time to look at another exam and this is differential equations third exam from Spring 2024 ah what a lovely exam we have ahead of for us we're going to be talking Thea transforms we're going to do a little bit of Series so it's going to be good times we've got six problems to do and of course we start by putting down our name because if we don't put our name we're not going to get the credit there we go let's begin number one we're asked to find the inverse appla of the following 2 s cubus 1 over s^2 * s^2 + 1 all right now what do we see we see polinomial upstairs we see polinomial downstairs we're going to be doing partial fractions so really most of this work on this problem is going to be partial fractions all right so let's get into that now we see our original form so we'll write that down again we say all right what do we see downstairs well we see s s and S2 + 1 which means we're going to break this up into three parts what are they well this s squar says we're going to have something over s and then something over s squar we have to work our way up and then here the S2 + 1 we're going to have something there so it's a constant over the S term a constant over the s s term and the S2 + 1 we're going to have constant S Plus constant all right four constants to solve for a b c d well where do we get those constants we need to start doing some work and what we do we start by clearing out our denominators okay so that means we're going to multiply everything here by our downstairs and we're going to do that on both sides and uh all right so we have S2 * S2 + 1 we'll do it over here as well S2 S2 + 1 and uh of course that just cancels and there that's a little bit more to do but that's okay we can do a little bit more and what do we get well we get that 2 s Cub minus one is equal to now the first term a single s cancels so we have a s S2 + 1 the second ter the s s cancels okay so that's B * S 2 + 1 and then in the last term the s s + one cancels we left with an s s so CS + D * S2 now what do we do well I like to say first see if you can make any good choices and then if you can't do something else is there good choice something which cancels stuff out well I think there is a good choice and that's because you see the S and you see this s squ here and so you say ah let's plug in s equals z all right well what happens if we plug in s equals z well uh turns out really easy to plug in zero so anything with s goes away so there's a negative 1 and then we get zero here we get zero on the end and here we get B so you say aha B is 1 that was quick now at this point not so many great choices now we could of course get exotic we could choose s equal I and yeah you'll actually get some information out of that uh but not a lot of people are comfortable with that so I'm going to not do that okay so I'm going to stick with things I'm comfortable with because you know I I want to make sure I get get it right right that's the goal we want to get it right so what can we do well at this stage what we can do is we can now expand out this right hand side group coefficients and then we'll compare coefficients on the left and the right okay so we say what do we have well again we have this 2 s s Cub - 1 here we'd have a s cubed + a s b s 2 + b c s cubed + D s² okay well Group by powers of s so the S cubes we see a + C * s cubed uh how about the S squs uh well we see a b and a d so plus b + D S2 all right good uh how about the s's uh well that's just that that's the only s term okay so plus a s and how about constants there's only one right there and so plus b now at this point say well hey we've got polinomial in s equals polinomial in s now that's only possible if all of the coefficients line up so we say okay now we compare the coefficients to the two sides so we say well the constant here that's going to match the constant here so that says hey B has to equal negative 1 we already knew that uh the coefficient of s cubed here two has to match the coefficient for S cubed here okay okay so a plus c has to equal two what about these there is no s s over here and there is no s over here well they're kind of is what it means is that these terms have to be zero and from this we quickly say hey a equals z now that we know a equals z that tells us wait a second C has to equal 2 we say well wait we also know that b is -1 and B plus d equal 0 so hey that tells us that D equal 1 and now we've solved a b c d good we've done the partial fractions so are we done with the problem no but we're done with sort of the hard part of the problem so what we can now do is we can do an update we can say well hey instead of the lla inverse of our original thing let's see what we're really going to do the LL inverse of so we have that the Lao inverse and we'll go ahead and write out our original problem 2 s Cub - 1/ s^2 * s^2 + 1 well that's the same as the llao inverse of now be careful this is a place where a lot of people make mistakes they they forget to put in all the constants in the right places so a is zero okay well that's good that means we can drop the Z over s B is -1 so