Differential Equation Using Laplace Transform + Heaviside Functions

all right in this video we're going to use a Laplace transform along with a he beside function to solve a differential equation so here we've got the differential equation y double prime plus three y prime plus 2y equals G of T we've got the initial conditions Y of zero equals zero and y prime of zero equals one in G of T is just going to be a piecewise function it's going to be equal to one if T is between zero and one and it will be equal to zero T is greater than or equal to one and we're going to assume here that T is only defined for values of zero and larger so okay so really what we're doing here is we're going to use the Laplace transform to solve linear constant-coefficient differential equations with piecewise continuous forcing functions of exponential order that's really what we're doing if you want a nice precise bit of terminology so the first thing I'm going to do are our function here G of T this is going to be a long problem so because I'm really going to try to go through a lot of the stuffs so a couple things we're going to use a lot of Laplace transforms so I would certainly have a table of Laplace transforms handy I'm going to jot some of them down but again I will be referring to those so here's our function G of T ok it's 1 if it's between 0 & 1 and that's 0 everywhere else so the first thing I'm going to do is I'm going to write that using one of our he beside functions so I've got videos on those you might want to check those out but notice our function here it's basically turned on between 0 and 1 and the way that we can write that in terms of the he beside function is that would be H of T minus H of t minus 1 and recall this is basically a function that turns on at T equals 0 and turns off at T equals 1 which is what this one is doing so again if you haven't seen he beside functions got some videos on those you might want to check those out so but you can also write this one equivalently as 1 minus H of t minus 1 but I think I'm just going to use this form on the left because that's what I've talked about in some other videos ok so we've got the function we're trying to solve here I'm just going to rewrite it we've got this function y double prime plus 3 y prime plus 2y that equals our function G of T but G of T again we can now write using our key beside functions so H of T minus H of t minus 1 and what I'm going to do now is I'm just going to take the Laplace transform of both sides so I'm going to take the Laplace transform of the left side and of the right and recall since its linear you can basically just bust it up so I'm going to take the Laplace transform of Y double prime I would take the Laplace transform of the term 3y prime you can just pull the three out front so I'm taking the Laplace transform of Y prime plus 2 multiplied by the Laplace transform of Y little Y and then I'm going to take the Laplace transform of the right side as well so the Laplace transform of this key the side function H of T minus the Laplace transform of H of t minus 1 so you can't see it here off to the side I've got a big table of Laplace transforms here let me let me show you just so you know that I'm not I'm not joking here so I've got a big table of Laplace transforms that I'm going to be using and again if I've never taught differential equations it's been a while since I've been a student I don't know how many of these you would be honestly expected to have memorized you know if you're taking a quiz or a test over this stuff certainly ask so you know if it was me you know being a student working on this stuff we definitely have you know some of this stuff I would have a table right here in front of me I guess is what I'm trying to say so okay so let me go back to my example where I want it to be where'd you go okay so we're taking the Laplace transform here of all this stuff and recall you know again if you can check the Laplace transform of Y double prime that's given by the formula that's going to be little s squared we use capital Y of s minus s times capital y of excuse me little Y of 0 see I'm already got to be careful here with our capital wise and our little wise so it's s squared capital y of s minus s times little Y of 0 so this is where our initial conditions are going to come into it minus y prime of 0 so I'm just using the formula for the Laplace transform of Y double prime that's this part and then we'll just keep going so then we have plus 3 times the Laplace transform of Y prime the Laplace transform of Y prime that's given by the formula s capital y of s minus y of 0 and I'm going to run out of room here plus 2 times the Laplace transform of Y and that's given by capital y of s now we have to take the Laplace transform of H of T minus the Laplace transform of H of t minus 1 so I'm just going to write that down because I'm going to compute that here off to the side so we took the Laplace transform of the we took the Laplace transform of the left side and we still have to do it on the right side but let me show you how to compute that part really fast so what are the formulas is let's see here we go so one of the formulas for Laplace transforms and he beside functions because that's what we're doing here so let's see little applause transform the formula that we're going to use here we've got the Laplace transform of some function f of t minus c multiplied by h of t minus c the Laplace transform of that is e