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Inverse Laplace Transforms

Inverse Laplace Transforms (PDF)

Inverse Laplace Transforms

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Lesson Overview 0:09
    • Laplace Transform L{f}
    • Run Partial Fractions
  • Common Laplace Transforms 1:20
  • Example 1 3:24
  • Example 2 9:55
  • Example 3 14:49
  • Example 4 22:03
  • Example 5 33:51

Transcription: Inverse Laplace Transforms

Hi and welcome back to the differential equations lectures here on, my name is will Murray and today we are going to talk about inverse Laplace transforms.0000

The idea of the inverse Laplace transform is that we are given the answer after somebody else is done the inverse Laplace transform and were going to try to reconstruct the original function f(t).0010

We begin in a function in terms of s and we have to figure out which function in terms of t would have given that by taking the Laplace transform.0023

Usually what happens is the function will be given will be a rational expression in terms of s, that would be a polynomial and s divided by another polynomial in s.0035

It will be a rational expression in terms of s and what we want to do is we want to run partial fractions on it in order to separate it out into simpler expressions in terms of s.0059

Once we get into simpler expressions in terms of s, we are going to use this table of common Laplace transforms, these were all of Laplace transforms that we worked out in the previous lecture.0075

If this do not look familiar what you might want to do is go back to the previous lecture here on and you will see a lecture on Laplace transforms.0087

You will see where each one of these terms comes from, we figured out that if you start with one then what you get when you take the Laplace transform is 1/s, t gives you 1/s^2, t^2 gives you 2/s^3.0096

In general, the Laplace transform of t^n is n factorial/s^n +1 and then we figured out the Laplace transform of e^at is 1/s - a, te^(at) is 1/s - a^2 and in general t^n e^(at) is n factorial/s - a^n + 1.0110

We figured out the Laplace transform of cos(bt) is s/s^2 + b^2 sin(bt) gives you b/s^2 + b^2 and then you can combine e^(at) cos(bt) and e^(at)sin(bt) to get s - a/s -a^2 + b^2 or just b/s - a^2 + b^2.0137

The idea is you really want to have this chart handy because we are going to run partial fractions on the expressions we are given and then we are going to compare our answers to this chart.0162

We get answers in terms of things like s - a and s^2 + b^2 we get answers that look like that, and then we will come back and look at this chart, and we will say that must have been the Laplace transform of e^(at).0176

That must have been the Laplace transform of cos(bt), that is how this works.0190

Let us try that out on some examples and you will see how it goes, the first example is to find the inverse Laplace transform of 7s +5/s^2 + s -2.0197

The first step here is to run partial fractions on the expression we are given, now partial fractions is something that you really learn about in calculus 2.0213

If you do not remember how to do partial fractions, what you really want to do is go back and look at the calculus 2 lectures, we do have some calculus 2 lectures here on

You will see there is a whole lecture on partial fractions where you can practice this technique a little bit before you start using it for inverse Laplace transforms.0242

What I'm about to do with partial fractions here is mysterious to you or you are rusty and have not done it in a couple years, maybe it is worth going back and watching that other lecture on partial fractions before you come and work through this one.0251

Let us go ahead and now I'm going to assume that you remember partial fractions, we are going to write 7S +5, I'm going to factor that denominator there, that s^2 + S -2.0265

I can factor that into S +2 times S -1 and then remember the way we do partial fractions is we write that as a/S +2 + B over S -1 and then you try to figure out what the a and b should be.0278

The way we do that is you clear the denominator, I'm going to multiply both sides by my common denominator and here my common denominator is S + 2 x S -1.0300

I get a common denominator and I get 7S + 5 is equal to a x S -1+ b x s + 2 and now want to figure out what A and B are and there is kind of a clever trick which is to plug in some good values of s.0313

The clever way to do that is to plug in S=1 and that will make that term dropout so s=1 and if I do that on the left-hand side, I will get 7s +5 that is 12 is equal to a x 0+ b x 3 and that tells me right away the b=4.0335

If I go back and if I plug-in S=-2 then 7 x s will be -14 + 5 is -9 ,we get -9 is equal to a x s -1 will be -3a, a x -3 + b x 0.0362

That is why I picked -2 was to make that b term dropout, I get -3a = -9, a=3, and my partial fraction expansion of 7s + 5s^2 + s -2 is my a was 3, that is 3/s + 2 + my b is 4/s - 1.0389