we have -1 * 1 / s^2 I I'll just put that constant in front all right then we have cs plus d okay well C is 2 and D is 1 so that's going to make it a 2 s + 1/ S2 + 1 all right good good now what can we do well uh first thing is we can say look let's break this up because we can handle s over S2 + 1 we can handle 1/ s s + one and because of algebra we have that plus upstairs we're allowed to break up the addition upstairs so we say all right so this is the same as you can say the llao inverse of and again we have our -1 1 / s^ 2 + 2 s/ s 2 + 1 + 1 / S2 + 1 and now because of the plus inverse is linear you just say hey do each part so the 1 / S2 s S2 + 1 the 1/ S2 + 1 so we're ready to write down our answer good times we have the minus the Laos inverse of 1/ s^2 is T So minus t + 2 the plus inverse of s/ S2 + 1 is cosine and particular here cosine of T because s^2 + 1 2 and inverse of 1/ S2 + 1 well that's sign and uh we're taking advantage of here this is one squar and that's a one so beautiful everything lines up and we found the Lost inverse which means we're done nice nice number two we're given uh the lass of a particular function and told that the lass of f of T is s/ S 4 + one and also that F of 0 and frime of 0 are both zero now we also have the Laos of a bunch of functions here and we're asked to fill in the box with what it is okay so you'll notice i' I've done a little bit of writing here and so what's happening is that this problem is testing rules so what I've done is for reference I've written down the important rules so these usually are found on the equation sheet but we're just putting them here for reference so we're going to say okay let's go through and let's figure out what these are okay so uh let's start with the Lao of f Prime of T now there's a rule that says how do you take Theos of a derivative or a second derivative or really actually any derivative so the rule for the lass of the second derivative is you take s^2 * L of the original function and there's some correcting terms here now notice our terms are zero so these parts go away so we say oh so we have to multiply our original term our original appluse by s s well since we start with s/ S 4 + 1 multiplying by s s should give us an S cubed or S 4 + 1 which we see there so our first answer is e all right good now our second question what's the lause of t f of T well there's a rule if you multiply by T the result is the same as taking a derivative and then adding a negative sign so we say okay well we see this s/ S 4 + one that's our function so we need to take the Dera of that all right well flashbacks to Cal one quotient rule so okay all right we got to do some work here so we're going to take the bottom S 4 + 1 * the of the top which is 1 minus the top s * the of the bottom which is 4S cubed all over the bottom term squared well if we simplify this we have S 4 minus 4 S 4 so that's - 3 S 4 + 1 over S 4 + 1 s now this is f Prime of s now but we want negative frime of s so we're looking for 3 S 4 - 1 over S 4 + 1 2 and there we go so the second one is M all right right well now for the next one the Lao of the integral from 0 to T F of to F of T minus to D to this one says you have to recognize what's going on well what we have is convolution so when you see ft ft minus to you know they're sort of going in opposite ways and it's a way to combine two functions so this is just other words say F convoluted with f now what's the rule if we have a convolution of two functions you multiply the results of each individual one so we say oh so this part right here this was going to become our F of s time F of s or F of s squar so we're going to say all right we want to look for s s over S4 + 1 squar okay s s that's good Ah that's not S 4 + 1 squ okay keep looking keep looking keep ah here we go s^2 S 4 + 1 2 that is L all right now our next one e 5 t f of T well the rule here we have e to some constant t f of T what happens is you shift so now instead of s It's s minus C so in particular here we say oh we're looking for s - 5 right cuz we have e to the 5 T So C is 5 so you are going to take this and everywhere we see an S we're putting s-5 so there should be an s-5 upstairs okay there's one place where there's an s-5 upstairs this has an s- 5 upstairs but that eus 5S no no that shouldn't be there uh oh yeah so so it is it has to be this one and you can see inde d s - 5 S - 5 4 + 1 so we have have F good good okay uh now one more so U of tus 5 F of tus 5 what's the rule if we see U T minus a f t minus a then it's adding an e to the minus a s f of s so we say oh so in our case we're because we have T minus 5 looks like a is five so we should say we're looking for E Theus 5S times what we have okay so we want need an eus 5S times our function so S e- 5S S 4 + 1 yes good now of course you'll notice there are some other places with eus 5S where there's this s minus 5 oh that's trying to like you're trying toit wait wait what what so so they're trying to sort of mix rules together to see if we can pay attention but of course we're paying attention we we are going to get these