Ultra-Mega Differential Equations Review Problem!!!!

here we're going to look at a nice linear second order differential equation and we're going to find both linearly independent solutions to it and we're going to do that using a lot of different strategies and i think this makes this problem and this solution to this problem an excellent review for about the first half of a differential equations class so here the differential equation that we want to solve looks like this we have 1 minus x squared times y double prime minus eight x times y prime minus 12 times y equals zero and then we know because there are zeros to this polynomial one minus x squared at one and negative one then any solution that we will find will only apply on certain intervals so there will be some solutions that apply on minus infinity to -1 some on the interval minus 1 to 1 and then some on the interval from one to infinity and that's because the roots of this polynomial split the real line up into those three spots so we're gonna look at the space when x is between minus one and one and we're going to use three main tools so the first is assuming that we have a power series expansion for our solution and then we'll do a combinatorial analysis on the coefficients for that power series and then once we have a single solution we will use able's identity to create a first order differential equation for the second solution abel's identity says if we have this type of second order differential equation so we've got y double prime plus p of x y prime plus q of x y equals zero then we can calculate the wrons scan two different ways first it is the exponential of the negative anti-derivative of p of x where that's our p of x function and then also it's equal to the determinant of a matrix made up by the two linearly independent solutions in other words y1 times y2 prime minus y1 prime times y2 where y1 are those linearly independent solutions so if we've already got one of those solutions let's say for instance we know what y2 is then that gives us a differential equation for y1 or we could name them the other way as well okay and then we're going to solve that first order differential equation using the first order linear differential equation solution strategy and that is if we've got a first order linear differential equation y prime plus a of x times y plus b of x then you first set alpha equal to the exponential of the anti-derivative of a of x so that's sometimes known as the integrating factor and that gives you a solution of this form so we've got y equals 1 over alpha and then we've got a constant plus the antiderivative of alpha x times b dx okay good so now we're all set up with the tools as well as the differential equation and we're ready to get to it first by finding a solution via a power series so let's go ahead and assume we have a solution that is a power series expanded at zero we're gonna expand it at zero because we're interested in this interval minus one to one so we wanna put something right in the middle of that interval so here that will look like this we've got the sum as n goes from zero to infinity of a sub n x to the n now what we want to do is calculate the first derivative of that the second derivative of that and plug that into our differential equation then we can mold that into some sort of recursion on the coefficients so let's see how to do that so here we'll have y prime equals the sum as n goes from 0 to infinity of n times a n x to the n minus 1 so that's just term by term differentiation i want to point out here that i don't really need to include the n equals zero and that is because the n equals zero term is zero but i'm going to leave it in there for now next we see that y double prime is equal to this sum as n goes from 0 to infinity of n times n minus 1 a sub n x to the n minus 2 again by term by term differentiation where we differentiated this one and the first two terms there are zero but i'm going to leave them in just for structure now what i'll do is i'll plug in this expansion of y y prime and y double prime into our goal differential equation so that's going to give us 1 minus x squared times y double prime minus 8x times y prime minus 12 times y so let's see what that turns into so we'll have 1 minus x squared y double prime is that object right there we've got this sum as n goes from 0 to infinity of n times n minus 1 a sub n x to the n minus 2 then we have minus 8x times y prime so that's going to be minus 8x and now we'll have this sum as n goes from 0 to infinity of n a sub n x to the n minus 1 there's y prime and then we have minus 12 times y so that's going to be minus 12. and then we've got the sum as n goes from 0 to infinity of a and x to the n and then all of this is equal to zero okay good so now what we're going to do is start to put all of this stuff together so i'm going to take this one minus x squared and distribute it on to every term from this sum so let's see what that gives us so distributing 1 onto that will give us this sum as n goes from 0 to infinity of n times n minus 1 a sub n x to the n minus 2 so that's multiplying everything in that sum by the number 1. and then we will have minus the sum as n goes from 0 to infinity of n times n minus 1 a sub n x to the n notice our x squared built our x to the n minus 2 back up to x to the n now a similar thing is gonna happen here and well that one's already okay so let's go ahead and just bring those down so we'll have minus 8 times the sum n goes from 0 to infinity of n a sub n x to the n our x to the n minus 1 was built back up to x to the n and then we have minus 12 this sum n goes from 0 to infinity of a sub n x to the n and then all of this is equal to 0. now i want to notice that these last three terms are indexed in a way so that i can combine them very easily notice they are all some polynomial in