That is my partial fraction expansion, I have to remember the relevant entries from a chart on inverse Laplace transforms what I remember is that the Laplace transform of e^at is equal to 1/s - a.0422

This is coming from the chart of Laplace transforms, that was not something I figured out on the spot, that was something according from the chart.0443

I'm going to use linearity here, I see 3 x s + 2, f(t) that means my a must be -2, that is 3 e^-2t + 4 my a must be 1 now, 4e^t.0461

That is a function where we took its Laplace transform, you would get this function of s that we started, let me recap how we did that.0487

First thing we did here was we expanded, we factor the denominator into S +2/S -1 and then expanded it out into partial fractions form.0497

If you are rusty on your partial fractions that was really a calculus 2 topic, in the beginning of calculus 2 we learned about integration techniques, check back at your calculus 2 notes.0507

Maybe watch the calculus 2 lecture on partial fractions here on and you will see how we did this, kind of the details of that but we clear denominators here.0516

I got 7s + 5= a x s - 1 + b x s +2 and we wanted to solve for a and b, in order to make certain terms dropout, we can choose good values of s and plug it.0527

Here we plug in S=1, which makes that term go to 0 and we quickly get 12 = 3b, b=4, plug in s= -2 which makes the b term go to 0.0541

We can solve for the a term is equal to three, that tells me what the partial fraction expansion of our original function is and then I go back and I look at my chart of Laplace transforms and I try to find something that matches these expressions here.0554

What I see is the Laplace transform of e^at is 1/s - a , that matches both of these expressions and the first one a= -2 and I got 3e^-2t, second one a=1, I just got 4e^t.0571

That is my inverse Laplace transform of the function that we started with, let us try another one here, we are going to find the inverse Laplace transform of -2s -3 / s^2 + 3s.0589

All hinges on partial fractions, you really want to be solid on your partial fractions before you mess around with inverse Laplace transform, we are going to go -2s -3 and I'm going to factor the denominator again.0604

s x s + 3 and then set up my partial fraction expansion which is a/s + b/s + 3 and I'm going to clear my denominators by multiplying both sides by s x s + 3.0621

On the left I get -2s -3 is equal to a x s + 3 + b x s and now I want to figure out what a and b are.0644

To do that, I'm going to choose good values of s that can make terms dropout selectively, I see that if I plug in s=0 that make the b term dropout and on the left I get -3, on the right I will get a x 3 + b x 0.0660

That tells me right away that a is -1, if I plug in s= -3, I'm going to make the a term dropout, I will get a x 0 + b x -3.0679

On the left if I plug in -3, I will get -2 x s, that is -2 x -3 that is 6 -3 is + 3 and that tells me that my b is also equal to -1 and now that I know what my partial fraction expansion is, I get -2s -3 /s^2 + 3 s= -1/s - 1/s +3. 0696

The chart of Laplace transforms is what I'm going to look at, that was back at the beginning of this lecture.0735

We had a chart of Laplace transforms if you want to have that chart somewhere near you whenever you are looking for inverse Laplace transform.0743

You do not want to figure it out from scratch, you want to look at the chart and see which kinds of functions correspond to which Laplace transforms.0752

In this case, I notice that L(1) is 1/s, that one matches that one, and L(e^at) is 1/s - a, that one matches that one and so my f(t) in order to produce this Laplace transform must have been -1 to get the -1/ S -1 over, not one over but e.0762

Since its s +3 that is like s -3, my a is -3, e^-3t, that is the function that we must have started with if we took the Laplace transform and got this function of s that we are given.0798

Let us see how that works out, again we went partial fractions which means factoring the denominator, setting up the partial fraction form, clearing the denominators by multiplying by the common denominator.0815

We get this expression in terms of s, we are trying to solve for a and b but we can do that by plugging in clever values of s, we plug in s=0, I get a nice, easy equation to solve for a.0829

Plug in s=-3 that makes the a term dropout, we get a nice easy equation to solve for b and we now have this partial fraction expansion of the original function.0841

We look at that and we then we look back at our chart of Laplace transforms, try find something that matches it and I can find something, I see L(1) is 1/s, L(e^at) is 1/s - a.0854

I plug goes back in and I see that in the first one I just got -1, in the second one my a must be -3, I got e^ -3 t, I know that the original function there, the inverse Laplace transform of what we started with is -1 -e^-3t.0868

