# Linear Differential Equation of Second Order

1. May 4, 2012

### SHISHKABOB

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$\pi$/2)

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

2. May 4, 2012

### 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 - $\alpha$1)

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

3. May 4, 2012

### sharks

You're correct.

4. May 4, 2012

### sharks

$$x(t)=e^{-3t} (-2\sin 4t)=-2e^{-3t} (\sin 4t)$$
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:
$$\cos (4t - \frac{3\pi}{2})=(\cos 4t)(\cos \frac{3\pi}{2})+ (\sin 4t)(\sin \frac{3\pi}{2})=-\sin 4t$$Proved!