WEBVTT mathematics/differential-equations/murray
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Hi, welcome back to www.educator.com, my name is Will Murray and this is the differential equations lectures.
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Today we are going to learn about the Laplace transforms, let us start with the definition, the Laplace transform of a function, so will write the function in terms of t.
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The Laplace transform by definition that is this calc and equal sign means, its definition is the integral from zero to infinity of each of the negative st x f(t)dt.
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Let me emphasize right away here the variable of integration here is t, I will be integrating in terms of t and then we are plugging in t=0 and we will take the limit as t goes to infinity.
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That s never gets anything plugged in to it, so your answer will be a function of s, all the Laplace transforms that we do today will start out with a function of t, we will run the Laplace transform and we will end up with a function of s.
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One really nice property about the Laplace transform is that it is linear which means that if we have a function f(t) and g(t) and constants a and b ,then the Laplace transform of a(f) + b(g).
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You can just do the Laplace transform separately, you can do L(f) and you can do L(g) and then you can just carry along the constants on the outside.
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That is very convenient, that means you can break up a function into its little pieces to the Laplace transform on each one.
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Just put them back together it is very safe to do that, let us try taking some Laplace transforms and see what kinds of things we end up with.
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The first example here is to find the Laplace transform of t⁺n where n is greater than or equal to zero, this is actually several problems in one because we want to find the Laplace transform for each different power of t.
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After we do, we will know how to find the Laplace transform of any polynomial, so let us start by picking low values of n's, we will start out with n=0 here and that just means that the f(t) is equal to t⁰ that is just 1.
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We are going to take the Laplace transform of one, let me just remind you about the formula for the Laplace transform L(f), remember by definition is equal to the integral from zero to infinity of e⁻st x f(t) dt.
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When f(t)=1, L(1) is by definition is the integral from zero to infinity of e⁻st and f(t) is just 1, so it is just dt.
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We have to integrate that remember we are integrating with respect to t the integral of e⁻st is just e⁻st and then we divide by the coefficient of t so that is -1/s and then we integrate that from t=0 to what we take the limit as t goes to infinity.
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Now when t goes to infinity remember this is a negative power of e, this is like saying 1/e to infinity that would just be zero, that is the the term when t goes to infinity, we would just give a 0 because we will have an e in the denominator.
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When t=0 this is minus -1/s +1/s e⁰ which of course is just one and so what we get for Laplace transform is 1/s.
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That is our Laplace transform of 1and now let us do our next power up is t, L(t) again by definition is the integral from zero to infinity of F(t) x e^=st, that is t x e⁻s(t) d(t).
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In order to do that we have to integrate by parts and if you do not remember how to do integration by parts we have another whole series of lectures from the calculus 2 series here on www.educator.com.
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You might want go check out the calculus 2 lectures here on www.educator.com and there is a whole lecture on integration by parts.
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Today I will just use the quick shorthand version of integration by parts where you make a little chart here t and e⁻st.
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The quick shorthand version of integration by parts, write down derivatives on the left, t goes down 1, 0 t goes to derivative of 1 is 0 integrals of e⁻st, that is -1/s e⁻st and +1/s² e⁻st.
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That is the integral of the line above and then we write these diagonal lines and little signs plus and minus on the diagonal lines, this is the shorthand version of integration by parts.
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We multiply along the diagonal lines and we get minus t/s e⁻st -1/s², e⁻st and then we have to evaluate this from t=0 to the limit as t goes to infinity.
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We are going to plug in t going to infinity into each of these terms and what we see was, we will have an infinite term in the denominator there but then we have an infinite term in the numerator with that t.
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And an infinite term in the denominator with e⁻st but a negative exponent on the e beats any polynomial, so this term going to zero beats this term going to infinity.
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If you want you can check this using L'Hopitals rule, that is how you could confirm this but since that is a calculus 1 topic, we are just going to assume it for here.
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The infinity terms give you 0 - 0 and then zero terms give us while because when t=0 so this was t going to infinity and now we will do t as 0, this has t=0, there is a zero there.
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And then +1/s² e⁰ and that all simplifies down to just 1/s² because e⁰ is 1, that means that the Laplace transform of t is 1/s².
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Let us do one more here, this is when n=1, we are figuring out the Laplace transform of t¹, when n=2 we want to find the Laplace transform of t², again by definition that is the integral from zero to infinity.
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t²(⁻st)dt and again we can use integration by parts on that, and integration by parts I'm going to use the shorthand version.
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You can go back and check the calculus 2 lectures are on www.educator.com if you are a little rusty on your integration by parts.
