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

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inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1@nincompoop , I'm waiting.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3write 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.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1y''+p(x)y'+q(x)y=0 I'll start with homogenous first? And find two linearly independent solutions?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3i 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

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Mm  I should've clarified and used constant coefficients. >.<

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1http://mathworld.wolfram.com/SecondOrderOrdinaryDifferentialEquation.html wait, lemme read this >.>

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3yep!! if you had constant coefficients, you can factor up the operator as (Da)(Db)y = f(x) let (Db)y = u then it becomes (Da)u = f(x) < this is first order linear, solve for 'u' let u = g(x) be it's solution then, you have (Da)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/

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1@experimentX , what i never understood is why can one factor the operator? is it only because you can relate it to the characterisitic equation?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3depending on those values 'a' and 'b', you either get decaying type or harmonic or damped harmonic solution.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3this is because the operators are commutative. (Da)[(Db)y(x)] = (Db)[(Da)y(x)]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3at least for constant coefficients .. since there is no 'x' factor to differentiate.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1@nincompoop , I don't need your copy pasting it's lagging my post.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1I see @experimentX  so in case of so in other cases you can't simply factor the operator?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3No ... check this out (D + a)(x^2D + b) y(x)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3(D + a)(x^2D + b) y(x) =/= (x^2D + b)(D + a) y(x)

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Actually 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3there are lot's of application ... usually where you get sinusoidal periodic solutions. like all sorts of pendulums, RLC, etc ,etc

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm 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.
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