Second-Order Differential Equation

  • #1
Char. Limit
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Homework Statement


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


Homework Equations



[tex]x^2+1 = (x+i)(x-i)[/tex]

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...

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
vela
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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.
 
  • #3
Char. Limit
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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...

[tex]\frac{m}{k} r^2 + 1 = 0[/tex]

[tex]\left(\sqrt{\frac{m}{k}} r\right)^2 + 1 = 0[/tex]

[tex]\left(\sqrt{\frac{m}{k}} r + i\right) = 0, r = -\sqrt{\frac{k}{m}} i = -\omega i[/tex]

[tex]\left(\sqrt{\frac{m}{k}} r - i\right) = 0, r = \sqrt{\frac{k}{m}} i = \omega i[/tex]

So, we get...

[tex]x(t) = A ( e^{\omega i t} + e^{-\omega i t}) = A cos(\omega t)[/tex]

Is that right?
 
  • #4
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Why not start with x(t) = Acos(wt), take m times the second derivative, and confirm that it is equal to -kx(t)?
 
  • #5
Char. Limit
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Because our physics teacher challenged us to do it the hard way.
 
  • #6
vela
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Yes, that's right. Any linear combination of the two solutions you found, eiωx and e-iωx, in particular cos ωx, will also be a solution.
 

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