Real quick question on 2nd order differential equation

[rcw110131]

Homework Statement

How do I go about solving [itex]d^2\theta/dt^2+ (g/L) \theta= g[/itex]? It's been 2.5 years since I had diff eq.

Homework Equations

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The Attempt at a Solution

I don't know. I've spent the past 2 hours going through old books and searching online and still can't figure it out

 

[denverdoc]

Heres a quick and decent review of homogeneous and nonhomogenous second order diffyq's

http://www.haverford.edu/physics-astro/MathAppendices/Differential_Eqs.pdf

 

[denverdoc]

also since this looks a lot like a pendulum problem, curious as to the problem that gave rise to it, ie the pendulum would be the homogeneous soln where right side eqn=0.

 

[rcw110131]

also since this looks a lot like a pendulum problem, curious as to the problem that gave rise to it, ie the pendulum would be the homogeneous soln where right side eqn=0.

It's the problem a few posts below about a father pushing a child on a swing. If you have any other help I'd greatly appreciate it.

 

[denverdoc]

i see, i'd just repost your query there. I'd help but got to earn a living today

 

[HallsofIvy]

Homework Statement

How do I go about solving [itex]d^2\theta/dt^2+ (g/L) \theta= g[/itex]? It's been 2.5 years since I had diff eq.

Homework Equations

^

The Attempt at a Solution

I don't know. I've spent the past 2 hours going through old books and searching online and still can't figure it out

This is more math than physics but here is how: first ignore the "g" on the right hand side to get the "associated homogeneous equation" [itex]d^2\theta/dt^2+ (g/L)\theta= 0[/itex]. It's "characteristic equation" is [itex]r^2+ (g/L)= 0[/itex] which has imaginary roots: [itex]r= \pm \sqrt{g/L}[/itex] and so the homogeneous equation has general solution [itex]Ccos(\sqrt{g/L}t)+ Dsin(\sqrt{g/L}t)[/itex].

Now, since the righthand side of the orginal equation was a constant, g, we "look for" a constant solution, y= A, to the entire equations. y'= 0 so the equation becomes 0+ (g/L)A= g and so A= L. The general solution to the entire equation is [itex]y(t)= C cos(\sqrt{g/L}t)+ D sin(\sqrt{g/L}t)+ L[/itex].

 

[rcw110131]

Thank you so much, I can't even tell you how much I appreciate your help! I finally got it; using the initial conditions, C ends up as 1-L, and then taking the derivative and using the initial conditions again, D has to equal zero, leaving only the cosine term and the L. Thanks again!