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I'm going to write derivatives first derivative is 2t, derivative of 2t is 0, derivative of 2 is 0, integral of e⁻st is -1/s, e⁻st the integral of that is positive 1/s², e⁻st.
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The integral of that is -1/s² e⁻st and I will going to write my diagonal lines again and put my alternating signs plus minus plus and I'm going to multiply along the diagonal lines.
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I get -t²/s e⁻st - 2/s² e⁻st - 2s³ e ⁻st and again I have to evaluate that from t=0 the limit as t goes to infinity.
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I have e⁻st on every term so even though that is getting multiplied by some positive powers of t having an e⁻st in the denominator will drag all of these terms down to zero when t goes to infinity.
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This is 0-0-0 when t goes to infinity, I'm kind of slurring over what is really L'Hopitals rule so if you are not so sure why these all go to zero.
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Just check them out using L'Hopitals rule which is something you can learn on www.educator.com from the calculus 1 lectures and now if we plug in t=0, first term here has a t² in the denominator so that is 0.
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Second term has a 2t in the denominator so that is zero, third term is 2/s³ x e⁰ by way, I have been putting plus on each of these because t=0 is the lower limits.
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We are actually subtracting these off but then we had these negative signs so we are subtracting a negative which makes it positive and this gives us e⁰ is 1, 2/s³ is our Laplace transform for t².
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We are starting to notice a pattern here, we are going to take one more power of t and then we are going to figure out this pattern for sure.
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Let us try n=3 so the Laplace transform of t³ is the integral from zero to infinity of t³ e⁻st(dt) and we are going to use parts again.
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t³ e⁻st and again I'm going to use my shorthand version of integration by parts so derivative of t³ is 3t², derivative of that is 6t, derivative of that is 6 and derivative of that is 0.
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Integral of e⁻st is -1/s e⁻st, integral of that is 1/s² e⁻st, integral of that is -1s³ e⁻st,the integral of that is 1/s⁴ e⁻st.
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Now were going to write diagonal lines on each term here and there is a plus minus plus minus and so we can write down what our answer is.
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This is the tabular integration method of integration by parts so we get minus t³/s e⁻st -3t²/s² e⁻st - 6t/s³ e⁻st and -6/t⁴ e⁻st and all of this has to be evaluated from t=0 2t going towards infinity.
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Once again, if we take the limit as t goes to infinity of each of these terms , we will have some infinities in the numerator but nothing strong enough to be the e⁻st which gives us infinity in the denominator.
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All the infinity terms will give us zero and the t=0 terms, most of them will be 0 too, because we got t³ and then t² and then a t, but then we have 6/ - that should have been s⁴ not a t⁴ up there.
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We have plus because we are subtracting a negative, 6⁴ e⁻st, let me write that a little more clearly, 6/s⁴ e⁻st and so those are all the terms we get by plugging in t=0.
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Actually when t=0 e⁻st just gives us e⁰, e⁰ is 1 so it just simplifies down to 6/s⁴ and that is the Laplace transform of t³, that is what we started out with.
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I think it is time to start noticing a pattern here what you can notice here is that 6 came from this 6 over here originally and this 6 came from 3× 2.
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What is going to happen when you take higher powers is are going to be multiplying on bigger and bigger numbers and that 6 really comes from 3 factorial/s⁴.
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What we notice here is that the Laplace transform of one was 1/s, that is what we figured out on the previous page.
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Laplace transform of t was 1/s² Laplace transform of t² is 2/s³ we are starting to build up this factorial pattern in the numerator.
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Laplace transform of t³ is 6, 2×3/s⁴ and that is because we are building up that factorial pattern in the numerator so in general our Laplace transform of t⁺n is n factorial.
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We are getting an n factorial and in the denominator, when we had t³, we had s⁴, when we had t², we had s³, each time we are getting s⁺n +1 so that is our general formula for the Laplace transform of t⁺n.
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Let me recap what we did there, basically what we did here was we had several different problems because we are trying to figure out the the Laplace transform of t⁺n for all values of n bigger than zero.
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What we did was we took the n's one at a time, each time we took a value of n plugged in t⁺n here and that meant we took the integral of t⁺n x e⁻st.
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That is coming directly from our original definition of Laplace transform which was the integral of e⁻st x f(t)dt, evaluated from zero to infinity.
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We worked out that integral but in all of these we had to do integration by parts and I did not really show you the details of the integration by parts.
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I use this short hand tabular integration method that we learned back in calculus 2 to do the integration by parts.
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We worked out the integration by parts and then we try to plug in t going to infinity and we figured out that since we had a negative power of e on each of these terms, that gives us an infinity in the denominator.