raised to the negative C multiplied by s and then we take the Laplace transform of whatever our function as little f and that gives us capital f of s okay so in this case we've got two five we've got compute two Laplace transforms we have to compute the Laplace transform of H of T and the way that I could write that is well the function that's being multiplied out front that's just you can think about that as being the constant function one and then I could write H of T is just well H of T minus zero and according to our formula we'll get e raised to the negative well 0 times s or e to the 0 power and then we have to compute the Laplace transform of 1 and e to the 0 that's just 1 the Laplace transform of 1 that turns out to be 1 over s so the Laplace transform of the he beside function H of T that's just going to be the function 1 over s and if we do the same thing for our other function we would have the Laplace transform again you can think about the function in front of H of t minus 1 just as being 1 so then we would have H of t minus 1 so in this case it says we'll get e raised to the negative power of C multiplied by s C in this case is just to be equal to one so we'll have negative 1 x s and again the Laplace transform of 1 that's just 1 over s so we'll have e raised to the negative s divided by s all right so I'm now going to go back I said this one's going to be pretty long I'm going to go back and I'm going to start simplifying the left side of my equation and then I'm also going to just plug in what we just found here into the right side so the Laplace transform of H of T minus the Laplace transform of H of t minus 1 ok so let's go back and use our initial conditions so we've got s squared capital y of s minus let's see so we said that Y of 0 and so Y of 0 was equal to 0 and Y prime of 0 is equal to 1 so I'm going to substitute those in so we would have minus s times y of 0 which we set at 0 minus y prime of 0 which is 1 plus 3 multiplied by s x capital y of s again we said Y of 0 that was just equal to 0 so I'm not going to write that down plus 2 times capital y of s and then on the right side again after taking those Laplace transforms we said that the Laplace transform of H of T that was just going to be equal to 1 over s and the Laplace transform of H of t minus 1 that's e to the negative s divided by s so let's just keep simplifying here so s multiplied by 0 you're gone what I'm trying to do now is just to solve for this term Capital y of s so we can do that let's get rid of this stuff ok so we have s squared times y of s let's see so we've got this minus 1 that doesn't involve Y of s so just a little bit of algebra I'm going to add that to the right side so we have one over s minus e to the negative s over s plus one let's see we would have plus 3 s multiplied by capital Y of s plus 2 multiplied by capital y of s and again now I'm just going to solve for y of s so I can factor the y of s out I would have Y of s equals let's see if you could let's see so we could factor that out we would have Y of s left over we would have s squared plus 3 s plus 2 again equals all this stuff on the right side now I'm going to start rewriting this a little bit so I'm going to get common denominators we could write you know plus 1 as plus s over s you notice they've all right got a denominator of s I'm going to run I'm going to break this up I'm going to write this as s plus 1 over s and then I'm going to have my other term e to the negative s over s and the reason is it's going to this this corresponds it's going to correspond to a form you know if you again look at your table of the Laplace transforms these are going to fit one of those forms that's why I'm doing that so again it's really just kind of looking at those tables and trying to make everything sync up okay so now we've got Capital y of s I can just divide both sides now by this s squared plus 3s plus 2 so I would have s plus 1 divided by s that would all get divided by s squared plus 3s plus 2 minus U to the negative s divided by s and again we have s squared plus 3 s plus 2 so we're just really kind of getting going here so again like I said this is going to be kind of a long problem and they gave let's let's take summary real quick what are we done all I've done was just wrote my my function G of T using he beside functions and then all I've done is just taken Laplace transforms of both sides using the table of Laplace transforms that's everything I'm doing here and then it's just again just doing some algebra to isolate this term Capital y of s that's what we've done so far to get to this point okay so now what we're going to do is we're going to take the inverse Laplace transform of both sides that's that's that's what we're going to do at this point so let me make sure that I haven't done anything too crazy here I don't think so so we want to take the inverse Laplace transform of both sides so again we can do this in pieces because this Laplace transform is linear and again all they mean by linear is you can just break it out okay so I'm just writing hey we're taking the inverse Laplace transform of both sides okay so now I'm just going to simplify these a piece at a time that's all I'm going to do so that I'm going to work on this Laplace inverse Laplace of s plus 1 over s multiplied by s squared plus 3 s plus 2 and at this point on the left side when we take the inverse Laplace transform of capital y of s this is just going to give us our function