n multiplied by a n x to the n so what i'll do is i'll combine all of these and have them equal some polynomial and n times a and x to the n i'm going to go ahead and take a minus sign out of this because notice there's a minus included in all parts so that's going to give us minus the sum n goes from 0 to infinity we have n times n minus 1 that's from this first term plus 8 n that's from the second term plus 12. so notice those last two signs changed because i factored a minus out of the whole thing good and then all of that is multiplied by a and x to the n and that just comes from the fact that everything in those first sums were multiplied by a and x to the n now what i'll do is use the fact that the first two terms of the sum are zero in other words the n equals zero and the n equals one term are zero because of this polynomial in n which means i can change this starting point from 0 to 2 and then re-index this so i've got x to the n so i can combine it with this thing that's down here so in order to re-index this so that it looks like x to the n i'm going to replace every instance of n with n plus two so let's see what that gives us that's going to move our starting point down two and that's because when n plus two equals two then n equals one and then every n i see here i'm gonna replace with n plus two so that gives me n plus two n plus one a sub n plus two x to the n and so now i've taken this differential equation with this assumed power series solution and that gives me this equation down here involving these two sums and now i can put these two sums together given the fact that they're both indexed in a way so that they have an x to the n term now before we combine everything i'm going to check to see if this polynomial and n can be simplified so i'm going to expand it first so if i expand it i get n times n which is n squared minus n plus 8 n so that's going to be plus 7n plus 12. but that's nice because that factors like n plus 3 times n plus 4 which has some nice structure to it so now that i've got everything in terms of x to the n i can go ahead and write this as a single sum so this is going to be the sum as n goes from 0 to infinity i have n plus 2 times n plus 1 times a to the n plus 2 that's from this first series and then i'll have minus n plus 3 times n plus 4 times a to the n that's from that second series but then this entire thing is multiplied by x to the n and we know it's equal to 0 because we have a homogeneous differential equation now next we'll see that since there's a 0 on the right hand side then that means the left hand side has to always be 0 for all values of x but if this is 0 for all values of x then this thing that i'm under bracing here in orange has to be equal to 0 for all n and that gives me a nice recursion on these coefficients so let's see if we can write that down that means we have a n plus 2 equals n plus 3 times n plus 4 all over n plus 2 times n plus 1 times a n so we've got this two-step recursion but notice since we've got a two-step recursion that doesn't say anything about a0 or a1 so that means that a0 and a1 are free which is typical when you're doing series solutions for second order differential equations okay so now i'm going to go ahead and bring that up and we're going to move on to the next step so in the last board we assumed that we had a power series solution and what we found out was that the first two coefficients were free variables and then we had this two-step recursion that said a n plus two was equal to n plus three times n plus four over n plus 2 times n plus 1 times a n now what i want to do now is focus on only the even terms so in other words the a 0 a 2 a 4 terms and so on and so forth so i'll do that by setting a0 equal to 1 and a1 equal to 0. so notice this means that i'm only looking at a specific solution but that's okay once i have that specific solution i can use the wronskian and able's identity to get the other solution i don't think this is the easiest way to do this problem but it is a way to do this problem in order to review as much as possible so if we set a zero equal to one and a one equal to zero that means that a3 a5 a7 and all of those are equal to zero which means all we need to worry about is a0 a2 a4 a6 and so on and so forth so let's go ahead and maybe write that recursion in terms of an even term so that would be a 2n plus 2 equals 2 n uh plus three times two n plus four all over two n plus two times two n plus one times a sub 2 n now what i want to do is take that a sub 2 in and replace it with the recursion involving it so i'm going to take this a sub 2 n and then replace it with 2 n plus 1 times 2 n plus 2 over 2 n times 2 n minus 1 and then now that's going to be a sub 2 n minus 2. so just to reiterate here i am putting the recursion within the recursion this term right here is a sub 2 and just using this formula up here but now what we can notice is that things are starting to telescope and so notice this 2 n plus 2 is going to cancel this 2 n plus 2 2 n plus 1 is going to cancel this 2 n plus 1 and so on and so forth this 2n will be canceled with something up there this 2 and minus 1 will be cancelled with something up there and so you might say well when does this finish well this is going to finish at the point where we have a sub 2 which is going to be equal to 3 times 4 over 2 times 1 a sub 0 which we're setting a sub 0 equal to 1 so we don't really need to worry about that so by the structure that we have we see that the 3 and the 4 are going to cancel but that 2 and 1 in the denominator will never cancel so that gives us a nice claim and that is that a sub 2 n plus 2 is 2 n plus 3 times 2 n plus 4 over 2 times a 0 but we're assuming a 0 is 1. but we can reindex that slightly to say that a sub 2 n is equal to 2 n plus 1 times 2 n plus 2 over 2 times 1 which is 2 times a 0 which is 1 so