverse of a product would be a convolution, not a product. You could, of course, actually carry out the convolution operation to get the correct answer.

Just use partial fractions; it is easier, and it avoids convolutions.

vela
Staff Emeritus
Homework Helper
Multiply both sides by ##s^2(s^2+\omega^2)##. You can then set s=0 in the resulting equation. All terms except for the one with A will vanish, so you can solve for A. The rest of the coefficients will take a little more work. Note that with ##s=\omega##, you get ##s^2+\omega^2 = 2\omega^2##, not 0, so setting ##s=\omega## doesn't help you.

So, I multiplied both sides by s2(s2+w2) which left me with:

1 = As2+Aw2

Solving for A I get

A = 1/(s2+w2)

Why was s=0?

Could I do something similar to find B, like use s(s2+w2) on both sides?

vela
Staff Emeritus
Homework Helper
I meant multiply this equation
$$\frac{1}{s^2(s^2+\omega^2)} = \frac{A}{s^2} + \frac{B}{s} + \frac{C}{s^2+\omega^2} + \frac{Ds}{s^2+\omega^2}$$ by ##s^2(s^2+\omega^2)##, and then set s=0 to solve for A.

Yes, I did that. You were right. The left side of the equation became 1 whereas everything else but A turned to 0

1 = A(s2+w2)

A = 1/w2

Did we choose w=0 because it as convenient for us, as it left us with only A?

What would be the best method to solve for B?

I had considered multiplying both sides by s(s2+w2) but this results in A being divided by 0 if s=0.

vela
Staff Emeritus
Homework Helper
The other terms don't turn into 0. Why do you think they do?

Ray Vickson
Homework Helper
Dearly Missed
Yes, I did that. You were right. The left side of the equation became 1 whereas everything else but A turned to 0

1 = A(s2+w2)

A = 1/w2

Did we choose w=0 because it as convenient for us, as it left us with only A?

What would be the best method to solve for B?

I had considered multiplying both sides by s(s2+w2) but this results in A being divided by 0 if s=0.

Why are you making such heavy weather of this question? You just need the partial-fraction expansion of
$$\frac{1}{s^2(s^2+\omega^2)}.$$
and surely you must have done lots of those in Calculus 101. If not, just Google 'partial fractions' to see lots of on-line sources showing you how to do it. It really is not that hard; it just takes a bit of practice.

Because after each term is multiplied by s2(s2+w2)

we're left with

1 = A(s2+w2) + Bs(s2+w2) + cs2 + Dss2

Which simplifies to

1 = As2 + Aw2 + Bs3 + Bsw2 + cs2 + Ds3

evaluating at s=0

We have

1 = Aw2

A = 1/w2

vela
Staff Emeritus
Homework Helper
$$1 = (B+D)s^3 + (A+C)s^2 + (B\omega^2)s + A\omega^2$$ and then match coefficients on the two sides. You should eventually get B=D=0 and C=-1/ω2.