# Second-Order Differential Equation

Gold Member

## Homework Statement

Prove that for the differential equation $$m x''(t) = -k x(t)$$, one solution is $$x=A cos(\omega t)$$, where $$\omega = \sqrt{\frac{k}{m}}$$.

## Homework Equations

$$x^2+1 = (x+i)(x-i)$$

## The Attempt at a Solution

Well, I haven't taken differential equations yet, but I understand that you have to turn this differential equation into a polynomial somehow. Then, the roots of that polynomial give you the solution. I need the details, though...

## The Attempt at a Solution

vela
Staff Emeritus
Homework Helper
Since you're given a solution, you can prove it satisfies the equation simply by plugging it in.

More generally, for a homogeneous differential equation with constant coefficients, you can assume a solution of the form y=erx. Plug that into the differential equation and see what you get.

Gold Member
Hmm... all right, let's assume that's the case...

if x=A e^(rt), then x'' = r^2 A e^(rt), and so...

m r^2 A e^(rt) = - k A e^(rt)

Divide both sides by A e^(rt), because that's never zero...

m r^2 + k = 0

Divide both sides by k...

$$\frac{m}{k} r^2 + 1 = 0$$

$$\left(\sqrt{\frac{m}{k}} r\right)^2 + 1 = 0$$

$$\left(\sqrt{\frac{m}{k}} r + i\right) = 0, r = -\sqrt{\frac{k}{m}} i = -\omega i$$

$$\left(\sqrt{\frac{m}{k}} r - i\right) = 0, r = \sqrt{\frac{k}{m}} i = \omega i$$

So, we get...

$$x(t) = A ( e^{\omega i t} + e^{-\omega i t}) = A cos(\omega t)$$

Is that right?

Mark44
Mentor
Why not start with x(t) = Acos(wt), take m times the second derivative, and confirm that it is equal to -kx(t)?

Gold Member
Because our physics teacher challenged us to do it the hard way.

vela
Staff Emeritus