we can just leave it like that and now just for practice let's go ahead and prove this by induction great so let's take as our base case we have a sub 2. so notice by the recursion up there a sub 2 is equal to 3 times 4 over 2 times 1. like the calculation that we did before times a0 which is 1. but now notice that that holds this format here so this is equal to 0 plus 3 times 0 plus 4 over 0 plus two times zero plus one where here we're thinking about this two as zero plus two so it fits this thing that we have up here so in other words our base case is satisfied so the next thing that we want to do is suppose this is true for some k bigger than or equal to 1. so in other words we're making the assumption that a sub 2 k is equal to two k plus one two k plus two all over two and then look at the next case so maybe we'll do that by saying and notice that we have a sub 2 k plus 2 is equal to well it's equal to something by this recursion up here so it's equal to 2k plus 3 times 2k plus 4 all over 2 k plus 2 times 2 k plus 1 times a sub 2 k so that's by our recursion that we came up with on the last board but now we can go ahead and replace this a sub 2 k with our induction hypothesis which is right up there so that's 2 k plus 1 2 k plus 2 all over 2 and notice all of the right stuff cancels so 2k plus 2 cancels with 2k plus 2 and 2k plus 1 cancels with 2k plus 1 which leaves a sub 2 k plus 2 in the correct format especially if we consider this to be 2 times k plus 1 plus 1 and this one right here is 2 times k plus 2 plus 2 good so we've got a closed form for our even coefficients okay so i'm going to go ahead and clean this up and then we'll get to finding a closed form for the power series given these even coefficients on the last board we arrived at our first solution and it was the solution with the even terms in other words the a sub two n terms we saw that y one that's what i'm calling this solution was equal to the sum as n goes from 0 to infinity of a sub 2 n x to the 2 n which was equal to half the sum as n goes from 0 to infinity of n 2 n plus 2 times 2 n plus 1 times x to the 2 n and that's because of the recursion we found on the coefficients along with this fact that we were taking an initial condition of a 0 equals 1. now technically we could just factor an a0 out of this and put it right there but let's just set a0 equal to 1 for now now what we can do is use something that we know about series to find the closed form of this sum so this is like a little bit tricky but there are a number of ways to do it what i want to notice is that we've got a descending product of numbers 2n plus 2 times 2 and plus 1. and then we've got an x to the 2n so to me this looks like a derivative of something so what does it look like the derivative of well this whole object right here to me looks like the second derivative of x to the two n plus two notice if we take the second derivative of x to the two n plus two then a two n plus two comes down a two n plus one comes down and then we've got x to the two n so that works out so that means we can rewrite this thing as one half and now we have the sum n goes from zero to infinity i'll write this as the second derivative of x to the two n plus two and then that's it now what we can do is use the linearity of the derivative to bring that outside of the sum so that's going to give us one half and now we have the second derivative of the sum as n goes from zero to infinity of x to the two 2n plus 2 but what i want to do to that is bring an x squared out so all that i have is an x to the 2n on the inside and the reason i want to do that is because now this looks like a geometric series where you have the common ratio of x to the 2n and you have a starting term of 1. so here again we're gonna in other words use the fact that the sum as n goes from 0 to infinity of u to the n is 1 over 1 minus u just a standard geometric series fact and i want to notice that um this is the convergent value of the geometric series because we're on the correct interval so let's see what that gives us that's going to give us one-half and now we have the second derivative with respect to x of this looks like x squared over 1 minus x squared good so let's just talk our way through that so the sum as n goes from 1 to infinity of x to the 2n became 1 over 1 minus x squared because of what we discussed over there in pink and then we've got this like x squared term that's in the numerator because we multiplied by this x squared then we need to take the second derivative of this object because by our argument up here we needed to take the second derivative of this whole power series okay so now let's see if we can simplify this a little bit so maybe you could obviously just like throttle this with the derivative rules maybe the quotient rule twice and come up with a solution but maybe we can use a little bit of a trick so the trick that i want to use is adding 0 in here so i break this into two parts that are a bit easier to take the derivative of so maybe i'll subtract one and then i'll add one and then use the linearity of the derivative so that's going to give me one half and now i have the second derivative of x squared minus 1 over 1 minus x squared so that's what i get this x squared minus 1. still have that denominator and then plus 1 over 1 minus x squared so something like that good but what we notice now is that all of this right here is equal to negative one so when we take its first derivative well that's just going to be equal to zero okay good so now let's go ahead and take the first derivative here so that's going to leave us with one-half and then after taking a first derivative we have one first derivative left over because we're taking the second derivative here and then maybe you want to think about this thing as one minus x squared to the negative one so that you can