Laplace transforms to solve linear ODE's

In summary: The standard partial fraction expansion is to break the fraction into its component parts and do a partial fraction decomposition on each term.
  • #1
schmiggy
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*edit* can't upload images from phone with app for some reason, finding a computer.

Using the app for the first time so hopefully this works out ok..

I've attached an image with the problem written in blue, and a complete attempt in gray. I have the answer to the question however it is completely different to mine..

Also, relevant equations are in the gray box on the right (the first line for each, the second is with the values subbed in).

I think it's correct up until the line where I first get L[y] on its own. But I could be wrong..
 

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  • #2
If you want help, you should make your post convenient for the helpers to read.

Homework Statement



Solve the following differential equation using Laplace transforms:
$$\ddot{y}-3\dot{y}+2y=4t$$ where ##y(0)=4## and ##\dot{y}(0)=3##.

Homework Equations



\begin{align*}
\mathcal{L}[\ddot{y}] &= s^2L[y]-sy(0)-\dot{y}(0) \\
\mathcal{L}[\dot{y}] &= sL[y]-y(0)
\end{align*}

The Attempt at a Solution



Taking the Laplace transforms of both sides of the equations, we get
$$s^2 L[y] - 4s - 3 - 3(s L[y] - 4) + 2 L[y] = \frac{4}{s^2}$$
$$L[y](s^2-3s+2) - 4s+9 = \frac{4}{s^2}$$
$$L[y] = \frac{4s^3-9s^2+4}{s^2(s-2)(s-1)} = \frac{4s-5}{(s-2)(s-1)} = \frac{A}{s-2} + \frac{B}{s-1}$$ This gives
$$4s-5 = A(s-1)+B(s-2).$$ Let s=1. Then -1 = -B, so B = 1.
Let s=2. Then 3 = A. So
$$L[y] = \frac{3}{s-2} +\frac{1}{s-1}$$ and ##y(t) = 3e^{2t}+e^t##.
You're right. You made an algebra mistake in simplifying L[y] when you got rid of s2 in the denominator. You can't do what you did.
 
  • #3
Hi, thanks for the reply, and sorry about the format, I honestly thought that would be ok having the question and attempt all in one place.

As for the mistake, I had a feeling I couldn't do that, however I don't really know how else I could do it.. my basic arithmetic/algebra skills are really holding me back.

My idea was to try and factorise it so that I could eliminate one of the denominators but I can't see the factor..
 
  • #4
schmiggy said:
Hi, thanks for the reply, and sorry about the format, I honestly thought that would be ok having the question and attempt all in one place.

As for the mistake, I had a feeling I couldn't do that, however I don't really know how else I could do it.. my basic arithmetic/algebra skills are really holding me back.

My idea was to try and factorise it so that I could eliminate one of the denominators but I can't see the factor..

The problem with your original post was that your picture was a) sideways and b) relatively small, making it hard to read your work. vela was just letting you know that that kind of post is not likely to receive replies because it's a fair amount of extra work for the homework helpers to read your work properly before they can help you. Not everyone is willing to retype out your entire problem like vela did. :)

For your problem, one method you could try is to first write your fraction as two terms like so:

$$\frac{4s^3 - 9s^2 + 4}{s^2(s-2)(s-1)} = \frac{4s^3-9s^2}{s^2(s-2)(s-1)} + \frac{4}{s^2(s-2)(s-1)}.$$

You can cancel the s^2 in the first term, but it sticks around in the second term. You can then perform the partial fraction decomposition on each term separately. Does that help?
 
  • #5
schmiggy said:
Hi, thanks for the reply, and sorry about the format, I honestly thought that would be ok having the question and attempt all in one place.
I recognize it's often more convenient for students to simply post an image of their work, but from the helper's perspective, it can be annoying because you often end up having to open the image in a separate window to make it big enough to read and then you have to switch back and forth between windows as you compose your reply. Sometimes I'll skip posts like those because I don't want to be bothered with doing that, and I know some of the other helpers around here do that as well. At the very least, you should at least type the problem in directly so we don't have to open an image just to see what the thread is about.

As for the mistake, I had a feeling I couldn't do that, however I don't really know how else I could do it.. my basic arithmetic/algebra skills are really holding me back.
You're at a point where you really need to have the basics down cold, otherwise you end up wasting a lot of time chasing down avoidable mistakes (as you already seem to recognize). It might not hurt to go back to spend a little time reviewing some of it. What seemed like random rules back when you first learned them might make more sense to you now.

My idea was to try and factorise it so that I could eliminate one of the denominators but I can't see the factor.
The standard partial fraction expansion is
$$\frac{4s^3-9s^2+4}{s^2(s-2)(s-1)} = \frac{A}{s-2} + \frac{B}{s-1} + \frac{Cs+D}{s^2}.$$ When you have a quadratic factor on the bottom, you need a linear term on top instead of just a constant.
 
  • #6
Mute said:
The problem with your original post was that your picture was a) sideways and b) relatively small, making it hard to read your work. vela was just letting you know that that kind of post is not likely to receive replies because it's a fair amount of extra work for the homework helpers to read your work properly before they can help you.
I agree completely. We homework helpers and mentors are all volunteers, helping out because we enjoy doing it, but when it becomes too much work to help someone, some of us just say to heck with it.
vela said:
I recognize it's often more convenient for students to simply post an image of their work, but from the helper's perspective, it can be annoying because you often end up having to open the image in a separate window to make it big enough to read and then you have to switch back and forth between windows as you compose your reply. Sometimes I'll skip posts like those because I don't want to be bothered with doing that, and I know some of the other helpers around here do that as well. At the very least, you should at least type the problem in directly so we don't have to open an image just to see what the thread is about.
Yes, 100%.
Mute said:
You're at a point where you really need to have the basics down cold, otherwise you end up wasting a lot of time chasing down avoidable mistakes (as you already seem to recognize). It might not hurt to go back to spend a little time reviewing some of it. What seemed like random rules back when you first learned them might make more sense to you now.
Good advice.
 

1. What is a Laplace transform?

A Laplace transform is a mathematical tool used to convert a function of time into a function of a complex variable, s. It is commonly used in engineering and physics to solve differential equations.

2. How does a Laplace transform help solve linear ODE's?

A Laplace transform can convert a linear ordinary differential equation (ODE) into an algebraic equation that is easier to solve. This is because the transform changes the equation from a time domain to a frequency domain, making it easier to manipulate and solve.

3. What type of ODE's can be solved using Laplace transforms?

Laplace transforms are most commonly used to solve linear ODE's with constant coefficients. They can also be used for systems of equations and non-constant coefficients, but the process may be more complex.

4. What are the advantages of using Laplace transforms to solve ODE's?

One advantage of using Laplace transforms is that they can solve ODE's with initial conditions, which can be difficult to solve using other methods. They also provide a systematic and efficient approach to solving linear ODE's.

5. Are there any limitations to using Laplace transforms to solve ODE's?

While Laplace transforms are useful for solving many types of ODE's, they may not work for all cases. They require the ODE to have a well-defined initial value, and they may not work for non-linear or time-varying systems.

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