[MUSIC] Okay, now the, just one last question in this section. To think about the second order differential equation, okay? That's xy double prime plus 3y prime -xy is equal to 0, okay? Then you can immediately see that the point X is equal to 0 is a regular singular point, okay? And then we would like to solve this differential equation for the positive real axis, okay? About the point X is equal to 0, okay? Regular singular point, okay? So we said s user, y is equal to sum of 0 to infinity of C of n, X to the n plus ry, okay? Plugging this expression into the differential equation, okay? Then you will get from the first, you will get some of these routine finished second derivative so, n+r, n+r-1 times the C of n, X to the n+r minus 2 plus 1 so it will be -1, okay? And plus 3 times of summation, okay? N+r, C of n, X to the n+r-1 as summation is from the n is equal to 0, right? And the minus xy, that is a minus of summation of 0 to infinity C of n, X to the n+r+1, okay? That must be equal to 0, okay? So there we need the the shifting of indices, right? Definitely. For the first sum, rays descend by 1, okay? So that it will be the X to the n+r. So raise this n by 1, so n+r+1, n+r, C of n+1, X to the n+r. Instead submissions should start from -1, okay? For the second part, okay? Same thing, right? Raise this n by 1, then 3 summation of n+r+1, C of n+1, X to the n+r, okay? And summation started from -1, right, okay? And for the last one, okay? We have to lower this n by 1, okay? So lower this n by 1, C of n-1, X to the n+r, instead raise it by 1. So we get this one, okay? So when n is equal to -1, okay? We have to such terms from the first and the second, right, okay? There you will get r times r+2 times this is 0, X to the n-1, okay? Okay? When N is equal to -1 from the first and the second, you get this. When n is equal to 0, when n is equal to 0 from the first and second again, you will get r+1 times r+3 times the C1, X to the r, okay? When n is greater than or equal to 1, okay? You get from 1 to infinity of n+r+1, okay? N+r+3, Cn+1- Cn-1 times X to the n+r, okay? You can simplify it in that way, okay? So for this series to be identical to 0, okay? If and only if, okay? All the coefficients of X to the distinctive powers will be equal to 0, right? So from the very first one, because we are going to assume that C0 is non zero so you have a so called the initial equation, okay? From the second, you will get r+1 times r+3 times C1 equal to 0. From the very last one you will get the reconciliation, okay? Say n+r+1, okay? N+r+3 time C of n+1- Cn-1 equal to 0 for all n greater than or equal to 1, okay? For the coefficients we get these three equations, okay? From this, we get two indigenous roots, larger one equal to 0, smaller one is equal to -2, right, okay? So first let's look at the larger in dissolute, r1 equal to 0, right? Also from this, yeah, this one before or the larger in dissolute r1 equal to 0, okay? When r1 equal to 0, the second equation becomes 3 times the C1 equal to 0, right? So what means, simply C1 equal to 0, right? And the two term reconciliation, okay? It becomes, okay? It becomes C of n+1 equal to 1 over n+1, and times n+3 times C of n-1, okay? It holds for n greater than equal to 1, okay? We get this relation, okay? This is a total reconciliation. Because C1 equal to 0, then C1 will determine C3, C3 will determine C5, and so on, right? Because C1 equal to 0 from this, you can immediately see that all the C1, C3, C5 and so on, k equal to 0, right? In other words, all the other coefficients, right? Will be equal to 0, okay? How about the even coefficients then? C of 2n, okay? From this one, you can read it as 1 over, okay? That's 2n times 2n+2, C of 2n-2, right? Okay, is it okay, right? That's the pattern we have from this one, okay? Who wants to further, okay? This is equal to 1 over, let me write it in this way. 1 over 2 squared, n times n+1 times C of 2n-2, right? This is 1 over 2 squared n times n+1 times, okay? Then C of 2n-2 will be 1 over 2 squared, okay? Okay? 1 over 2 squared n-1 times n, C of 2n-4, right? By the pattern, right, okay? You can keep going, and finally you will get the following things, right? This will be 1 of the 2 squared n times n+1, times 1 over 2 squared n-1 times n, dot dot dot, 1 over finally, 2 squared times 1 times 2 times C0. And this relation holds for all n greater than equal to 1, right, okay?. Can you read it, right? Of course we can write it, let me have yeah, this is equal to, right, okay? How many 2 squared do we have, okay? Read it, 1, the very next time will be 2 and so on n, okay? So that means that you have 1 over 2 squared n, 2 to the n, 2n, n times n-1 times and finally 1 that is n factorial. N+1 times n times n so on times 2 that it will be the n+1 factorial and the C0. It holds for all n greater than or equal to 1, okay? Test the formula for the even coefficiencies 2ny, okay? So, okay? Because if we may take C0 to be any non zero number, right, okay? So, I will take the C0 is equal to 1, okay? As the user, okay? Take C0 is equal to 1, okay? Then we get the very first solution y1 will be, okay? This will be sum of this root infinity of C of 2n, X to the 2n, right? Because all the other coefficients vanish so we only have even coefficients, right, okay? And the value C2n is given by the formula. So we can immediately write it as a summation from zero to infinity