like easily use the generalized power rule of the chain rule or wrote it whatever you want to call it and notice we're going to be left with 2x over 1 minus x squared quantity squared so that's what you get in the end when you take this first derivative now we're left with this object and i'm going to be honest this object it's a little bit trickier to use one of these tricks in order to take the derivative so i would say maybe just do this on your own with the quotient rule and you'll get the following so you'll get that this is equal to 3x squared plus 1 over 1 minus x squared quantity cubed and that's easy to check just using the quotient rule or whatever so in other words we've got one of our solutions is all taken care of we've got y1 is equal to this thing now we're going to move on to the next step which is using abel's theorem to give us a differential equation for the second solution so let's go ahead and do that okay let's see where we are so we used a power series solution in order to find the first solution to this second order linear differential equation so not only did we find a nice recursion on the coefficients of that power series but we used some tricks to find a closed form for this for that power series and this was the closed form so just to reiterate we've got one of our solutions which we'll call y1 is equal to 3x squared plus 1 over 1 minus x squared quantity cubed and off camera i calculated the derivative of this solution as well because that's going to be useful for our next step it turns out that y 1 prime is equal to 12 x cubed plus 12 x over 1 minus x squared to the fourth and that's just doing the quotient rule on this guy right here it's not super pretty but it's straightforward okay now we want to use abel's identity right here which says if we have a second order differential equation this actually works for a higher order differential equation too but if we have a second order differential equation then we can write the wronskian down two different ways so by the definition which is this y1 times y2 prime minus y1 prime times y2 where y1 and y2 are your two linearly independent solutions to this differential equation but if you don't know the solutions you can calculate the wronskian right from the differential equation and that is the exponential of minus the antiderivative of p of x dx where p of x is the coefficient of y prime when you have molded your differential equation so that the coefficient of y double prime is equal to 1. so now let's get to constructing a first order differential equation for our next solution using able's identity so let's say we need to mold this thing so that the coefficient of y double prime is one so that's fairly straightforward so that'll give us y double prime minus eight x over one minus x squared y prime and then minus 12 over one minus x squared y equals zero good so that wasn't too bad we just divided by one minus x squared so now in the language of abel's identity over here this thing this means that my function p of x is equal to minus 8x over 1 minus x squared good which tells us that our wronskian is going to be equal to the exponential of the antiderivative of negative p of x so that means we're going to have 8x over 1 minus x squared dx so something like that now what i want to do is take that antiderivative so i can do that with u substitution so let's go ahead and notice that if i set u equal to 1 minus x squared then that's going to make d u equal to minus 2x dx but i don't have a minus 2x dx i have an 8 dx so that means that my 8 times d my 8x dx is actually minus 4 d u and i get that by dividing by negative 2 obviously okay so that means i have the exponential of the antiderivative of minus 4 over u d u where u was by that substitution right there so this gives me the exponential of minus 4 times the natural log of u but next i can see that that is the exponential of the natural log of u to the minus 4 where i use my exponent rule for natural logs but finally that is going to be equal to u to the minus 4 or 1 over u to the fourth so i've got 1 over u to the fourth but u was 1 minus x squared so i've got 1 minus x squared all to the negative fourth or in the denominator to the fourth so that's my value of the wronskian using this part of abel's identity and now the wronskian via the definition let's see what that is so we also have the wrons scan will be equal to something like this so we'll have y one y prime i'm going to rename y2 just equal to y because it's like an unknown so y1 we know but y prime we do not know so it's going to be y1 times y prime minus y1 prime times y so in other words we have 3x squared plus 1 all over 1 minus x squared cubed times y prime that would be like this term and then minus well y1 prime which is all of that over there we've got 12x cubed plus 12x all over 1 minus x squared to the fourth times y so that's my wronskian this other way so let's see what we've got we've got a wronskian via abel's identity and then a wrong scan via the definition which gives us a differential equation for our next solution so i'll go ahead and clean up the board and we'll start with that differential equation so in the last board we used able's identity to form a first order differential equation for the second of two linearly independent solutions to our goal differential equation so we found the first solution via a power series and then finding the sum of that power series using some tricks then this differential equation was found by setting able's wronskian so what i mean by that is abel's formula for the wronskian in terms of the differential equation equal to the definition of the wronskian where we take y1 to be one of our solutions and y to be our second unknown solution so check it out this is a first order linear differential equation so that means we've got a strategy for solving it we