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That is stronger than any of the infinities in the numerator, that is kind of short hand way of getting around L'Hopitals rule but to check it formally and confirm it you would run L'Hopital's rule that shows you the all the infinity terms go to zero.
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We plug in t=0, most of these terms still go to zero because when t=0 there were 0 in the numerator of most of these terms but this last term did not have a t in the numerator.
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We got 6/s⁴ x e⁰ which is where we got our Laplace transform for t³.
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We did this for each power of t until we noticed a pattern, and the pattern is that we had this factorial these factorial is building up in the numerator, we always had s +1 in the denominator .
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That is our generic formula for the Laplace transform of t⁺n is n factorial/s⁺n +1, in our next example we have to find the Laplace transform of f(t) equal e^(at) assuming that s is bigger than a.
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Let us work that out, again from the definition of Laplace transform, let me mention first that s bigger than a, that tells us that s - a is going to be a positive number and that is going to be useful as we do our integration.
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Let us remember the definition of Laplace transform L(e⁺at) by definition of Laplace transforms, this is using our original definition, is the integral from zero to infinity of e⁻st times whatever your f(t) is.
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In this case, it is e^(at)dt and what I'm going to do is I like to combine these exponents the (st) and the (at) and I'm going to factor out the negative sign.
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We have the integral from zero to infinity of e⁺negative, now this is s - a, because I factored out a negative sign times (t)dt, the key point there is that s - a is positive.
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We figure that out at the beginning using our assumption, if we integrate that, that is e⁺negative - (at)/(-s - a) and we are going to evaluate that from t=0 to the limit as t goes to infinity.
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That means were going be plugging in t going to infinity in here, but remember s minus a is positive, that is what we set up here s minus a is positive.
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e⁺s - a is a negative power of e, when t goes to infinity we got a negative power of e, that means it is going to 0 and when t equal zero, were subtracting the t=0 term and that is negative so it turns into a positive.
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We get e⁰ because we are plugging in t=0/s -a r and it simplifies down to 1/s - a, remember e⁰ is 1, we are done with that one.
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Let me recap what we did, we started out the original definition of Laplace transform, that is the integral from zero to infinity of e⁻st times whatever the f(t) is.
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We took that f(t) and drop that into that integral formula, then I combined the exponents into s - a while negative s - a.
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What I'm really doing is changing that positive a there into a (-a), it turns out to be convenient when we do the integration.
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The integral of e⁻s -at is just the same thing divided by negative s - a and then we plug in t goes to infinity so that is a negative power of e.
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That is why we got zero here, it is e⁺negative infinity which is the same, remember 1/e to the infinity and then t=0 because we had t here that gives us e⁰, that is just 1/s - a.
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On our next example, we are going to final the Laplace transform of f(t)= cos(at), we are going right from the definition of Laplace transforms.
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The Laplace transform of cos(at) just by definition is the integral from zero to infinity of e⁻st times whatever your function is.
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In this case, its cos(at) dt and this integral would be kind of ugly if you encounter this in calculus two class, you would probably use integration by parts twice.
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Integrate by parts twice, alternately you could use a computer algebra system or an online integration tool, you can use that to solve this integral.
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Or you could use a chart of integrals which are probably find inside the back cover of your calculus book there is lots of charts of integrals there and this will definitely be one of them.
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Because it is a little cumbersome I do not want to go through the details here, I'm just going to read the answer off, I got this from a chart of integrals but you could also find it by any of these other methods.
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It is negative s/a² + s², I'm reading the integral of e⁻st x cos(at) tuns out to be -s/s² + a² x e⁻st cos(at) + a/a² + a², e⁻st sin(at).
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What we are supposed to do with this is evaluate it from t=0 to the limit as t goes to infinity.
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We got to plug our limits t going to infinity and t=0 in here, what we notice is that we have an e⁻st, remember that is going to have an e in the denominator, whenever we plug in t=infinity or t approaching infinity this will be 1/e⁺infinity.
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Will this be one to the infinity and so both of those terms go away to zero and when we plug in t=0 the sin term is going to be zero, let us remember that sin(0) is equal to zero and cos(0) is equal to 1.
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The sin term goes to zero the cos term goes to 1, so minus s/a² + s² x e⁰, that is when we plug in t=0 and it is minus all of these because it is the lower limit here.
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All these terms dropout except for this term that e⁰ is 1 and the two negatives cancel so we get s/a² + s² and it is positive because the two negatives cancel each other, that is our Laplace transform of cos(at).
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We get a function of s there should not be any t's left over in these when you are done with taking the Laplace transform so let me recap how we got that.
