Classical Path using Lagrangian and EOM

[spacetimedude]

Homework Statement

Show that the classical path satisfying $\bar{x}(t_a) = x_a$, $\bar{x}(t_b) = x_b$ and $T = t_b-t_a$ is
$$\bar{x}(t) = x_b\frac{\sin\omega (t-t_a)}{\sin\omega T} + x_a\frac{\sin\omega (t_b-t)}{\sin\omega T}$$

Homework Equations

The Lagrangian: $L = \frac{1}{2}m(\dot{x}^2-\omega^2x^2)$
The EOM: $\ddot{\bar{x}}+\omega^2\bar{x}=0$

The Attempt at a Solution

The initial step to this problem is confusing me. I have only been exposed to SHM problems in which the solution to its differential equation is $x(t) = A\cos(\omega t) + B\sin(\omega t)$. But in this particular question, the solution starts with
$\bar{x}(t) = A\sin\omega (t-t_a) + B\sin\omega (t-t_b)$.

My original thought was that it had something to do with the trig identities and rearrangements, but I could not get to the solution.
It would be great if someone can lead me to the process of solving this differential equation or linking me to a site with an explanation. I have only taken a course in differential equation that was integrated into my physics course and have not taken a standalone maths course.

 

[Orodruin]

You can start with your approach too. However, once you have solved for your constants, you should be able to rearrange it to the same form using trigonometric identities. The thing is that both $\sin(\omega(t-t_a))$ and $\sin(\omega(t-t_b))$ are linear combinations of $\sin(\omega t)$ and $\cos(\omega t)$ and so a linear combination of them is a linear combination of $\sin(\omega t)$ and $\cos(\omega t)$ as well. As long as $\omega T \neq \pi n$, the linear combination is also general.