In the next example, we got find the inverse Laplace transform of s + 4/s^2 + 4s + 5, now I like to run partial fractions on this one which would mean factoring the denominator.0890

The problem is that denominator does not factor, there is no way to break this up using partial fractions instead what I'm going to do is another old algebraic technique and I'm going to complete a square on the denominator.0905

It does not factor if it did factor, I would not be bothering to complete a square, s^2 + 4s + 5= I'm looking at that s^2 + 4, remember how complete a square, you take the coefficient of s and divided by 2 and square it.0923

4/2 is 2 and 2^2 is 4, I write s^2 + 4s + 4 and then I see that in order to make this equal, I had to add another one here.0945

The whole point of that was that gave me a perfect square (s + 2)^2 + 1 and my expression here is now s + 4/(s + 2)^2 + 1.0962

Instead of running partial fractions I completed the square in the denominator and now I'm going to look back at my Laplace transform chart and see if I see anything that matches that.0983

Laplace transform chart and what I notice there is the Laplace transform of e^at cos (bt), according to the chart is s - a/(s - a)^2 +b^2 and L(at) sin(bt) is b/(s - a)^2 + b^2.0996

I got a couple terms here that seem to match what I have for my function pretty well but I see that in order to use this cosine term, I got to have that s - a matching what is in the square there.1043

I do not have that right away here and I have s + 4 not s +2, I'm going to write this as s+ 2/s +2^2 +1 + now since I have s + 4 before, I have to add on another 2/s + 2^2 + 1.1061

I look at that and this first one is exactly tailor-made for cosine transformation with a=-2 and b=1 and this second one is ready for assigned transformation again a= -2 and b=1.1087

There is a slight incongruence there because here we have b and here we have 2, but I will just write that as 2 x 1, we will think of that 2 as being a coefficient, I will put hat on the outside.1109

Our f(t) from that first term we rig that up to be a cosine transformation, our a was -2 so that is e^-2t x cos(t), the cosine of (bt) is 1 and our sine transformation on the other one is e^-2t x sin(t).1125

We had a 2 in the numerator and we are just supposed to have a b there and our b was 1, we put a 2 out here.1152

Just think of that 2 as a coefficient and that is the f(t) whose Laplace transform would give us that function of s that we started with.1158

Let us recap how we figure that out there, first thing we try to do is we are hoping to do partial fractions on this function we started, we are hoping to factor that denominator but it does not factor.1175

Instead of factoring it, we are going to complete the square, here is what I'm doing is completing the square here, I take 4/2 and I square it and I get 4, that is where that 4 comes from.1189

Since I had a 4 on the inside here, I had to have a 1 here to balance out this 5, I write this is as s + 2^2 +1, that is the first thing I do there is create this denominator S +2^2 +1 then I glanced at my Laplace transform chart.1205

You really want to do this all in your head, you probably want to be looking at Laplace transform chart while you are solving these problems and I look for something similar and I see e^at cos(bt) and e^at sin(bt).1224

Give me expressions that are quite similar to what I have but the cosine one, the sin one is just a constant, the cosine one has s - a which matches the s - a in the denominator.1240

I did not have that right here, I split the s + 4 up in the s + 2 and 2, now that S + 2 matches the denominator and it is perfect to set up the cosine transformation where my a is -2, my b is 1.1255

That is where I got this e^-2t cos(t), now the sin transformation, I see b/s - a^2 + b^2, because that b1 has to be 1, which means the two does not quite match.1273

If I write that two was a coefficient that 2 will just come down on the outside and then it is okay to use b=1, I get e^-2t x sin(t).1289

What we found there was a function that if you took its Laplace transform, you would have gotten the original function that we started with, what we found was the inverse Laplace transform of that function of s that we started with.1300

The next example we got find the inverse Laplace transform of 4s^2 -11 s +11/s - 1^3, once again this is kind crying out to run partial fractions on it, let us go ahead and set up the partial fractions here.1323

The denominators are already factored, that is quite convenient, I'm going to set up my partial fractions format for one I have cubed in the denominator, 4s^2 - 11s + 11/s - 1^3 is equal to- remember you go a/s - 1.1346

You can not just go +b/s -1 + c/ s -1, again because that would just combine into a + b + c/s - 1, just a single s - 1, you will never get S -1^3.1372

