# Yet another differential equation

Hello,

I would appreciate any help solving the following equation:

\begin{align} y''[t] + \dfrac{d}{m}y'[t] + \dfrac{k}{m}y[t] = G \\ \end{align} subject to: $$t[0] = t0$$ $$t'[0] = 0$$

This is not HW, this is for a website where I try to simulate object attached to a spring falling, as I presented in this question; I just forgot to add damping (d) force so that the object will stop its movement after a while.

Thank you!

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You can ask wolfram alpha for such questions – Norbert Aug 25 '12 at 22:38
This is a damped harmonic oscillator (cont'd.). – Rahul Aug 25 '12 at 22:39

Multiply through by $m$ to give $my'' + dy' + ky = Gm.$ First, find the complementary function by solving the homogeneous equation $my'' + dy' + ky = 0.$ Use the trial function $y(t) = e^{\lambda t}$. We get $(m\lambda^2 + d\lambda +k)e^{\lambda t} = 0.$ Since $e^{\lambda t} > 0$ we need to solve $m\lambda^2 + d\lambda +k = 0$ for $\lambda.$ Whence:

$\lambda_{\pm} = \frac{1}{2m}(-d \pm \sqrt{d^2 - 4mk}) \, .$

The complimentary function is then $y(t) = ae^{\lambda_-} + be^{\lambda_+}$ for any $a$ and $b$. Next, we need to find the particular integral. Since the right hand side of $my'' + dy' + ky = Gm$ is constant, we try $y(t) = \alpha$ where $\alpha$ is a constant. Putting that into $my'' + dy' + ky = Gm$ gives $k\alpha = Gm$ and so $\alpha = Gm/k.$ The general solution is then:

$y(t) = ae^{\lambda_-} + be^{\lambda_+} + \frac{Gm}{k} \, .$

Finally, we impose the initial conditions that $y(0) = t_0$ and $y'(0) = 0.$ You will get two simultaneous equations in $a$ and $b$. Solve them gives:

$a = \frac{m\lambda_+(Gm-kt_0)}{k\sqrt{d^2-4mk}} \, ,$

$b = \frac{m\lambda_-(Gm-kt_0)}{k(d^2-4mk)} \, .$

Provided I've not made a silly mistake, I get:

$y(t) = \frac{m}{k}\left[ G + \frac{(Gm-kt_0)}{\sqrt{d^2-4mk}}\left(\lambda_+e^{\lambda_-t} + \lambda_-e^{\lambda_+t}\right) \right] \, .$

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And if $d^2-4mk\lt0$, you might find it more convenient to express the answer in terms of sines and cosines, instead of complex exponentials. – Gerry Myerson Aug 25 '12 at 23:29
In reality the solution will be a product of a phase shifted cosine and a real exponential. But we need to know the numbers first. – Fly by Night Aug 25 '12 at 23:45
Wow thank you so much! I'll try it and I really got to take again that math course... thanks again :) – Israel Aug 25 '12 at 23:56