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We use the straight definition of Laplace transform which says you do the integral of e⁻st times whatever your function is, we dropped in cos(at) now that is the integral that would have been a bit of a headache in calculus 2.
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One way to do it is to do integration by parts twice, that is if you are going to do it by hand.
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Another way do it is just drop it into a computer algebra system that will work out for you or an online integration system which there are several these days.
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You can also use an integral chart if you still got your calculus book just check in the back cover you will see all kinds of charts of integrals and this is will be one of you them.
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e⁻st x cos(at) you might have slightly different variables, it might have a and b, it might have x instead of t but basically you will see that integral.
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And you will see it expanded out into -s/a² + s² e⁻st cos(at) and then a/a² + s² e⁺sc sin(at).
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We try to plug in the limit as t goes to infinity everywhere, but when we do that, these e terms we had negative exponents on the e.
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We get each of the infinity in the denominators all the all those terms turn into zero and dropout.
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It is just a matter plugging in t=0 when that goes into sin we just get zero here so that term drops out.
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We plugged in the cos(0) is 1 and we have to keep that term, we get an e⁰, that gives us 1 and finally we just get s/a² + s², of course it is negative from this negative right here, that came out of doing the integral.
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Since it is the lower limits, we have another negative sign here and so those two negatives gives us a positive.
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We are finally just left with s/a² + s², so example 4 is quite similar to example 3 will do it in kind of a similar fashion, we are going to find the Laplace transform of f(t) is sin(at)
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We are going to use just the straight definition of the Laplace transform L of sin(at) by definition were using the definition of Laplace transform.
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Integral from zero infinity of e⁻st times whatever function you are talking about, that is the sin(at) in this case dt.
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This is kind of a nasty integral, just like in example 3, we would use parts integration by parts twice or we would use a computer algebra system or an online integration system or an integral chart.
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Any of these other methods should give you the answer to this integral, what it should tell you is -a/a² + s², e⁻stx cos(at) + s/a² + a² x e⁻st x sin(at).
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We have to evaluate that from t=0 to the limit as t goes to infinity, let us look at what happens when we plug in t=infinity.
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Here we got a negative exponent to the e, I wil have e⁺infinity, which is the same as 1/e⁺infinity and we have the same thing over here.
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That means when t goes to infinity, we will have these two terms that both go to zero.
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When we plug in t=0, the first term is cos(at), that is cos(0), we will have -a/a² + s², e⁰ x cos(0) + s/a² + s² e⁰ x sin(0).
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That is what we get plugging in t=0, the sin(0) goes to 0, that means this entire term drops out,cos(0) gives you 1, e⁰ gives you 1, and we have -a/s² + s².
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Those two negatives cancel and gives us a positive a/ a² + s².
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We are done with that Laplace transform, let us recap and see how that worked out.
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We started out with the definition of Laplace transform, that means e⁻st times whatever function you have x f(t), we dropped that and integrate that (dt).
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The integration is messy because it got an e in the sin term, if you did that in a calculus 2 class, you would be doing integration by parts twice and then solving around for the original function.
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You could also throw the whole thing into a computer algebra system software package or an online integration tool, there are several of those that you can use.
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Or you can look at the integral chart which is probably located on the inside back cover or maybe front cover of calculus textbook, almost all of them have integral charts.
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What you will see is a formula for e⁻st sin(at) and when you integrate, turns into this complicated expression with e⁻st(c0s) e⁻st (sin).
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We want to evaluate that from t=0 to t going to infinity,the top term going to infinity turns into 1/e⁺infinity for each of these terms, each of those is 1/e⁺infinity, which goes to 0.
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All the infinity terms dropped out to 0,when you plug in t=0 we got cos(0) is 1, e⁰ is 1, you will get -a/a² + s², plug in t=0 on the other side we get sin(0)=0, that whole term drops out.
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We have minus, this is minus because we are subtracting the two limits, t goes to infinity - t goes to 0, that is coming from there.
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This negative came form that negative right here, which just comes out of what the integral chart gives you as the answer to the integral.
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These 2 negatives cancel each other out and we end up with a/a² + s².
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On our final example, we have a very complicated function f(t) is 3(cos) 4t - 2(sin) 5(t) + e⁺2t + 3t² + 7t -2.
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We want to find the Laplace transform for that, if we use the definition of the Laplace transform, by definition we have to do the integral form 0 to infinity of e⁻st x f(t)dt.
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Our f(t) would be this enormous expression, we have to plug all of that into f(t) and we just get this horrific integral and we really do not want to do that.
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There is a much better way, we are going to use linearity and we are going to exploit the fact that we already know the Laplace transform of the various pieces of this function r.
