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

1. MathSofiya Group Title

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. MathSofiya Group Title

the 36 kinda threw me off

3. Algebraic! Group Title

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

4. MathSofiya Group Title

no

5. MathSofiya Group Title

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

6. Algebraic! Group Title

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. MathSofiya Group Title

I can write it all out of you want me to

8. Algebraic! Group Title

it's ok...

9. Algebraic! Group Title

let me catch up

10. MathSofiya Group Title

ok

11. MathSofiya Group Title

I don't know anymore

12. MathSofiya Group Title

@Outkast3r09 watcha think?

13. Algebraic! Group Title

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. MathSofiya Group Title

we've got something there!

15. Algebraic! Group Title

and a sign changing term...

16. Algebraic! Group Title

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

17. Algebraic! Group Title

k=1,2,3...

18. MathSofiya Group Title

but would the k! work?

19. MathSofiya Group Title

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

20. Algebraic! Group Title

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

21. Outkast3r09 Group Title

definitely an alternating series interp

22. MathSofiya Group Title

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

23. MathSofiya Group Title

@Outkast3r09 @Algebraic!

24. MathSofiya Group Title

for the even numbers

25. MathSofiya Group Title

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. Outkast3r09 Group Title

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. Outkast3r09 Group Title

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

28. Outkast3r09 Group Title

would you agree?

29. Outkast3r09 Group Title

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

30. MathSofiya Group Title

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

31. Outkast3r09 Group Title

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

32. Outkast3r09 Group Title

i believe this is your series =]

33. Outkast3r09 Group Title

This is right lol i overlookd my own self haha

34. Outkast3r09 Group Title

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

35. MathSofiya Group Title

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. MathSofiya Group Title

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

37. Outkast3r09 Group Title

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. Outkast3r09 Group Title

write it as 1*3*5*7

39. MathSofiya Group Title

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