inkyvoyd 2 years ago Solve the general second order linear ODE.

1. inkyvoyd

@nincompoop , I'm waiting.

2. experimentX

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.

3. inkyvoyd

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

4. experimentX

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

5. inkyvoyd

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

6. inkyvoyd

http://mathworld.wolfram.com/Second-OrderOrdinaryDifferentialEquation.html wait, lemme read this >.>

7. experimentX

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/

8. inkyvoyd

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

9. experimentX

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

10. experimentX

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

11. experimentX

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

12. inkyvoyd

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

13. inkyvoyd

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

14. experimentX

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

15. inkyvoyd

?

16. experimentX

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

17. inkyvoyd

ohhh

18. inkyvoyd

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

19. experimentX

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

20. inkyvoyd

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.

21. experimentX

sure