right and so there we go e m l FH or of course the word no no it's not a word it's not a word uh but it is right answer and that's another problem down let's keep going number three we have F of T which is a p wise function is 4 T if T is less than 2 and t^2 + 4 if T is greater than or equal to two and we have two parts Part A what's the lause of f of T now f is pawise and so the first step into solving this is to say we've got to rewrite this function in a way which is peace wise friendly and so we say okay well how do we do that well we say what's our initial Behavior 4T say all right so it's 4T and now when do we change well at two so we're gonna have an onoff switch at two so U of T minus 2 now what are we turning on well that's what comes after two so that's our T ^2 + 4 so we have our onoff switch what's on subtract what goes off and now it's what we came before to so subtract 4T all right well uh next thing is we really want to make this part T ^2 + 4 - 4 t look like an expression involving tus 2 if they're nice it won't be too bad I think they may have been nice because notice what do you see this is T ^2 - 4T + 4 or that's the same as tus 2^ 2ar so you say aha this is really 4T plus our onoff switch U of T minus 2 time t- 2^ 2 now that's not our answer but what we've done is we've Rewritten it so that we can get an answer we've Rewritten it in the right format so that our rules apply so here we go we have our llao of f of T well that's equal to Theo of this expression because that's F it's just a different way to write it so Lao of 4 t+ U of T minus 2 * t- 2^ 2ar now say okay what's the L of 4T well the four is four and then Lao of T is 1/ s² U of T minus 2 well that says we're going to have an eus 2s so whenever we see an onoff switch the lios transform is going to have an e to the minus a s wherever the switch occurs at a so we have e Theus 2s and now remember what happens here when we're doing our transform we say well it's not tus 2^2 we look at this and say ah I should think of this and say what's the real function here well this is our F of tus 2 is t - 2 s so I I say well that's really t^2 and so when I'm going to come down down here I want to say okay what's the luse of t^ 2 well the L of t^2 would be 2 factorial and we'll go ahead and write as two factorial we'll simplify that in a second over S 2 + 1 so that's the of t^ s okay well uh that we can write we have 4/ s^ 2 and then here 2 factorial is 2 so 2 e - 2 s/ s cub and there we go there's part A Part B find the llao of the integral from0 to T F of to / the square root of T minus to DT okay now uh this looks like it's going to be complicated or is it uh what can we do well notice here this inside we can rewrite this as the integral 0 to T of f of to and then the square of T minus to I can bring that upstairs as T minus to to the minus a half and now our great Aha and our great aha is to say weit wait a second we can handle this we have the technology because what we see this is really the function f convoluted with the function t to the minus one2 so we say Ah convolution that's what's going on so we're looking for the L transform of that well that's going to be the same as we take the Lao transform of f times the lass transform of tus a half now here's some good news we know the llao transform of f we just found it all right so we say great this will become 4/ s^2 + 2 e- 2 s over s Cub now the pl transform T the minus a half well because it's not a whole number we use gamma instead of factorial so we say okay so that's going to be gamma Min - half + 1 over s - a half + 1 well uh this is gamma of a half and Gamma of a half turns out it's square of Pi and so we'll say our final answer well we have 4 s^ 2 + 2 e- 2 s over s cubed and we times it by < TK piun over s^ 12 because minus half + 1 this to the 1/2 and we're done that's it all right good good so uh yeah it's not too bad the main thing is to say hey we recognize convolution we don't actually try to figure out this integral we say let's use our tools and using tools is a great great way to solve problems number four a mass spring system has a mass of m = 1 a spring constant k = 4 and is initially at equilibrium so there's no motion and we're at rest right we're in the neutral position at time t equal 3 the mass has an external impulse force and that means you're hitting it of Two and A time T = 5 and X impulse pulse is 7 so we're going to be hitting at time three and at time 7 now impulse means just one hit one and done now that's the key word impulse we we know that so what does that translate into in terms of our terminology well that says these are going to be Deltas so we have Deltas here okay so these aren't sustained forces these are are just momentary forces and you'll notice that they emphasize the word external okay so what do we do part a set up a differential equation that models the position X of T of the mass well we know we're in a spring Mass system so we say well there's there's a generic formula for spring Mass systems it looks like this it says MX prime plus CX prime plus KX equal your X external