just need to put it in this correct form y prime plus a of x times y equals b of x which means not only do i need to clear the denominator here but i also need to divide by this 3x squared plus 1. so let's start by clearing this denominator so i'm going to multiply everything by 1 minus x squared cubed that's going to get rid of my denominator so that's going to give me 3x squared plus 1 times y prime minus so this is going to be x cubed plus 12 x that should have been a 12x over 1 minus x squared that's times y and then next i'll have 1 over 1 minus x squared so all of that stuff got cancelled a little bit now we're in good shape i'm going to go ahead and divide by this 3x squared plus 1 term that's going to give me y prime minus so this is going to be 12x cubed plus 12x all over 3x squared plus 1 times 1 minus x squared that's multiplied by y and then this is going to be equal to 1 over 1 minus x squared times three x squared plus one again because i divided by this three x squared plus one now i'm gonna clean this up a little bit i'm gonna take this minus sign and pull it through to this one minus x squared to change that to an x squared minus 1 just so i don't have a hanging minus sign right here so that means my alpha function is going to be the exponential of the antiderivative of this coefficient of y given my strategy for solving a first order linear differential equation so that means we need to find the anti-derivative of this thing which i will box in purple so how do you find the anti-derivative of such a thing well the best way to do it is probably to use a partial fraction decomposition so let's maybe set up our partial fractions over here so that we can maybe put the calculation of the antiderivative down here so we're going to want 12x cubed plus 12x all over 3x squared plus one and then i'm gonna write this x squared minus one as x plus one x minus one so notice this will split up into a over x plus one plus b over x minus one plus c x plus d over three x squared plus one notice that three x squared plus one is an irreducible quadratic next thing that i will do is i'll multiply by something that will cancel all of the denominators and that will be this product over here on the left hand side in the denominator that gives me 12 x cubed plus 12 x equals so that's going to be a times x minus 1 times 3x squared plus 1 and then i have plus b times x plus 1 times 3x squared plus 1. and then finally i have cx plus d times x squared minus one i'll just push that x plus one and the x minus one back together now what we want to do is notice i've got a cubic polynomial on the left hand side after all is said and done i also have a cubic polynomial on the right hand side which means i have constant terms coefficients of x coefficients of x squared and coefficients of x cubed so i'll set all of those equal to each other from both sides of the equation so i'll start with my coefficients of x cubed then my coefficients of x squared my coefficients of x and then my constants which i'll just denote as 1. and instead of expanding that whole right hand side i'll just kind of reason my way through it so what are the coefficients of x cubed well from this first term i have x times 3x squared times a so in other words i have 3a from this second term i'll have 3b for essentially the same reason and then for my third term i will have c and that's it plus c and then over there on the left hand side of the equation i have 12 so all of this is going to be equal to 12. now let's look at the coefficients of x squared on both sides of the equation so for this term i will have minus 3a that's from the a times -1 times 3x squared so i have minus 3a so for the second term i'll have 3b so that's going to be plus 3b and then for the third term let's see what i'll have i'll have d so that's going to be plus d and then over here on the left hand side of the equation we see that there's no x squared term so that means that that is all equal to zero so let's see what we have for the coefficients of x on each side so from this first term i have an a and that's going to be actually so for this first term i have an a for the second term i'm going to have a b and then for this third term for the x term i will have a c so that's going to be plus c and that's going to be equal to 12 because i've got a 12x on the left hand side now what about the constants so for this first term i'm going to have a minus a for the second term i'll have plus b and then for the third term i'll have minus d and that's going to be equal to 0 because there's no constant on the left-hand side of the equation i'll leave it to you guys to solve this system of equations any way that you want but i'll just write the solution right here so when all is said and done we have a equals b equals three c equals minus six and then d equals zero so now we can go ahead and calculate our alpha function so our our alpha function will be the antiderivative so our alpha function will be the exponential of the antiderivative of this thing right here which is in purple but this thing right here in purple can be ripped apart into these three rational functions with those coefficients down there by all the work that we just did so let's see we've got alpha of x is equal to the exponential of the antiderivative of so we've got three over x plus one so that's like our first term and then we've got plus three over x minus one that's our second term and then finally we have minus six x over uh three x squared plus one and then all of that is dx and all of that is in the exponent so each of these has fairly simple anti-derivatives so let's just jump to that so we're going to have the exponential of 3 times the natural log of x plus 1 plus 3 times the natural log of x minus 1 and then minus 2 times the natural log of 3x squared plus 1 like that great so now we can mash all