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This linearity of the Laplace transform is going to be extremely useful here.
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We already know what the various pieces of the function do when you take their Laplace transform, L(1) we figured this out, this is back in example 1.
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You might want to go back and check at example 1 if you do not remember this, it was 1/s, L(t) was 1/s².
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L(t²) was 2/s³, I'm looking at the various pieces of this function here and I'm trying to remember what the Laplace transform of each one was.
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These came from example 1, if it is been a while since you worked through example 1, maybe go back and take a peek at that and you will see where this come from.
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L(e⁺at) is 1/s-a, I believe that was example 2, you might want to go back and check that out if that does not look familiar.
00:36:17.000 --> 00:36:48.000
L(cos) of (at) worked that one out in example 3, and that was s/ a² + s² and finally L(sin)(at) was a/a² + s².
00:36:48.000 --> 00:36:58.000
We worked that one out in example 4, we are not really going to do any new math in this example.
00:36:58.000 --> 00:37:04.000
We are just going to exploit all the work we did in all the previous examples and that should be pretty quick.
00:37:04.000 --> 00:37:27.000
I'm going to look at this f(t), I see 3(cos)4t, here is my Laplace transform for cos(at), the Laplace transform of f is from the 3(cos)4t, I'm going to get 3.
00:37:27.000 --> 00:37:56.000
cos(4t) is going to give me s/16 + s², because the a is 4 there, -2 x sin(5t), sin gives me a/a² + s², a is 5 so 5/25 + s² + e⁺2t.
00:37:56.000 --> 00:38:24.000
I can read that one right here, that is plus 1/s - 2 + 3t² + 3, t² gives me 2/s³, 3 x 2/s³ + 7t + 7 x 1/s² - 2 x 1/s.
00:38:24.000 --> 00:39:07.000
The Laplace transform of 1 is 1/s, maybe I can clean that up a little bit, that is 3s/s² + 16 -10/ s² + 25 + 1/s - 2 + 6/s³ + 7/s² - 2/s.
00:39:07.000 --> 00:39:13.000
That was my Laplace transform of this large complicated function here, let me recap how we worked that out.
00:39:13.000 --> 00:39:24.000
We did not want to go back to the definition of the Laplace transform, that would involve writing the integral of e⁻st x f(t), where f(t) is this enormous function.
00:39:24.000 --> 00:39:31.000
We have been to a horrible integral to work out from scratch, we are going to use linearity and we are just going to break this function up to its pieces.
00:39:31.000 --> 00:39:37.000
Each of these pieces is something out I figured out, the Laplace transform earlier.
00:39:37.000 --> 00:39:51.000
Example, the first time I look at this polynomial, we have the basic pieces are the 1 and t and t², and we figured out the Laplace transform for each one of those in example 1.
00:39:51.000 --> 00:40:03.000
Laplace transform of one is 1/s, t gives us 1/s², t² give us 2/s³, I got the Laplace transform of each of those.
00:40:03.000 --> 00:40:16.000
e⁺at in example 2, I figured out its Laplace transform is 1/s-a, for the cos and sin, I figured out the Laplace transform of those in example 3 and 4.
00:40:16.000 --> 00:40:28.000
Those were s/a² + s² and a/a² + s², what I did was I took each of those functions and just plug them back in here and then attach the right coefficient.
00:40:28.000 --> 00:40:49.000
3/ transform of 3 + cos(4t), my a here was four, that is where I get that 16 + s², for sin my a was 5, that is where I got that 5, that 25.
00:40:49.000 --> 00:41:06.000
Plugged in that -2 as a coefficient here, e⁺2t my a was 2 there, plug that in there as 2 and I get 1 - s/2 and then t² gave us Laplace transform is 2/s³.
00:41:06.000 --> 00:41:19.000
There it is there, Laplace transform of t and the Laplace transform of 1, I just cleaned everything up, combine the 3 s, 2 x 5 is 10, combine 3 x 2 x is 6.
00:41:19.000 --> 00:41:28.000
Combine everything together and finally I got that Laplace transform of that big, horrible function without ever actually having to do any new integration.
00:41:28.000 --> 00:41:34.000
I just relied on what I had worked out in the previous examples.
00:41:34.000 --> 00:41:45.000
That is the end of our lecture on Laplace transform, in the next lecture we are going to learn about inverse Laplace transforms, learn how to go backwards from the answer of a Laplace transform back to the original function.
00:41:45.000 --> 00:41:52.000
That is our next lecture in the differential equation series here on www.educator.com. My name is Will Murray, thanks for watching.