## anonymous 4 years ago Use power series to solve the differential equation $y''+xy'+y=0$ $y=\sum_{n=0}^{\infty}$ $y'=\sum_{n=1}^{\infty}a_nnx^{n-1}$ $y''=\sum{n=2}^{\infty}n(n-1)a_nx^{-2}$ $\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n=0$ $\sum_{n=0}^{\infty}x^n\left[(n+2)(n+1)a_{n+2}+a_nn+a_n\right]=0$ $a_{n+2}=-\frac{a_n}{n+2}$

1. anonymous

I can't seem to figure out the pattern $n=0:a_2=-\frac{a_0}{2!}$ $n=2:a_4=\frac{a_0}{3!}$ $n=4:a_6=-\frac{a_0}{36}$

2. anonymous

the 36 kinda threw me off

3. anonymous

did you forget the 'x' in x*y' ?

4. anonymous

no

5. anonymous

because before the x it was $y'=\sum_{n=1}^{\infty}a_nnx^{n-1}$

6. anonymous

bit hard to follow your work since things are missing. I honor you for putting so much info into your question statement though:) refreshing change from " solve y=4x +5"

7. anonymous

I can write it all out of you want me to

8. anonymous

it's ok...

9. anonymous

let me catch up

10. anonymous

ok

11. anonymous

I don't know anymore

12. anonymous

@Outkast3r09 watcha think?

13. anonymous

factorial of the evens, not sure how to express that... n=0 1/2 n=2 1/(2*4) n=4 1/(2*4*6) n=6 1/(2*4*6*8)

14. anonymous

we've got something there!

15. anonymous

and a sign changing term...

16. anonymous

could also do $\frac{ 1 }{ 2^{k}k! }$

17. anonymous

k=1,2,3...

18. anonymous

but would the k! work?

19. anonymous

I think I've done enough math for tonight...g'night!

20. anonymous

$\frac{ 1 }{ 2^{1} 1!}, \frac{ 1 }{ 2^{2} 2!}, \frac{ 1 }{ 2^{3} 3!},\frac{ 1 }{ 2^{4} 4!}$ $= 1/2, 1/8, 1/48, 1/384$

21. anonymous

definitely an alternating series interp

22. anonymous

$y=\sum_{n=0}^{\infty}a_nx^n=a_0\sum_{n=0}^{\infty}(-1)^{2k}\frac{x^{2k}}{2^kk!}$

23. anonymous

@Outkast3r09 @Algebraic!

24. anonymous

for the even numbers

25. anonymous

this is what I get for the odds $n=1:a_3=-\frac{a_1}{3}=-\frac{a_1}{3\cdot1}$ $n=3:a_5=\frac{a_1}{15}=\frac{a_1}{5\cdot3\cdot1}$ $n=5:a_7=-\frac{a_1}{105}=-\frac{a_1}{7\cdot5\cdot3\cdot1}$

26. anonymous

best way to do this is just to do all these together and then let a1=1 and a2 = 0 $0+\frac{-1}{3*1}x+0+\frac{1}{5*3*1}x^3+0+\frac{-1}{7*5*3*1}x^5$

27. anonymous

so your x corresponds to $x^{2k-1}$

28. anonymous

would you agree?

29. anonymous

the negatives are constant with each n so you can write that as $(-1)^k$

30. anonymous

I would agree, but how do we put the denominator in terms of k's? (2k+1)!

31. anonymous

$\sum_{n=0} ^\infty \frac{(-1)^kx^{2k-1}}{1*3*5*7*.....(2k+1)}$

32. anonymous

i believe this is your series =]

33. anonymous

This is right lol i overlookd my own self haha

34. anonymous

however i'm thinking that your zero term is outside not sure =/

35. anonymous

so when we say (2k+1)! Let's try it for k=3 $\sum_{n=0} ^\infty \frac{(-1)^3x^{2\cdot3-1}}{1*3*5*7*.....(2\cdot3+1)}$

36. anonymous

$\frac{-x^5}{7!}$

37. anonymous

what 1*3*5*7*.....(2k+1) means is that you continuously keep the one before it so for n=0 you get 2(0)+1 = 1, next do n=1 1*(2*1+1)=3 so now you have for n=1 , 1*3 I wouldn't write it as 7! because that is 7*6*5*4*3*2*1

38. anonymous

write it as 1*3*5*7

39. anonymous

oh ok, so I'm allowed to write it as $1*3*5*....(2k+1)$ I thought I had to write the whole denominator in terms of k, but that makes sense now