of those together and that's going to give us this exponential of so we're going to have the natural log of x plus 1 quantity cubed times x minus 1 quantity cubed but that's x squared minus 1 quantity cubed and then we'll have a denominator because there's a minus sign there and we'll have three x squared plus one quantity squared then we have a natural log inside of an exponential so those are going to cancel each other out leaving leaving us with x squared minus one cubed all over three x squared plus one quantity squared so there is our alpha function okay so let's maybe go ahead and clean up the board and then we'll pick up right after this step so we've used lots of tricks to get down to this point so we've got one solution to our goal second order differential equation we know our second solution satisfies the following first order linear differential equation the integrating coefficient was calculated to be this i want to say on the last board i had all of this stuff in the denominator squared but that was a little bit of a typo there are probably already comments about it um but i want to say that it's fixed now that should be just be three x squared plus one now we're going to use this strategy for solving a first order linear differential equation so we know that our next solution which we'll just call y will be equal to 1 over alpha so that's going to be 3x squared plus 1 over x squared minus 1 cubed times the quantity so i'm going to leave this constant off because notice that choosing this constant equal to -1 just gives us back our original solution so it doesn't give us any new information so in other words i'm just going to set this constant equal to zero because all we really want is two linearly independent solutions then we can write our arbitrary solution as a linear combination of those so that means i've got this which is our one over alpha times the anti-derivative of alpha times b so that's going to be this thing right here times this thing right here so notice that this x squared minus 1 cubed is going to cancel with this 1 minus x squared down here but we're going to pick up a minus sign so i'll go ahead and just put the minus sign in front of the whole thing so maybe we'll put it like this and then we'll have x squared minus 1 quantity squared over 3x squared plus one quantity squared dx so we have something like that so now all we have to do is take this antiderivative we multiply it by this function out front and then we will have our second of two linearly independent solutions to our differential equation the first one we found up there so since this is going to be kind of a difficult anti-derivative i'm going to box this thing in yellow so that i can rewrite it down here just as yellow box equals so let's look at this kind of thing so it's a rational function and in addition it's got this thing in the denominator which hints us towards probably using trigonometric substitution so let's notice that this hints towards the substitution x equals 1 over the square root of 3 times tangent theta and that's because that turns 3x squared plus 1 into tangent squared theta plus 1 which is equal to secant squared theta but notice the way that 3x squared plus 1 appears in our integral is being squared so that means we can go ahead and square all parts of this which means in fact what we really have is a secant to the fourth theta in the denominator now let's go ahead and calculate our dx component so we know the derivative of tangent is secant squared so that tells d us dx is 1 over the square root of 3 secant squared theta d theta so let's see what we get from inserting all of this stuff into our yellow boxed anti-derivative so we're going to have yellow box equals the anti-derivative of so let's see our numerator right here is x squared minus 1 squared so that's going to be one 3 tangent squared theta minus 1 quantity squared so that's what we get from this being inserted in here that's good then we've got dx but notice that dx is 1 over root 3 secant squared d theta so i'm going to write the 1 over root 3 right here then i have secant squared theta d theta right here then let's look at our denominator we calculated our denominator in full over here and that gave us secant to the fourth theta so it's pretty clear to see that we can take this secant squared cancel this down to a secant squared and then keep in mind that secant squared is the same thing as 1 over cosine squared theta so we can change this division by 1 over cosine squared to multiplication by 1 over cosine squared so let's go ahead and do that that's going to give us 1 over the square root of 3 then we have the antiderivative of cosine squared theta times 1 3. i'm going to rewrite tangent squared theta as sine squared over cosine squared so sine squared theta over cosine squared theta and then i have minus 1 quantity squared and then d theta so i have something that looks like that okay so this is shaping up now next thing that i'm going to do is notice that this object right here this cosine object is squared and then this entire sine squared over cosine squared minus one thing is also squared so i can push those together and then square them all at once so let's see what that gives us that gives us 1 over the square root of 3. i have the antiderivative of so that's going to be when i bring this cosine inside it won't be squared anymore so that's going to give me 1 3 sine squared over cosine so notice the cos cosine squared came inside and became a cosine canceled one of those cosine squared in the denominator and then i'm going to have minus cosine theta so i have something like that and then this entire thing is being squared and then i've got d theta that's still my antiderivative here so just to reiterate when this thing entered this squared binomial you know we have to say that it's not squared anymore