of 1 over 2 to the 2n n factorial, n+1 factorial, okay? And actually 2n, okay? And that is the sum of infinity of 1 of 1 factorial, n+1 factorial times X over 2 to the 2n, right? That's the solution, okay? Okay? Now let's look at the smaller industry route, say alpha is equal to -2, right, okay? Now, therefore the smaller in ratio route, okay? Again, the recurrence relation we found, okay? Says that C1 equal to 0, and the total reconciliation, okay? It will be reduced to 2, okay? The reconciliation it will be at the n-1 times n+1 times C of n+1, and minus C of n equal to 0 for all and greater than equal to 1, okay? The reconciliation becomes this one, okay? When n is equal to 1, okay? For n is equal to 1, it becomes 0 times of C2 plus minus C0, Is equal to 0, so that, You will have C0 is equal to 0, okay? Yeah, but C0 is equal to 0, but C observatory, okay? And from that on, okay? When n is greater than equal to 2, Cm+1 is equal to 1 over m-1, m+1 times C of n-1 for all m greater than or equal to 2, right, okay? For all the others, right? Again, C1 equal to 0 means C3 will be equal to 0. C3 equal to 0 means that C5 will be equal to 0. So again from this and that, right? It implies that C1 equal to C3 equal to C5 equal tends on, they're all equal to 0, right, okay? Okay, we do not have any other tumble, right, okay? And look at the recurrence relation again, this reconciliation, okay? So for the even terms we have C0 is equal to 0, C2 is arbitrary, and all the other the even coefficient sees up to n will be determined by C2, by this reconciliation, right? How C of 2n is equal to from look at this pattern, right? That is the 1 over, okay? 2n minus 2 times 2n, C of 2n minus 2, right? In other words this is equal to 1 over 2 squared n-1 times n and C of 2n-2, right?. Do the same thing as before, right? This is the same as 1 over 2 squared, n times the C of, no, I'd like to go the one step further then. This will be 1 over 2 squared, n-2, n-1, C of 2n-4, right, okay? You will get 1 over 2 to the 2n -1 and n-1 factorial, and n factorial, times the C of 2 for all n greater than equal to 2, okay? But if you plug the n is equal to 1, okay? N is equal to 1, this will be C2. When n is equal to 1, the coefficient is 1. So that is the C2, So, okay? This relation, okay? Which holds also for and it's equal to 1, okay? So what it means then for the the for the solutions right. Can't miss correct responding to the smaller individual route all is equal to minus two, okay. The following is top solution we are expecting will be, okay. Y two becomes okay. Extra T -2 the very first equipment and C zero. Do you remember C zero C zero is 0 right? 0 times 1 plus, okay. What is C one? C one equal to 0. Right. Zero times X. Right. Plus then possibly some non-0 coefficients will remain, from C2 and C4 and so on. Right. So see it's up to N X to the 2 N. And started from one to infinity. Right? That's the solution, we already found what is C 2 N. 6 of 2 N is given by this 1. Okay, so plugging that one then this is equal text three minus 2, okay, summation from 1to infinity Instead of seats of 2N you will get 1/2 2 d two n -1 and -1 factorial A and factorial times, okay see to write C2 It is a common so take it out over there. Okay. And that's 32 n. Right, Okay. That's the solution we're expecting for the smaller industry route are there r2 is equal to minus 2.. Right? What is it? This is the C 2 times summation 1 to infinity of 1/2 22 and minus 1 and minus 1 factorial and Victoria and this is actually 2 N minus 2 Right? Okay. Look at this. Okay. Here's the final touch. Right? Raise dissent by 1 raise dissent by 1. And this is the sea to some of 1/2 to the 2N, because I'm raising this end by 1 and factorial and plus 1 factorial to times X to the 2N and instead of raising this end by 1, you must lower this range by1. Okay. So some nations will be from 0 to Infinity. Right. Okay. Now I think you can recognize it. What is this pot? That's exactly the force of solution we get for the larger individual route. Right, Okay. So, therefore our conclusion is the word for the smaller Asian industry lose the solution we obtain is just the constant multiple of the first one. Okay. Is it Okay. Right. That means what, we have, let's go back to the, roots of the industrial equation what was zero and -2? Right. We have so, In summary, we have, For the larger industrial to 0 we obtained a Solution Y 1. Okay. This is flow venous type solution For the smaller industry route r2 is equal to minus 2 we get solution Y 2 Is it simply a constant multiple of Y1, C2 Y1 of X. Right, okay. They are trivially linearly dependent. Okay. So, we can conclude that the frivolous method of finding the city solution. Right? Okay. Failed to give us 2 linearly independent solutions. Okay. So in this case, okay, we can conclude gifts only 1 okay. Only one Lynn linear independent solution, right? Of tai, sum of 0 to infinity of next to the M plus survey okay. But because the given differential equation, the second order, finding a general solution of the secondary differential equation means we need to linear independent solutions, but we have only one. Okay, so 2nd linear independent solution, right? Not of this type, but we need the Y two K which is a linear independent from this, the very first one. This is unknown yet. Okay, so that will be the task we have to do in the next section. Okay?
