
Classical Mechanics
I'm starting on classical mechanics in university (second year) and I skipped first year physics. I do have quite a good mathematical knowledge but I'm not getting this question:
Verify by direct substitution that the function $\displaystyle \phi(t) = A \sin (\omega t) + B \sin(\omega t)$ is a solution to the second order differential equation $\displaystyle \ddot{\phi} = \omega ^2 \phi$
(Since this solution involves two arbitary constants  the coefficients of the sine and cosine functions  it is in fact the general solution)
So I was just thinking of substituting the first equation into the second to get:
$\displaystyle \ddot{\phi} = \omega ^2 (A \sin (\omega t) + B \sin(\omega t))$
And then distribute and integrate twice to get the differential?
I'm really confused about the question.

Re: Classical Mechanics
No, substitute on the left side also. In other words differentiate twice and see if the two sides of the equation are really equal.

Re: Classical Mechanics
Actually I just realised typed in the equation wrong. The first one is supposed to be: $\displaystyle \phi(t) = A \sin (\omega t) + B \cos(\omega t)$ but I don't think it really matters.
So something like this??
$\displaystyle \phi(t) = A \sin (\omega t) + B \cos(\omega t)$ \\EQUATION 1
$\displaystyle \dot{\phi}(t) = A \omega \cos (\omega t)  B \omega \sin(\omega t)$
$\displaystyle \ddot{\phi}(t) =  A \omega ^2 \sin (\omega t)  B \omega^2 \cos(\omega t)$ \\EQUATION 2
So I substitute equation 1 onto the right hand side and equation 2 onto the left hand side of $\displaystyle \ddot{\phi} = \omega ^2 \phi$?
That'll give me:
$\displaystyle  A \omega ^2 \sin (\omega t)  B \omega^2 \cos(\omega t) = \omega ^2 (A \sin (\omega t) + B \cos(\omega t))$
And yes if I distribute the righthand side, it is clear that they are equal.
So that's the answer?