because it entered the house where it's already squared okay and then distributing that cosine onto both of those terms gave us this thing right here fantastic so now let's multiply this out and see what we get so we get 1 over the square root of 3 then we have the antiderivative of so this thing is going to be 1 over 9 and then we have sine to the 4th over cosine squared so i'll write this as sine to the fourth theta over cosine squared theta and then let's see my cross terms so i'll have minus two thirds this times this but that's going to be minus two thirds sine squared minus 2 over 3 sine squared theta and then finally plus cosine squared theta and then that's what i get when i multiply all of this out okay so now the next thing that i maybe want to do is take this sine squared and rewrite it as 1 minus cosine squared so if i rewrite that as 1 minus cosine squared then i can combine it with the cosine squared in the next part so let's see what that gives us we're going to have 1 over radical 3 still and then 1 9 sine to the fourth theta over cosine squared theta so that part i'm not changing yet and then next i have uh minus two thirds that's kind of by itself and then plus two thirds cosine squared plus three thirds cosine squared so that's going to be plus five over three cosine squared theta and then all of this is d theta so how did we get to that well we got to that by adding this cosine squared which is multiplied by two thirds to that cosine squared which is multiplied by one okay great so now maybe i'll go ahead and bring this up here and we'll keep chugging along so let's see where we are we've got this yellow boxed integral is one over root three and then we've got the antiderivative of one over nine sine to the fourth over cosine squared minus two thirds plus five thirds cosine squared so that's what we ended up with on the last board now what i want to do is write this sine to the fourth over cosine squared as sine squared over cosine quantity squared so that may not seem super helpful but it will be helpful because we can take this sine squared and rewrite it as 1 minus cosine squared theta so let's see what that gives us so we've got 1 over root 3 then we have the antiderivative of one over nine and then we're going to have one minus cosine squared theta over cosine theta and then all of that is being squared and then next minus two thirds plus five thirds cosine squared theta d theta okay so we're there now we can simplify this thing a little bit notice we can simplify this by rewriting it as secant theta minus cosine theta quantity squared notice we've got 1 over cosine and then cosine squared over cosine so that's actually a nice way to write that now let's see what we get we can multiply this thing out we have one over root three anti-derivative of one over nine then we'll have secant squared theta minus two times secant times cosine secant times cosine is 1 so that's going to be minus 2 and then minus cosine squared theta sorry i meant that should be plus cosine squared theta okay fantastic and now next we can see that we have minus uh two thirds and then plus five thirds cosine squared theta and then all of that is d theta so that's where we are now so now let's see if we can combine anything so notice we've got this 1 over 9 cosine squared and then over here we have this 5 over 3 cosine squared so let's maybe bring this up and then we'll start putting things together here and finishing it off i brought the last thing that we had up and i expanded it all out which really just involved multiplying through by this 1 9. now we're going to give something some common denominators and start combining so i'll give this thing a denominator of nine that's going to turn it into six over nine i'll give this thing a denominator of nine as well that's going to turn it into 15 over nine that's so we can combine it easily with this stuff right here so let's see we're going to get this is equal to 1 over the square root of 3 then we'll have the antiderivative of 1 over 9 secant squared theta and then let's see what we've got we've got minus two over nine minus six over nine that is minus eight over nine and then we've got one nine plus fifteen over nine that's going to be plus 16 over nine cosine squared theta and then all of this is d theta fantastic now this guy right here secant squared is pretty easy to find the anti-derivative eight ninths is pretty easy to find the anti-derivative the 16 9 cosine squared is a bit tricky but not too bad if you use the exponential version of cosine squared i should say the complex exponential version of cosine squared so we'll use the fact that cosine squared theta is the same thing as e to the i theta plus e to the minus i theta over 2 quantity squared so let's see what we get when we multiply that out so when we multiply that out we're going to get 1 over 4 so that's what we get from the denominator and then that is going to be multiplied by e to the i 2 theta that's e to the i theta squared and then next it'll be plus two that's because e to the i theta plus times e to the minus i theta is one and then finally plus e to the minus uh two i theta so maybe this is two i theta as well okay so we've got something like that so now we can take the 16 over 9 cosine squared take the 16 and have it cancel with this 4 and that's going to give us 4 over 9 and then we'll have e to the i 2 theta plus 2 plus e to the minus i 2 theta or minus 2 i theta whatever order you want to put that in so now notice some things are going to cancel we've got a minus 8 over 9 and then here we have a plus 4 over 9 times 2 so that is going to cancel with that so just to reiterate this is the complex exponential expansion of this 16 over 9 cosine squared so let's see what that leaves us with we've got 1 over the square root of 3 anti-derivative of 1 9 secant squared theta notice our