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### Course: Differential equations>Unit 3

Lesson 4: The convolution integral

# Using the convolution theorem to solve an initial value prob

Using the Convolution Theorem to solve an initial value problem. Created by Sal Khan.

## Want to join the conversation?

• I feel like Sal said a whole bunch of stuff about other things he would add, including more context on what Laplace transforms mean and what a convolution represents, but this is the last video in this set, and Youtube has it as being from 2009. Any chance that we'll actually be getting any more videos any time soon?
• Sal mentioned there would be some future videos to solve the convolution equation in the end of this video. However a few years left since this video was uploaded, there have not been any future video in this section. How long do I need to wait for from now?
• I don't know mate... He also said there would be a future video to prove the convolution theorem...
• I didn't understand why do you have to take the convolution of sin(at) * e^-t sint ?
and not just the inverse and its done?
• It is ok if you just take the inverse Laplace of the expression at . That would be the answer. It's probably a complicated expression.
But its inverse Laplace of that IS NOT the product of the Laplace inverses. Its the convolution of the inverses (the theorem Sal showed in the video before), which leads to the integral at the end.
• I tried to enter the answer into a definite integral calculator, however, the calculator wasn't able to compute it. If we were to evaluate this integral, what would it be?
• I made a couple of substitutions and used Wolfram-Alpha. The substitutions were:
tau = x, t = c, and alpha = b. I then entered:
integral((e^(x-c))*sin(c-x)*sin(b*x))
into Wolfram-Alpha, I got a complicated indefinite integral. When I tried:
integral((e^(x-c))*sin(c-x)*sin(b*x)) from 0 to c
it ran out of computing time. I attempted to evaluate the integral by hand, but didn't get very far in reducing it. It would seem that this one would be best relegated to a numerical solution on a computer. IMHO.
• This appears to be the end of Math but somehow I don't feel like I know everything yet. Can anyone recommend a good next step to learn more math?
• what if y'' was 2y'' and the rest are the same .. do we multiply 2 to (Y(s) - ,, -,, ) or we just ignore the 2 .. !
• He lost me at the :20 second mark up until the second mark.
Duh fq did I just watch?
I'll stick to my elementary algebra :/ Unless someone can explain where he got the extra letters from.

p.s I do have a legit question, when he used the "s" symbol, that's the symbol for "sine" right?
• Hey man, what you just watched was Sal solving a second order differential equation (with initial values for y(0) and y'(0)) using the Laplace transform. Preforming the Laplace transform actually takes your original function, which is a function of time ( e.g., f(t) ), and transforms it to a function of s ( e.g. f(s) ). So that means the "s" actually represents that the function is in a new space.
If you truly do only have knowledge up to elementary algebra, I would suggest calculus I and calculus II as your next step (or even pre-cal). The study of differential equations is an extension of concepts from those two classes and should therefore only be approached following their completion. Goodluck in all your future endeavours!
• I know Sal is trying to show viewers how to use the convolution integral, but is it possible to use the guess and check method he explained before in the video titled: "Inverse Laplace Examples".
• I'm not 100% sure but I'm going to say No.
I think the reason he introduces the convolution integral is because that's the only way (or the only way we know at this point) to solve an inverse Laplace transform that's a product of 2 Inverse Laplace functions.

So for example if
L(y) = 1/(s^2 + 1) * s/(s^2 + 1)
then you CAN'T say
y = sin t cost
but you can say
y = sin t * cos t (the convolution)
(1 vote)
• I'm not familiar with the notation at . What is the meaning of s²Y(s)...?
(1 vote)
• `s²Y(s)` is equivalent to `s^2` times the Laplace transform of `y`.

As Sal mentions, Y(s) is the Laplace transform of y (Y as a function of s). It's equivalent to L[y], if you prefer that notation.

I'd assume that if you're this far into the differential equations content, the variable s is familiar by now. If not, it's just the typical variable we use after applying the Laplace transform to a function. I'd go back and check out the earlier videos on Laplace transforms.

Hope that helps.
• It is really not clear to me what is said at this moment, the subtitles does not make much sense to me. Could someone please explain? Thanks

EDIT: Now I think I understand what he wants to say: if we imagine Y as the integral of y then in every step we are decreasing the power of s by one and increasing the order of derivative of y by 1, maybe its clear from the video, but not to me..

I am not sure if the subtitles are correct at this moment