y of T so at this point all we have to do is just figure out this inverse Laplace transform figure out this inverse Laplace transform put it together and we'll have our solution ok so so let's start off by finding this inverse Laplace transform of s plus 1 divided by s multiplied by s squared plus 3 s plus 2 so what you would have to normally do in this case is to bust it up you would break it up and you would want to use to use your table of Laplace transforms as you basically do partial fractions on this so it's just going to be a lot of simplifying using partial fractions at this point so using partial fractions and the first one is kind of nice because if you factor the bottom if you factor s squared plus 3s plus 2 that factors as s plus 1 multiplied by s plus 2 so ok this one's set up to work a little bit nicely ok so we can cancel out the the s plus 1 and the s plus 1 so we're left trying to find the inverse Laplace transform of we've got 1 over s multiplied by s plus 2 so now this is where we're going to do partial fractions to basically put this in a good form so that we can use our table ok so again kind of off to the side here ok so recall with partial fractions these are both linear terms in the denominator we could write 1 over x s plus 2 we can write that as a over s plus B over s plus 2 and our goal is now to figure out these constants a and B so again if you forgotten partial fractions I've got tons of videos on these this is normally something again you see in calculus too when you're learning all your different integration techniques so if you're watching this that means you're in differential equations so I'm probably going to go through this a little bit faster we're multiplying both sides now by s multiplied by s plus 2 I'll do the first one so s multiplied by s plus 2 on the left side will just be left with 1 on the right side when we distribute we would have a multiplied by s plus 2 plus B multiplied well there the S Plus 2 s would cancel and we would just be left with s again we're trying to figure out a and B notice if you let if you let let's see s equals 0 if we let s equals 0 we would have 1 on the left side we would have a multiplied by 2 or 2 a notice the term involving B times s would be gone so if we solve we'll get 1/2 will be the value of a likewise notice we could let s equal negative 2 if you let s equal negative 2 well you still got one on the left side this first term involving a is now gone you have negative 2 multiplied by B so negative 1 half will be equal to B ok so I'm trying to again find this inverse Laplace transform of 1 over s multiplied by s plus 2 but instead what I'm going to do is find the inverse Laplace transform we said 1 over s times s plus 2 it's a we said we could write that as a over s well a is equal to 1/2 we said B was equal to negative 1/2 and again now we can just use our table of values so hopefully you can see mine well enough let's check here so forget about the constants for a second forget about the 1/2 on top and the negative 1/2 really you can factor those out just worry about the 1 over s and the 1 over s plus 2 it says if you have 1 over s can you read that probably not if you got 1 over s it's the very first one the inverse Laplace transform of that is 1 if it's 1 over s minus a it's e to the a T so so we've got 1/2 multiplied by the inverse Laplace transform of 1 over s minus 1/2 and then we would have the inverse Laplace transform of 1 over s plus 2 so now I'm just getting the stuff against straight from a table just straight from a table so the inverse Laplace transform of 1 over s that's just 1 the inverse Laplace transform of 1 over s plus 2 that's going to be e raised to the ok so normally it's in the form s minus something so you could write this as s minus negative 2 so we get e raised to the negative 2 multiplied by T ok so 1/2 minus 1/2 e to the negative 2t ok so what have we done at this point what we've done is we found the the inverse Laplace transform of the term s plus 1 divided by s multiplied by s squared plus 3 s plus 2 so this part is going to be useful that's going to be part of our solution and now I'm going to go back and I'm going to do the same thing I'm just going to figure out the inverse Laplace transform of the other term so now we're going to figure out the inverse Laplace transform of this other term so now let's find the inverse will plus of e to the negative s divided by s times s squared plus 3 s plus 2 okay this one is going to take a little bit more word and the thing to notice in this case is that this is of the form so this is of the form we're trying to find an inverse Laplace transform we've got e we could write this as e to the negative CS multiplied by some function f of s so in this case we're trying to find what would be trying to find we've got our our C value is negative 1 our function f of s our function f of s is going to be this 1 over s multiplied by s squared plus 3 s plus 2 ok now there's a formula for these in this case so the formula we're going to use so the formula that we're going to use in this case it says if you if you've got an inverse Laplace transform of a function of the form e to the negative where do I have it here e to the negative s times f of s we can I'm just using a table here that gives us a function of the form f of t minus 1 times h of t minus 1 and again that's because of my value up here of negative 1 ok so again this is all just table stuff so the next thing we