Here is what you do in partial fractions, you go b/s - 1^2 and c/ S -1^3, again I'm relying on the fact that you remember your partial fractions from calculus 2.1389

If you are little rusty on that, we got a lecture on partial fractions in the calculus 2 series over here on

In the mean time and let us keep moving forward, we are going to clear our denominators here, that means multiplying everything by S -1^3.1410

That means on the left we get 4s^2 -11s +11= a now we got an s -1 in the denominator and s -1^3 in the numerator.1422

Lined up are just a x s -1^2 + b x s - 1 because to of the powers cancel, just +c as a constant, that is very nice there, we want to solve for a, b and c.1435

This is going to be a little bit messy but if we do something clever, we can plug in s=1, if we plug in s - 1 right away, on the left hand side we will get 4 -11+ 11, that is 4 = a on the right hand side a x 0 + b x 0 + c.1451

That will tell us very quickly that c=4, that is nice because it will mean that later on we will not be solving for three equations and three unknowns, we just are a and b now, we know what the c is.1478

Let us keep going with expanding that right hand side, we get 4s^2 -11s + 11 is equal to a x s -1^2 is s^2 - 2s + 1 + b x s - 1.1492

Now +c, I already figured out that is 4, this is (as)^2 - 2(as) + a + (bs) - b + 4 and if I look at this and start preparing coefficients from the s^2 term of both sides, I see that I get 4 is equal to a.1510

That is very nice I know immediately that a=4, Let me go ahead and use the s term, the s term on the left hand side will give me -11, on the right hand side will give me, I have s terms here and here.1538

-2(as) + b, I already figured out that a was equal to 4, -11 is equal to -2 x 4 is 8 + b, if I move that a over to the other side, I get b is equal to -11 + a is -3.1556

I figured out my a, b, and c here, that means I'm done with my partial fractions expansion and let me just summarize what we figured out just from our partial fractions from our algebra here.1585

We got for 4s^2 -11s +11/ s -1^3 is equal to a/S -1, there is my a right there is a = 4, 4/s -1 + b/s -1^2, I see that b is -3, I change that plus to minus -3 /s -1^2 and +c /s -1^3, my c I figured out was 4.1601

4/S -1^3 and I'm going to use my Laplace transform chart, let me copy a couple of entries from the Laplace transform chart which I think will be useful here.1642

This chart I posted it in one of the slides at the very beginning of this lectures so you can rewind and check that out if you do not remember where this coming from but I'm looking for terms that look like these ones that I have in my expression for S here.1671

What I see is that L(e^at) is 1/s - a, that is certainly will be useful L(te^at) is 1/s - a^2 and in general L(t^n) x e^at is n factorial over s - a^ n +1.1687

I see down here that I'm going to have s - a^3, that means I'm going to be dealing with L(t^2/e^at), let me go ahead and plug in an n = 2 here.1734

2 factorial is just 2/s -a^3 because it is n + 1, I'm going to compare these terms with the Laplace transform that I have been given, 4/ s -1, I'm looking up here at 1/s - a, this is 4 looks like a=1 on all of these.1749

This is 4e^t and then -3 x te^t because I'm working up here at 1/s - a^2 and now for the 4 x s -1^3, I see that up here I have 2/s - a^3, I going to write this as 2 x 2/ s -1 cubed.1781

This is + 2 x t^2 e^t, my a is one all those and that is the function for which if I took the Laplace transform I will get this rational expression that we started with, I will give you a quick recap of how we did that.1810

We started out with this nasty rational expression and we ran partial fractions on it, the partial fractions expansion when you have S -1^3 in the denominator this is set as s - 1, S -1^2 and S -1^3 .1834

Then we are clearing the denominators by multiplying both sides by S -1^3 here and that gave us 4s^2 -11s +11 is equal to a x s -1^2, that is canceling a couple powers of s -1 to get b x s - 1.1851

And cancelling all the s -1 to get the c there and then I noticed that if I plug-in a good value of s=1, I can make the a term and the b term dropout quickly figure out that c=4.1873

It gets a little harder after that, I expanded out my right hand side S -1^2 is S^2 -2s + 1 and then I distributed the a through all the terms here and I sorted out into powers of s.1886

I sorted out the s^2 term and the in the S terms, the s^2 is just a 4 here and an a over here ,I get 4=a, the S terms I see -11 over on the left and a -2a and a b on the right.1901

