Linear Differential Equation of Second Order

[SHISHKABOB]

Homework Statement

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.

Homework Equations

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\pi/2)

as the answer...

so I must be doing something horribly wrong, but I have no idea what

 

[SHISHKABOB]

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 - \alpha₁)

so I guess it's just a matter of converting it

 

[DryRun]

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 - \alpha₁)

so I guess it's just a matter of converting it

You're correct.

 

[DryRun]

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