have to do is now we need to find the inverse Laplace transform of 1 over s multiplied by s squared plus 3 s plus 2 well like again can write s squared plus 3 s plus 2 those s plus 2 multiplied by s plus 1 and again I'm going to do the same thing I'm going to do partial fractions so partial fractions so we've got 1 over s multiplied by s plus 2 multiplied by s plus 1 again these are all linear I can write this as a over s plus B over s plus 2 plus C over s plus 1 if I multiply both sides by the denominator s times s plus 2 times s plus 1 I'll be left with 1 equals a multiplied ok the essence would cancel we would have s plus 2 times s plus 1 plus B multiplied by s times s plus 1 plus C multiplied by s times s plus 2 now I'm not going to go through you know again you could let s equal 0 we could let s equal negative 1 and then you can also let s equal negative 2 to solve for a B and C and when I did that I got a to be equal to 1/2 I got C to be equal to negative 1 and I found B to be equal to 1/2 so again that's kind of skipping those steps but you start to work a little bit to find those so at this point okay let's say where we add here so if we now take the inverse Laplace transform of a over s so it's going to be 1/2 over s plus B that's 1/2 over s plus 2 plus C so that's plus negative 1 over s plus 1 and again this inverse Laplace transform of all this function that's going to be my capital F of s ok so I need that function that's my capital f of s here ok so if I compute that let's see what have we got here so we said the 1/2 comes along the inverse will cost transform of s that's just 1/8 plus 1/2 the inverse Laplace transform of s plus 2 again that's going to be e to the negative 2t then we've got our minus 1 the inverse Laplace transform of s plus 1 that's going to be e to the negative T ok so we're almost almost there now almost there now what's the one part that we need so if we compute this inverse Laplace transform now of where to go this inverse Laplace transform of e to the negative s multiplied by 1 over s times s plus 2 times s plus 1 ok so we've got our function here so so it's going to get multiplied by H of T minus 1 in this case our our function here that we just found where'd you go this is going to be our function little F of T so we said that this is going to be equal to f of t minus 1 times H of T minus 1 again just using our table so we figured out that F of T II was equal to this function so now I'll just replace all the T's with T minus one so in that case if we replace all the T's with t minus one we would have one half plus one half times e to the negative two times t minus 1 minus 1 e to the negative let's see e to the negative let's put our parentheses in there t minus 1 and again that's getting multiplied by h of t minus 1 whoo ok long stuff here again I'm just using just using our formula here and the last thing we need to do now is just put all of this stuff together so where did everything go ok so in summary so the inverse Laplace transform of Y of s the inverse Laplace transform of Y of s that's just going to be our function y of T and we said the inverse Laplace transform of this function s plus 1 divided by s multiplied by s squared plus 3 s plus 2 we computed all of that I think I've lost the piece of paper that it was on where'd it go so when we computed the Laplace transform of this term when we computed that we got what did we get we have 1/2 minus 1/2 e to the negative 2t and then it says we have to subtract away the inverse Laplace transform of this e to the negative s divided by s multiplied by s squared plus 3 s plus 2 that's what we just figured out that's all of this stuff so let's see that's going to be 1/2 plus 1/2 e to the negative 2t minus one minus one D to the negative T minus one power all being multiplied by H of T minus one so I've now put everything together grab all my parentheses in the right place I think so and that to me now looks like that is going to be our solution trying to make sure I didn't leave anything out here all looks good to me so who again long problems long problems so so what did we do we use T beside functions we took a lot of Laplace transforms just using formulas you know we plugged in our initial conditions are Y of zeros on our Y Prime's and then it was just using partial fractions to put again things in a good form to match up with our table we had to use this this function involving key beside functions but other than that I don't think there's anything else that that we left out so again it's just a lot of putting things in the correct form so that you can use your tables so I'm sure there will be questions pill feel free to please post and ask if I can't help you hopefully somebody else can put you in the right direction if you would like to see another one of these by all means let me know I can make some more and yes we'll go from there so alright friends again not an easy procedure you know I mean in the sense this is really long and it's just using tables so again I would have at the table of Laplace transforms open right in front of me as I went through these so all right again I hope this helps and good luck

Channel: patrickJMT

Views: 196,599

Keywords: math, university, college, laplace, transform, heaviside, function, forcing, discontinuous, differential, inverse, table, mechanics, solve, solving, equation, linear, constant, coefficient

Id: mdj26_bUpBY

Length: 30min 27sec (1827 seconds)

Published: Sun Nov 06 2016