Question

Series Solution Help y" + sin(x)y = 0

Answer 1

Find the fundamental set of solutions for the following equation\[y"+(\sin(x))y = 0\]

Answer 2

Actually, just the first four nonzero terms.

Answer 3

i will write down the list ov zeros

Answer 4

Assume that y = a₀ + a₁x + a₂x^2 + a₃x^3 + a₄x^4 + ...
==> y'' = 2a₂ + 6a₃x + 12a₄x^2 + ...

Substituting this into the DE (along with the series for sin x) yields
[2a₂ + 6a₃x + 12a₄x^2 + ...] + [x - x^3/3! + ...] [a₀ + a₁x + a₂x^2 + a₃x^3 + a₄x^4 + ...] = 0
==> [2a₂ + 6a₃x + 12a₄x^2 + ...] + [a₀x + a₁ x^2 + (-a₀/3! + a₂) x^3 + ...] = 0, via multiplication
==> 2a₂ + (a₀ + 6a₃) x + (a₁ + 12a₄) x^2 + ... = 0.

Equating like coefficients (assuming that a₀ and a₁ are arbitrary):
2a₂ = 0 ==> a₂ = 0
a₀ + 6a₃ = 0 ==> a₃ = -a₀/6
a₁ + 12a₄ = 0 ==> a₄ = -a₁/12.

Hence, y = a₀ + a₁x + 0x^2 + (-a₀/6)x^3 + (-a₁/12)x^4 + ...
..............= a₀ [1 - (1/6)x^3 + ...] + a₁ [x - (1/12)x^4 + ...].

I hope this helps!

Answer 5

well do it help??????????????????????????

Answer 6

do it helppppppp

Answer 7

Wonderful. Yes, then I am deriving it correctly. Thank you for your response. So, every time we have a product of summations, we can just multiply term by term?

Answer 8

yes u can

Answer 9

Click Best Response

Answer 10

I take it that you're studying differential equations, Mr. Bauer (or is it Ms. Bauer?).
I'd classify y'' = y*sin x as a SEPARABLE differential equation. Real Mee has come up with an elegant approach involving series. As an alternative approach:

I'd rewrite y'' as (dy/dx); thus, (dy/dx) = y*sin x.
Separating variables, (dy/y) = (sin x)dx.
Integrating, ln |y| = -cos x + c or -cos x + ln C

Evaluating e^ln|y| = e^(-cos x + ln C), we'd get |y| = (e^-cos x) times some other constant.

Thus, the general solution of the d. e. would be y = e^(-cos x) times some constant.

Answer 11

So, we write y" as (dy/dx) and not (d^2y/dx^2)?

Answer 12

What do you think, @OOOPS ?

Answer 13

I am with @TheRealMeeeee

Answer 14

but not completely satisfy with the answer. However, I don't know how to solve it yet. I need look up my note

Answer 15

if the coefficient of y is a polynomial, we expand that way, yours is a trig, cannot do the same

Answer 16

Yes. There is some initial conditions involved and you also have to carry out a couple more terms to find out the relations of a_3 and a_4.

Answer 17

Actually, never mind. We are fine as far as that goes.

Answer 18

oh yea, @TheRealMeeeee did it for you, he wrote the Maclaurin series in polynomial form, my bad, I carelessly read it

Answer 19

@mathmale please clarify your reasoning.

Answer 20

I must apologize for having written the differential equation incorrectly. You are indeed correct: y'' can be re-written as (d^2 y / dx^2). Let me know whether you want to continue this discussion. I believe those who tried to help you find a series solution to the d. e. were more on target, but my more direct approach would at least give you a way to check any answer derived from a series solution.

Answer 21

Yes, we can plug our solution back in to test our answer. So far, it is my understanding that separable equations were only for first order differential equation. It's interesting to consider whether or not there are separable equations for higher order D.Es as well.