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(a) Seek power series solutions of Airy differential equationy''−xy = 0 about x = 0; find the recurrence relation.(b) Find the first fo...

Question

(a) Seek power series solutions of Airy differential equationy''−xy = 0 about x = 0; find the recurrence relation.(b) Find the first four terms in each of two solutionsy1 and y2.

(a) Seek power series solutions of Airy differential equation y''−xy = 0 about x = 0; find the recurrence relation. (b) Find the first four terms in each of two solutions y1 and y2.



Answers

a) Find all solutions of the recurrence relation $a_{n}=2 a_{n-1}+2 n^{2} .$
b) Find the solution of the recurrence relation in part (a) with initial condition $a_{1}=4$ .

In part a were given a linear, homogeneous recurrence relation. We were asked to find the characteristic roots of this relation. The relation is a n equals and minus four. So we have that The characteristic equation is are to the fourth minus one equals zero. Who's tells us that are the forced equals one were that r squared is equal to plus or minus one and we have that If r squared is equal to positive one that we have that fire is equal to plus or minus one. And if our square is equal to negative one, then our is equal to plus or minus I. So we have four characteristic roots, plus remind us one and close reminds I in part B whereas to find the solution to this recurrence relation with given initial conditions, a zero equals one, a one equals zero, a two equals negative one and a three was one. So because the characteristic roots are all distinct, the general form for a solution is Alfa one times one to the end, plus Alfa tube times negative one to the end, plus out for three times I to the end plus out before times negative I to the end. This can be written also as Alfa one plus Alfa two times negative one to the end plus Alfa three times I t n plus Alfa four times negative I to the end Where Alfa One girl for Ford are some constants. Okay needs more initial conditions. We have a system of equations one is equal to and is equal to zero We get for one and native one to the zero is still once after one plus off it too Plus Alfa three plus Alfa four we have zero She's a one is equal to Alfa one minus self too. Plus I offer three minus I off before we have the negative one which is a to is doubtful one plus Alfa to minus out for three and then negative I squared It's going to be minus Alpha Ford And finally one which is a three is equal to outfall one minus healthy too minus Alfa three I or I offer three and then it becomes plus I former so you obtain the system of equations. So to solve this system I want to use calc and elimination. So I have the matrix 1111 one. Next row is one minus one. I negative, I zero. The next row is 11 Negative one negative one negative one in the last row is one negative one Negative. I positive I one and so first else attract Row one from rose. 23 and four. So I have 1111 one. And the next row. I have one minus 10 Negative one minus one is negative. Two. I'm minus one. Is I minus one and negative. I minus one. This negative. I minus one euro minus one is negative. One, They're too real. I have one minus 10 one minus 10 Negatively. Minus one is negative. Two negative. Minus one is naked too. And negative. One minus one is negative. Two and finally in the fourth row one minus 10 I could have one. Minus one is negative. Two negative. I minus one is negative. One minus. I and I minus one is negative one plus I one minus 10 This could be simplified to 1111 one second row Divide by negative too. To get zero one and then negative one over. Negative two is 1/2 minus I over to and negative one minus. I were negative. Two is 1/2 plus I to and negative one, divided by negative two is just 1/2 in the third road, about the negative to again against 00 11 one in the fourth row. Divide by negative to get zero one and then 1/2 plus I over to and don't have minus I ever to and zero over. Negative two is just zero, and now subtract wrote to from row one and subtract two from a four. So I get one. Minus zero is one one minus one is zero one minus 1/2 minus. I over to. He's going to amass half is 1/2 and then minus negative my to his plus iron, too. And one minus 1/2 plus I over to is 1/2 minus. I over to, and one minus 1/2 is 1/2 the second row. I still have 01 1/2 minus I over to 1/2 plus I over to 1/2 in Row three. I still have 00 11 one, and in Row four has your A minus 00 one minus one is zero 1/2 plus I over two minus 1/2 minus I over to yes, I over two plus I ever to just simply I and 1/2 minus Iver to minus 1/2. Plus I over to is negative iron to minus iron, too, which is negative. I and zero minus 1/2 iss negative 1/2 simplify further. I'll subtract I Times Row three from Road for and with the weather additions, so I'll subtract 1/2. Plus I ever to times wrote three from roll one. So I get 10 1/2 plus I were to minus one have plus I over to which is zero and 1/2 minus eight to minus 1/2 plus ever to is negative I ever to minus cyber to just negative I and lastly it 1/2 minus 1/2 times 1/2 plus I ever to this is the scene as when half minus the fourth, which is 1/4 minus. I have before So 1/4 minus cyber for row two. Elsa tricked when half minus I ever to, but for a three. So I get zero zero. Sorry. 01 and then when half minus I ever to minus 1/2 minus I ever to zero and would have plus I over to minus 1/2 minus over to is ever to plus I retune judges I and 1/2 minus made a mistake here. This should be instead of 1/4 minus I ever. Four. This is 1/2 minus when half plus I ever to just simply negative I ever to. And here we have 1/2 minus 1/2 minus Iver To which is positive I ever to for three is going to stay the same 00111 you know for I'll subtract I time 03 So I get 00 I minus I zero negative I minus eyes negative to I and negative on half minus I It's simply new Given half months, I you'd notice that defied the last row by negative to I So this becomes one then negative 1/2 minus I over a negative to I for most played. I am by I we get negative 1/2 I plus one over the opposite may get to, which is just too So this becomes 1/2 minus ire before and now I will add row four times I to row one. So I get 100 and negative I plus I zero native I to plus i times when half minus i ever four. This is the same in its native Iver. To pull cyber to that become zero and then negative I squared before is simply positive. 1/4 we grow to on going to subtract items road for so get 010 i minus eyes Europe and then I ever to minus high times 1/2 minus I ever for it's the ire to minus the high times when half zero and then the positive I square before becomes negative 1/4 and I'm going to subtract Row four from Road three. So I get 0011 minus 10 and then one minus 1/2 minus. I, before minus 1/2 is 1/2 and then minus negative. Either for its plus I have before and finally roof or is the same. And so I obtained system of equations solutions out for one equals 1/4. Health of two equals negative. 1/4 offer three equals 1/2. Plus I ever for and Alfa four equals 1/2 minus. I have before and so plugging these back in we got the solution. These initial conditions is 1/4 minus 1/4 times negative one to the end. Plus when Half plus I over for times I to the end plus when half minus I over for times negative I to the end. This is our answer.


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