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Linear Differential Equation of Second Order

by SHISHKABOB
Tags: differential, equation, linear, order
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SHISHKABOB
#1
May4-12, 03:28 PM
P: 614
1. The problem statement, all variables and given/known data
I have the equation mx'' + cx' + kx = 0

where m = 2, c = 12, k = 50, x0 = 0, x'(0) = -8. x is a function of t, and primes denote derivatives w.r.t. t.


2. Relevant equations



3. The attempt at a solution

so the equation is 2x'' + 12x' + 50x = 0, which I simplify to x'' + 6x' + 25x = 0.

Characteristic equation is r2 + 6r + 25 = 0

using the quadratic formula, I find roots r = -3 4i

therefore the general solution is x(t) = e-3t(c1cos(4t) + c2sin(4t))

I then solve for the unknown constants, c1 and c2:

x(0) = 0 = c1

so x(t) = e-3t(c2sin(4t))

x'(t) = -3e-3t(c2sin(4t) + e-3t(4c2cos(4t))

x'(0) = -8 = 4c2

so c2 = -2

∴ x(t) = e-3t(-2sin(4t))

but... the back of the book gives

x(t) = 2e-3tcos(4t - 3[itex]\pi[/itex]/2)

as the answer...

so I must be doing something horribly wrong, but I have no idea what
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SHISHKABOB
#2
May4-12, 03:38 PM
P: 614
oh whoops, I just realized that, for the set of problems that I'm working on, it asks me to write the function in the form

C1e-ptcos(ω1t - [itex]\alpha[/itex]1)

so I guess it's just a matter of converting it
sharks
#3
May4-12, 03:49 PM
PF Gold
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P: 836
Quote Quote by SHISHKABOB View Post
oh whoops, I just realized that, for the set of problems that I'm working on, it asks me to write the function in the form

C1e-ptcos(ω1t - [itex]\alpha[/itex]1)

so I guess it's just a matter of converting it
You're correct.

sharks
#4
May4-12, 04:18 PM
PF Gold
sharks's Avatar
P: 836
Linear Differential Equation of Second Order

[tex]x(t)=e^{-3t} (-2\sin 4t)=-2e^{-3t} (\sin 4t)[/tex]
Now, you try to obtain the above from the given answer, to check for equivalence:
[tex]x(t)=2e^{-3t} (\cos (4t - \frac{3\pi}{2}))[/tex]
Comparing the two answers above, what you need is to prove that:
[tex]\cos (4t - \frac{3\pi}{2})=-\sin 4t[/tex]
Use the cosine subtraction formula:
[tex]\cos (4t - \frac{3\pi}{2})=(\cos 4t)(\cos \frac{3\pi}{2})+ (\sin 4t)(\sin \frac{3\pi}{2})=-\sin 4t[/tex]Proved!


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