y'' +(x^2)y = 0 x_0 = 0

i need to find the 13 term using the power series..

Question

y'' +(x^2)y = 0         x_0 = 0

i need to find the 13 term using the power series..

amistre64 · Answer

assume that a power series fits

y = sum an x^n ; n=0 to inf

y' = an n x^(n-1) ; n= 1 to inf

y'' = an n(n-1) x^(n-2) ; n=2 to inf

and fill it in

amistre64 · Answer

$$\sum_2^{inf}a_nn(n-1)x^{n-2}+x^2\sum_{0}^{inf}a_nx^n=0$$

adjust to get like terms and summations

$$\sum_2^{inf}a_nn(n-1)x^{n-2}+\sum_{0}^{inf}a_nx^{n+2}=0$$

$$\sum_2^{inf}a_nn(n-1)x^{n-2}+\sum_{0+4}^{inf}a_{n-4}x^{n+2-4}=0$$

$$\sum_2^{inf}a_nn(n-1)x^{n-2}+\sum_{4}^{inf}a_{n-4}x^{n-2}=0$$

$$a_11(1-1)x^{1-2}+a_22(2-1)x^{2-2}+\sum_4^{inf}a_nn(n-1)x^{n-2}+\sum_{4}^{inf}a_{n-4}x^{n-2}=0$$

$$a_11(1-1)x^{1-2}+a_22(2-1)x^{2-2}+\sum_4^{inf}[a_n(n^2-n)+a_{n-4}]x^{n-2}=0$$

amistre64 · Answer

got my forst ns alittle off in the front; shoulda been 2 and 3, not 1 and 2

amistre64 · Answer

$$a_22(2-1)x^{2-2}+a_33(3-1)x^{3-2}+\sum_4^{inf}[a_n(n^2-n)+a_{n-4}]x^{n-2}=0$$
$$2a_2+6a_3x+\sum_4^{inf}[a_n(n^2-n)+a_{n-4}]x^{n-2}=0$$

now we can define an in terms of an-4

$$a_n(n^2-n)+a_{n-4}=0$$
$$a_n(n^2-n)=-a_{n-4}$$
$$a_n=-\frac{a_{n-4}}{(n^2-n)}$$

amistre64 · Answer

gotta check my maths :)

amistre64 · Answer

but, do you see how it works?  any questions?

amistre64 · Answer

$$a_4=-a_0\frac{1}{12};\ a_5=-a_1\frac{1}{20};\ a_6=-a_2\frac{1}{a};\ a_7=-a_3\frac{1}{b}$$since a2 and a3 = 0; all the a6+4k and a7+4k parts go zero and we can ignore them

this leads us to defining the solution in terms of 2 unknowns a1 and a2 and a rule to define their denominators if one can be found

amistre64 · Answer

a0 and a1 are the unknowns ..... had another typo

amistre64 · Answer

when i work this out to the 13th term I get:
$$a_{28}=-a_0 \frac{1}{4(3)8(7)12(11)16(15)20(19)24(23)28(27)}x^{26}$$

amistre64 · Answer

assuming im reading it right to begin with :)

amistre64 · Answer

$$a_{2n-1}=a_0\frac{(-1)^n}{4(8(n-1)^2+3(n-1)+3)}x^{4n-2}$$for all the even terms :)  and with that im done

anonymous · Answer

awesome thanks! my mistake was I was i tried to move n= -2.. idk if you can do that, but i tried and it didn't work out so much! so quick question..  the 13th term is not $$a _{13}$$? i am just learning this stuff and i wasnt sure if that was what they meant by 13 term.

amistre64 · Answer

i think the 13th term is regarded after removing the zeroed out portions.

since my construct has has counting ns as:  4,5,8,9,12,13,16,17 ...  then the 13th term would be a28

anonymous · Answer

Oh ok thanks for your help!!