That is why I get the -2 a and b, I already figured out that a was 4, I plug that in, get 8 there and then I find out that the b is -3, that means I figured out my a was here, my b was here and my c was here.1921

I plug each one of those back into this generic partial fractions formula, got the a, b, and c now, that is where I got the 4 -3 and 4 there and now I want to look back at my chart of Laplace transforms.1938

And see if I can find something that matches the approximate form that I have and when you go back and look at that chart I see 1/s - a and that is the Laplace transform of e^at, s - a^2 is the Laplace transform of te^at.1955

For higher powers this s - a^3, I figured out there is a generic Laplace transform of t^n t^at, in order to get S - a^3 here I have to take my n=2, I plug in n=2 there and I get 2 factorial is 2.1974

In order to match that 2 in the numerator, I had to factor this 4 into 2 x 2 and that is why instead of 4 here, we just get that 2 became 2t^2 e^at, but a was just 1.1993

That came in as my exponent in the e^at in then here I have 3t^at and 4e^at and putting all those terms together gave me the original function whose Laplace transform was this rational expression that we were given.2011

Let us try one more example here, we are going to find the inverse Laplace transform of 4s^2 + 4s/s +2 x s^2 + 4, our first step here is partial fractions and try to break this up into more manageable pieces.2030

I'm kind of relying on you to remember the partial fractions from calculus 2, got a lecture on partial fractions in the calculus 2 lectures here on, if you are still rusty on that.2052

Remember we want to set our generic partial fractions expansion form S + 2/s^2 + 4 and while we try to factor the denominator but we can not factor it any more than it is already been given to us.2063

We got an s^2 + 4 in the denominator that does not factor anymore, we just got to write this as a/ s + 2 and then we have a term over s^2 + 4 and we are going to have to call that BX + c.2086

Our goal is to find the a, b, and c as the goal of partial fractions and we are going to multiply both sides by our common denominator, in order to get rid of the denominators s + 2 x s^2 + 4.2107

When we do that, on the left hand side we get 4s^2 + 4s, on the right hand side we get a x s^2 + 4 + bx + c, I wrote x of course I was thinking in terms of calculus when we always use x as the variable.2125

In this case it is an s, b x s + c and we got s + 2 and now essentially we want to solve for a, b, and c, it is a little bit messy but I see something quick that I can do and find out one of my coefficients right away.2143

That is if I plug in S = -2, that will make all these terms dropout, I'm going to go ahead and do that, S = -2 and on the left side 4s^2 will be 16, it is 4 x 4 + 4 + s - if s is -2, minus 8 is equal to a x s^2 + 4 is 8.2165

The whole point here is that the (bs) + c terms get multiplied by zero and therefore they dropout and we get a x 8=8.2194

Right away we figure out the a is equal to 1 and that is as far as we can take it in terms of cleverness, now we have to actually expand out that expression on the right and it is a little messy but it is not too bad.2209

4s^2 + 4s, I already figured out that my a is one, I'm going to go ahead and write s^2 + 4, I'm going to expand out this (bs) + c x S + 2, that is (bs)^2 + 2(bs) + (cs) + 2c and I'm going to sort that out into powers of s's.2224

I see I got 1 + b x s^2 and takes care of that term and that term, now my S term are going to be 2b+ c is going to be my s term and my constant terms is going to be for 4 + 2c, this is still equal to 4s^2 + 4s.2254

And from my s^2 terms assuming that 4=1 + b , I think I'm going to skip to the constant terms because that looks a little simpler, constant terms tell me that on the left I have zero as a constants.2282

On the right I have 4 + 2c and that is probably enough that I can solve b and c, I would not worry about the S terms because I do not need it, but I see a 4= 1+ b, then b=3 and if 4 + 2c is equal to zero then my c= -2.2301

That means that I figured out all my coefficients a, b, and c and I can complete my partial fraction expansion 4s^2 + 4s/ s+ 2 x s^2 + 4 is equal to, it was a/s + 2 that is reading up here.2325

My a was one, 1/s +2+ (bs) + c, reading here might be as 3, 3s - 2, c is -2 and I got an s^2 + 4 here, so that is as far as I can go with the partial fractions.2351

I want to look back in my Laplace transform chart and see if I can find any functions that generally match the terms that I got here.2373