well we know our Mass it's one okay so that becomes X Prime now C is your damping constant but there's no damper involved in this problem so that's zero so that middle term goes away our spring constant is four so we say Okay Plus 4X and now we say our external well we have two external forces Nam we have those two pulses to the system all right so we say well there's a a force of two at time five so that's two oh sorry time three there's POs to two time five there's POS s so so there's two at three so that's two delta T minus 3 and 7 delta T minus 5 and that's our differential equation okay good nice now Part B use the llas transform to solve for x of D it's like okay we've got a nice differential equation and let's use Lao transforms to solve so the way we do it is we start by taking the Lao transform of both sides so Lao transform X Prime + 4x we'll start with that well Theo transform of XP Prime is going to be s^2 capital x minus s x of 0 uh minus X Prime of 0 that's first part then + 4 capital x now we know our initial conditions say that the X of0 and from Z are zero con and so what we end up with is we end up with s + 4 capital x okay so that's the left hand side if we feed it through that LL Machinery now we're going to repeat this with the right hand side okay so we say the Lao of 2 Delta tus 3 + 7 Delta tus 5 and these are nice very simple just a Delta function and we know how to take the Lao of Delta functions that introduces the e to the minus in this case 3T and then 7 e minus 5T Okay so we've taken the LA of the left hand side we've taken La of the right hand side now we set them equal so we get that s² + 4 * X is = to 2 e - 3 t + 7 e - 5 T and at this stage we divide by S2 + 4 because we're trying to solve for x and so we say aha X is and uh we'll have say uh well we'll write as 2 over s^2 + 4 we'll put the E minus 3T to the side plus and uh we're going to do something a little bit suspicious here I'll put the seven out in front 7 s² + 4 eus 5T all right now why did I do this and doesn't quite look right well there's something I want to do here we see the E to minus 3 T eus 5T we know that that means there's going to be a shift U of T minus a so what we want to do is we want to have what the to it be ready to do the plus inverse of so when you see S2 + 4 downstairs you're saying oh s or cosine you look upstairs if you see s you go cosine route if you see a constant you say s route now here's where we have to be careful 4 is 2^2 so what do we want upstairs well we want a two upstairs which is why it's like ah I have a two I put a two same same thing here four is 2 squ what do we want upstairs two not seven two so okay so how do we get a two up there well the simple answer is we put it see watch boom see now there's a two up there of course I just can't arbitrarily put a two so I I have to compensate so there's really a 7es * 2/ s^2 + 4 and now we're in good shape you might say wait don't we stop here well no this is capital x we didn't write it but this is a function of s we want Little X which is a function of T but we're almost there all right so here we go Little X I'll emphasize of T this is our Lelo inverse of capital x and then we say okay now we see our our rules coming into play Just the Two s^2 + 4 this should be our sine of 2T and again same with this but what do we need to do we need to account for the shifts and not forget the constants and okay let me move down a little bit here so e to minus 3T means there's an onoff switch happening at time 3 U of T minus 3 and then we take the S of 2T but we replace T by tus 3 so that becomes s of 2 t-3 this term we have that constant in front 7 Hales all right and now e the minus 5T onoff switch happening at time five so you have T minus 5 and instead of s of 2T It's s of 2 tus 5 so s 2 tus 5 and that's it we're done Tada we found it the motion of the spring and you'll see that it's at rest so it does nothing until time three and then at time three it says Whoa and it starts vibrating and then at time uh five it's like whoa second hit and so it'll switch up its vibration a little bit more and there we go that's the answer ah good well uh that's it for the plus but time for us to get more series number five given that Y is the sum from 0 to Infinity of a subn x to the N is a solution to this differential equation Y Prime - 2x y Prime + y = 0 then we're asked to find a recurrence relationship for the a subn okay now the key word here is we see recurrence relationship now what does that tell us we're going to do it tells us we're really going to try to keep everything as a sum all right well so we say Let's uh write down our sums we know what Y is we also need to understand what y Prime is and what y dble Prime is so I'll write down y again for reference so that's the sum 0 to Infinity a subn x to the n y Prime well that's the sum 0 to Infinity n a subn x nus one right because we bring the power of n down we take derivatives term by term y double Prime well that's going to be a sum 0 to Infinity of n n -1 a sub n x n-2 now I'm not going to worry about shifting I'm not going to