# Use Laplace to solve this differential equation

• Northbysouth
In summary: You also need to evaluate the equation at s=0.This makes sense, as without doing so, we wouldn't be able to find A.A is not important because you can evaluate ##x(t) = A e^{\omega t}## at t=0 and determine the value of A. You can't do that for B or C, so you need to find them more carefully.
Northbysouth

## 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?

Northbysouth said:

## 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

Yup, looks right.

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

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?

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

Northbysouth said:
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.

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?

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.

The other terms don't turn into 0. Why do you think they do?

Northbysouth said:
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

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

we're left with

1 = A(s2+w2) + Bs(s2+w2) + cs2 + Dss2
Right. This is a lot different than saying that you get ##1=A(s^2+\omega^2)##. I hope you can see that. (If you're going to say you were just setting s=0 in the other terms, then I'd just ask why didn't you set s=0 in the first term then?)

To get the rest of the constants, you can collect the terms:
$$1 = (B+D)s^3 + (A+C)s^2 + (B\omega^2)s + A\omega^2$$ and then match coefficients on the two sides. You should eventually get B=D=0 and C=-1/ω2.

## 1. What is Laplace transform?

The Laplace transform is a mathematical technique used to solve differential equations. It transforms a function of time into a function of complex frequency, making it easier to solve the differential equation.

## 2. When should I use Laplace transform to solve a differential equation?

Laplace transform is particularly useful for solving differential equations that involve initial conditions, discontinuous forcing functions, or systems with multiple inputs and outputs.

## 3. How do I use Laplace transform to solve a differential equation?

To use Laplace transform, you need to first take the Laplace transform of both sides of the differential equation. Then, manipulate the transformed equation to solve for the desired function. Finally, use the inverse Laplace transform to obtain the solution in the time domain.

## 4. What are the advantages of using Laplace transform to solve differential equations?

Laplace transform simplifies the process of solving differential equations, especially for complex systems. It also allows for the use of algebraic techniques instead of calculus, making it easier to solve the equations.

## 5. Are there any limitations to using Laplace transform to solve differential equations?

While Laplace transform is a powerful tool, it may not be applicable to all types of differential equations. It is also important to note that the solution obtained using Laplace transform may not always be in a closed form, requiring further manipulation to obtain the final answer.

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