What is the solution to a differential equation in trigonometric form?

[Sparky_]

Homework Statement

This is a solutuion to a differential equation I've solved:

-cos(sqrt(2)t) - 2sin(sqrt(s)t)

I've been asked to put it in the form:

Asin(wt + phi)

Homework Equations

A= sqrt(-1^2 + -2^2)

phi = sin^-1phi = \frac{c1}{A}
phi = cos^-1 phi = \frac{c2}{A}
and
phi = tan^-1 phi = \frac{c1}{c2}

The Attempt at a Solution

A = 5

phi = sin^-1phi = \frac{-1}{sqrt(5)}
phi =

phi = cos^-1 phi = \frac{-2}{sqrt(5)}
phi =

and
phi = tan^-1 phi = \frac{-1}{-2}
phi =

 

[CompuChip]

Note that if s inside the sine is not equal to 2, then you cannot write it as a single sine at all. So the frequencies have to be the same, and you can just read them off (\omega = \sqrt{2}). Then the amplitude is indeed A = \sqrt{5} (you gave the correct formula, but forgot the root in the later result. Also watch out to write (-1)^2 = (-1) \times (-1) = 1, which is something different than -1^2 = -(1 \times 1) = -1). Then just plug in t = 0 and you get an equation for phi, which you can solve, giving \phi = \sin^{-1}(1/\sqrt{5}) + \pi (if you want a value between 0 and 2 pi).

 

[Sparky_]

I did not mean to "post" it yet - I meant to hit "preview".

Here are some more details and clean-up and my official question:

A = 5

phi = sin^{-1}phi = \frac{-1}{\sqrt(5)}
phi = -0.4636

phi = cos^{-1} phi = \frac{-2}{\sqrt(5)}
phi = 2.678

and
phi = tan^{-1} phi = \frac{-1}{-2}
phi = 0.4636

Can you help with the phase angle? (3 different answers - none of which are correct)

The book has 3.6052 which I see is pi + 0.4636.

Thanks
-Sparky

 

[CompuChip]

So you have found A and omega, and now you want
A \sin(\omega t + \phi) = - \cos(\sqrt{2} t) - 2 \sin(\sqrt{2} t).
If you look at the point t = 0, you get the equation
A \sin(\phi) = - 1
from which you can solve \phi. So indeed, you get the equation
\phi = \operatorname{arcsin}(- 1 / \sqrt{5})
which you can take to be either -0.4636, or 0.4636 + pi. By graphing both you see that the latter is correct, and it is also neater, to express phi between 0 and 2 pi.