I will look at my Laplace transform chart, remember I posted that chart at the beginning of the lectures, we can flip back a few slides and you can see the chart posted back there and let us see what we have here.2385

Something that matches these terms that we have, I see the Laplace transform of e^at is 1/s - a, the Laplace transform I see for cos(bt) is S/s^2 + b^2, I picked that one because it looked like it might match the second term of our partial fraction expansion here. 2407

And Laplace transform of sin(bt) is equal to b/s^2 + b^2 and I'm just going to make a note here I see I got s^2 + 4 which means b is going to be 2. 2440

If b is equal to 2, then this will be equal to 2/s^2 + 4 and our cos(bt) would be just S/s^2 + 4, I think that is going to be enough to figure out my inverse Laplace transforms, this last term here, I'm going to separate it out and write this as 3s/s^2 + 4 -2/s^2 + 4.2462

I think I can write down my inverse Laplace transform f(t), I see 1/S +2 and I figured out the Laplace transform of e^at is 1/s - a, I guess my a there will be -2, this will be e^-2t, now I have 3s/s^2+ 4.2498

That is 3cos(bt), b is 2, cos(2t) -2/s^2 + 4, 2/s^2 + 4 is sin(bt), so minus sin(2t), it is not 2 sin of 2t because this 2, we do not want to think of it as a coefficient, that 2 becomes that 2 right there.2521

We do not have another 2 outside the sine, that is it, we got our inverse Laplace transform of the function that we started with, let me go back over that quickly and remind you each of the key steps there.2549

The first key step is remembering how to do partial fractions, remembering what you learn in calculus 2 about partial fractions, we want to expand this function out into its partial fraction expansion.2568

That is a/ s + 2 and since we can not factor s^2 + 4, its expansion is just (bs) + c, to figure out the a, b, and c we want to clear denominators, that is what I'm doing here is multiplying both sides by the common denominator.2580

That gives us on the right eight a x s^2 + 4, (bs) + c x s + 2 and I notice I'm really trying to solve for three variables but I can get one of them quickly by plugging an s=-2.2599

I noticed that by looking at that, if I plug in s=-2 that term will dropout, s=-2 on the left gives us 16-8, on the right made this one term drop out and it gives us a x 8 , quickly we figured out that a=1.2613

The other coefficients took more work, I took the a=1, I plug it back in up here and we got s^2 + 4 and then I expanded out (bs) + c x s + 2.2630

That is how I got all of these terms , I sorted them out into terms with s^2, terms with s, and terms with the constant and I equated like terms on both sides, on the left I first looked at s^2.2644

On the left I got 4, that 4 right there and on the right I got 1 + b, it told me quickly that b was 3, I skipped over s because it looked a little more complicated, I skipped right to the constants.2663

On the left I got zero, that is coming from that zero there, on the right I got 4+ 2s, that is where that came from and that was equation I can quickly solve for c being -2.2677

I took that a, b, and c and I plug them back into my original guess a, b, and c, that is how I figured out this partial fraction expansion, that is the end of the calculus 2 portion.2691

Then I looked back at my Laplace transform chart and the goal of that is to find some functions that match this expansion, when I looked at my Laplace transform chart, I saw 1/s -a coming from e^at and I also saw s/s^2 + b^2 coming from cos(bt).2702

b/s^2 + b^2 coming from sin(bt) that look pretty hopeful and I noticed since I had a 4 here, I must be using b=2 because it is always s^2 + b^2 and if I plug in b=2, for the sin I will get 2/s^2 + 4.2725

For the cosine I will get s/s^2 + 4 and in order to invoke those I separated this out into the 3s part and the 2, the 3s part gave me 3 cos(2t) and the 2 gave me sin(2t) not 2 sin(2t) because there was already a built in 2 here.2744

That 2, I used it to match that and I did not have another 2 left over as a coefficient, meanwhile this 1/s + 2, I see that I had a =-2 here, that just turns into e^-2t.2768

I put all those parts together and I get the original function of t for which if you took the Laplace transform you would have got this function of s that we started out with.2783

What I really found here was the inverse Laplace transform of that function of s and my answer is a function of t.2794

That is the end of this set of lectures on Laplace transforms and inverse Laplace transforms, in the next lecture here we are going to see how we can use Laplace transforms to solve differential equations and solve the initial value problems.2802

That is the next lecture in this series on differential equations here on My name is Will Murray and thanks for watching.2818