worry about changing my starts yet I'll do that when I need to do that for right now I just simple y Prime y prime one derivative two derives work it out now the next step is to say now that we have these individual pieces we're going to start with our expression and and replace y Prime by y Prime replace y Prime by y Prime and replace y by y okay what's our long-term goal here our long-term goal is to say we're going to make everything polom because this thing of it a series is a big polinomial that's our good mindset it's a big polinomial and what do we do with polins we like to group coefficients so we're grouping coefficients that's our mindset here all right so we have y Prime that's our sum n = 0 to Infinity we have n n min-1 a subn x nus 2 - 2x sum n = 0 to infinity and a subn x nus1 and then plus r y which is our sum and = 0 to Infinity of our a subn x to the n and all of this equals z now our goal is to group coefficients so we want to make all the powers look the same so next thing we look for is to say okay should can we bring in any powers of X well there is one place we can bring in a power of X and that's right near the start you see how we have this x here we can bring it in and then we have X x * x^ nus1 which means that this will look like an x to the N which is good cuz this looks like an x to the end bad news this does not look like x to the N not yet but we'll get to that okay so let's go ahead we'll rewrite that and uh you might say oh there's a lot of writing on these well yeah but that's okay in some sense it's just a matter of pay attention don't lose any of your terms and uh mostly it's mechanical now that might sound like boring but you know you have to pay attention and uh but work it out you'll be okay all right so I move that X inside okay now I'm going to go through a process my goal is to make all these powers of X look the same so Step One is I'm going to do shifting after we shift step two is I'm going to say hey let's make sure our starts agree so that might be called peeling I'm going to peel off terms as I need and then I'm going to smoosh it all together okay so what do we do well let's uh come here we like this power great so we're just going to change to a different symbol I I'll use K and we like this power great no problem now here this is where things get like so we say okay well let's just do a little bit of change so here k = n minus 2 or if you like that's Sim as n is equal to k + 2 okay so what happens to our first sum well this becomes the sum now n starts at zero so where does K start at well it's 0 minus 2 so we start at -2 and I know that seems weird but we'll fix it Infinity minus 2 is still Infinity that's great now what happens to n well n is k + 2 so we have k + 2 K +1 because n minus1 is k + one a sub n is k + 2 and then x to the K that's the first one now the other two sums the only thing we have to do is everywhere where we had an N we put a k so k a sub k x to the K and then plus our sum K = 0 to Infinity a sub k x to the K and there we go all right now the good news is all our powers look the same nice our starts not so much but Z is good Z is good what about this well what happens when you plug in -2 k + 2 makes it 0 I when you plug negative 1 it's zero so what is that telling us well it's telling us that if we were actually to start this we we can peel off the -2 peel off the1 and we say well it would be a 0 which is -2 + 0 1 plus the sum from 0 to Infinity of k + 2 * k + 1 a sub k + 2 x the K so this is the pel it's like pull off the first few terms now in this case the first few terms are zero not always so you want to be careful well great now we say hey we can ignore the zeros and all the sums start the same place all have the same power nice okay so that says one sum right one sum to rule them all so we have the sum K = 0 to Infinity of k + 2 * k + 1 a sub k + 2 minus 2 k a sub K plus a sub k k x the k equal Z right so we're just putting it all into one big series and because all our powers look the same they all look like X to K and all our STS look the same we're able to pull this off now this is not quite our answer but we're right there we're right there how do we finish up well what we do is we say what does this mean it means we've got a big polom equ equal Z well how could a polinomial be zero well the only way a polinomial can be zero is if all the individual pieces along the way are zero and so that means that this is zero for each for each choice of okay right this equals z for K greater than or equal to zero and that is the relationship we're after so in particular our answer is the following uh namely k + 2 * k + 1 a sub k + 2 and here since they both have a subk we can pull that out so this becomes - 2 k minus1 a subk = z and we'll put a little side note this is true for K greater than equal to zero it didn't specify when the recurrence relationship needs to hold but we can write that down sometimes we like to solve for sort of the the larger a sub something so if you wanted to you could say this is also the same as a subk + 2 is equal to well you move this to