Question

Solve the general second order linear ODE.

Answer 1

@nincompoop , I'm waiting.

Answer 2

write the DE in terms of operator, as you assumed the linearity, factorize it ...
let (first factored operator) y = u
(second factor) u = your input
solve the second for u, then use it to solve for first.

Answer 3

y''+p(x)y'+q(x)y=0
I'll start with homogenous first? And find two linearly independent solutions?

Answer 4

i don't think you can generally solve for the systems containing p(x) or q(x)
even if there is specific values for p(x) and q(x), they are not so easily solvable
Eg.
http://mathworld.wolfram.com/HermiteDifferentialEquation.html

Answer 5

Mm - I should've clarified and used constant coefficients. >.<

Answer 6

http://mathworld.wolfram.com/Second-OrderOrdinaryDifferentialEquation.html

wait, lemme read this >.>

Answer 7

yep!! if you had constant coefficients, you can factor up the operator as
(D-a)(D-b)y = f(x)
let (D-b)y = u

then it becomes (D-a)u = f(x) <--- this is first order linear, solve for 'u'
let u = g(x) be it's solution

then, you have (D-a)y = g(x) <--- again linear in y.

The e^(kx) are characteristics of linear equations. There are better methods, but this is only for concept/

Answer 8

@experimentX , what i never understood is why can one factor the operator? is it only because you can relate it to the characterisitic equation?

Answer 9

depending on those values 'a' and 'b', you either get decaying type or harmonic or damped harmonic solution.

Answer 10

this is because the operators are commutative.
(D-a)[(D-b)y(x)] = (D-b)[(D-a)y(x)]

Answer 11

at least for constant coefficients .. since there is no 'x' factor to differentiate.

Answer 12

@nincompoop , I don't need your copy pasting it's lagging my post.

Answer 13

I see @experimentX - so in case of so in other cases you can't simply factor the operator?

Answer 14

No ... check this out
(D + a)(x^2D + b) y(x)

Answer 15

?

Answer 16

(D + a)(x^2D + b) y(x) =/= (x^2D + b)(D + a) y(x)

Answer 17

ohhh

Answer 18

Actually I had originially asked this question to make @nincompoop stop being so annoying - I just realized that it has practical applications in RLC circuits though x.x

Answer 19

there are lot's of application ... usually where you get sinusoidal periodic solutions. like all sorts of pendulums, RLC, etc ,etc

Answer 20

Hmm I'ma close this and open a new question, actually more of a physics question. I'm guessing it'll pop up a first order ODE, I just want to formulate one I guess.

Answer 21

sure