constants are gone now and we're gonna have plus four over nine e to the two i theta and then plus four over nine e to the minus two i theta and now all of this is d theta so we've got something like that now let's go ahead and take the antiderivative which we can do now so this is going to be 1 over the square root of 3. the antiderivative of secant squared is tangent so we've got 1 over 9 tangent squared theta so that's pretty good and then we've got plus 4 over 9 and then let's see what we get for each of these parts so this first part will be e to the i 2 theta over 2 i so that's what we get for the first part and then this second part is going to be e to the i 2 theta over 2 i but then it picks up a minus sign because of this minus sign right here great so now what i want to notice is that this gives us 1 over the square root of 3 we have 1 9 tangent squared theta plus 4 over 9 and then the rest of it is the complex exponential version of sine 2 theta so we've got all of this is equal to sine 2 theta so that's good so let's go ahead and maybe bring that up and then finish it off so we've come a long way we've got this yellow box thing has been integrated down to this stuff right here but this is in terms of theta whereas our original variable was x but we can take care of that using the fact that x equals 1 over the square root of 3 times tangent so that means that this is x over 9 because notice we have 1 over square root of 3 times tangent times 1 over 9. so that part's easy then we're going to need to use a double angle formula for this sine in order to make it work out nicely so let's recall that sine of 2 theta is the same thing as 2 sine theta cosine theta so that's going to give us 8 over 9 sine theta cosine theta now we need to take care of this sine and this cosine and we're going to do that by completing a triangle out of this thing right here so let's go ahead and notice that this is equivalent to saying that tangent of theta is equal to x times the square root of three but we can complete a right triangle out of that so if we let this be angle theta tangent is opposite over adjacent so that's gonna be x times the square root of three and then one so that's going to tell us that this thing right here is the square root of three x squared plus one good but from that we can extract what sine and cosine are pretty easily so notice that that tells us that sine of theta which is opposite over hypotenuse is equal to x times the square root of three over square root of three x squared plus one and then cosine theta which is equal to adjacent over hypotenuse so that's going to give us 1 over the square root of 3x squared plus 1. great then i should say we still need a square root of 3 here that comes from distributing that 1 over the square root of 3 out front so that means we can take the product of sine and cosine and replace this product of sine and cosine with x times the square root of three over three x squared plus 1. so let's see what we get in the end we're going to have this is x over 9 and then we're going to have plus 8 x over 9 times three x squared plus one so that's what we get in the end for this anti-derivative now let's maybe mash this stuff together a little bit to see if we can simplify it at all so let's give these things a common denominator we can do that by multiplying this by three x squared plus one in the numerator and in the denominator so three x squared plus one like that so let's see that's gonna give us uh 3x cubed plus x plus 8x all over we're going to have 9 times 3x squared plus 1. so let's see this x plus 8x will give us 9x then we can factor a 3x out of the numerator that's going to leave us with 3x and then x squared plus 3 is what's left over over 9 times 3 x squared plus 1. so notice that 3 and that 9 will cancel and leave us with just a 3 in the denominator so we have x x squared plus three over three x squared plus one good so now i'll go ahead and bring that here and then we'll combine it with what we know right there and then we'll be essentially done so we finally got the anti-derivative of this yellow box that doesn't mean that we have our second solution our second solution is that yellow box box times this thing right here so let's maybe go ahead and write that down so our second of two linearly independent solutions maybe we'll call this y2 is going to be yellow box times this thing but since i've got a homogeneous differential equation i can really take any constant multiple of that i might as well take the constant multiple where i don't have a minus sign there so notice that this 3x squared in the numerator is going to cancel with that 3x squared in the denominator i'm going to be left with x times x squared plus 3 over x squared minus 1 quantity cubed so that's my second solution recall my first solution was up there which i boxed in green so now we've got two solutions to this differential equation which means my most general solution will be an arbitrary linear combination of these two so in other words it'll be y equals c one times this thing so three x squared plus one over one minus x squared quantity cubed and then plus c two times this thing so that's going to be x x squared plus 3 over x squared minus 1 quantity cubed so notice we can maybe put this stuff together if we want to well maybe we can change this x squared minus 1 to a 1 minus x squared that's equivalent to absorbing a minus sign into the c2 and then we notice we've got a common denominator already so we're actually good to go we've got our most general solution is 3x squared plus 1 times c1 plus x x squared plus 3 times c 2 all over 1 minus x squared quantity cubed and so that's a good place to stop

Channel: Michael Penn

Views: 44,062

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: yncPeXiRdck

Length: 60min 56sec (3656 seconds)

Published: Thu Oct 15 2020