the other side and divide by the coefficient 2 K minus1 Over k + 2 * k + 1 a sub K either one of those is great they both show the recurrence relationship and there we go that's it not too bad actually for a recurrence relationship and uh all right that's number five one more to go number six our final problem we need to find some numbers b0 B1 B2 B3 given that we have this Y which is the sum B subn xn is a solution to a differential equation E the X Y Prime - x y Prime + 2 y = 0 and we have some initial conditions now we have four terms to find two of them are freebies all right where are our freebies at well they come from saying ah if you think about Y and you plugged in zero what would you get you just get the constant term which is our B 0 so Y of0 is b0 well what about uh the Y Prime is z well that's if you took a derivative and then plugged in zero what would you get in fact let's go ahead and do this we'll we'll say what does y look like it looks like B 0 plus b1x plus B2 x^2 plus b3x cubed plus dot dot dot and I'll stop here because I'm R after B3 uh so y Prime looks like B1 + 2 B2 x + 3 B3 x^2 plus dot dot dot now plug in zero all the terms disappear except for that B1 so that says hey this y Prime is zero that's B1 and Y double Prime what would that be what B2 B2 + 6 b3x dot dot dot all right so B B1 are freebies how do we find B2 B3 well that's where the differential equation comes into play and now we sort of have our mindset okay we know that we found B 0 B1 and uh okay so these are from I'll put IC for initial conditions now where does B2 come from so B2 that's going to be looking at our x to the zero term if we were to expand everything out in this differential equation and B3 is our x to the one term all right good so I just need to understand the constant and X terms in this expression so with that mindset where we say okay well let's uh put in what everything looks like for the start we just care about the first so we're not going to do recurrence relationships we're just after the very beginning as the song says we'll start at the very beginning a very good place to start let's talk about e to the X now we know there's this Su for it K = 0 To infinity x the K Over K factorial but we only need the start only need up through the X term so it starts as 1+ X Plus dot dot dot okay so here we go e to the x 1 + x dot dot dot y Prime 2 B2 + 6 b3x Plus dot dot dot right here we go Min - x * y Prime well B1 and now notice because of this x here that we're already up to an X term if I look at two b2x that'll get me to an X squ term so I actually can stop right there cool and then plus 2 and then we have B 0 plus b1x plus dot dot dot and all that equals zero now do you need the dot dot dots yes you want to keep the dot dot dots uh that's good notation and we love good notation well I don't know if we love it but of course people who grade it they love good notation and we want to make them happy all right so now let's uh gather we'll gather our constant terms so looking here the way to make a constant we have the 1 * 2b2 and uh that's the only way to make a constant term right because other otherwise we get X's involved no constant term there now here we have a 2B 0 all right 2 B2 + 2 b0 and that's our constants okay well next up uh let's look at our X terms okay our X terms well how can we make an X well we can take 1 * 6 B3 okay so there's 6 B3 or we could take this x * 2 B2 so 2 B2 we could take this x * B1 don't forget the minus minus B1 and we can take this two times our B1 so that's plus 2 B1 and those are all our X terms all right great and uh then what well and then there's our higher order terms which we don't need to worry about okay so there we so what have we done we said we're after this differential equation we said how does it start in terms of our B's so we say well if you expand it all out and gather coefficients the constant term looks like 2b2 plus b 2B 0 the coefficient of x looks like 6 B3 + 2 B2 minus B1 plus 2 B1 and then there's other stuff we don't need to worry about because it's not going to help us get our answer we're after an answer and we're going to do what we need to do to get that all right well now our great aha we say well hey this part equals z and this part equals zero because the right hand side is zero and we know B 0 and B1 so let's look at this first expression well uh this says 2 B2 + 2 B 0 = 0 the twos cancel that says B2 is b0 B 0 is 1 so B2 is minus B1 okay so now we have B2 all right well from this other part we say okay well great 6 B3 + 2 B2 and Min - B1 + 2 B1 is plus B1 = 0 but we know things right yeah KN is great B2 is1 that's -2 B1 is 3 + 3 = 0 Well - 2 + 3 is 1 that says 6 B3 is equal to -1 and that says B3 is -1/ 6 so recapping we have B 0 = 1 we have B1 = 3 B2 is1 and B3 is-16 and that means we found our B 0 B1 B2 B3 which is exactly what we were asked to find and so we're done nice nice all right well another test done hope you did well or if you're studying for a test I hope you're going to do well and uh thank you for joining me and maybe maybe I'll get to see you again bye