DE undetermined coefficients De problem $$x'' + 196x = 40\sin (14t)$$
I have the general solution, I'm just struggling with the algebra with the undetermined coefficient. I set my guess as: $At\cos (14t) + Bt\sin (14t)$, took the derivatives and plugged in, but I'm not making any head way with getting values for $A$ and $B$. 

Your choice for a particular solution is correct.
We have:
$$x_p'' = t (-196 a \cos (14 t)-196 b \sin (14 t))-28 a \sin (14 t)+28 b \cos (14 t)$$
So, 
$$x_p'' + 196 x_p = 28 (b \cos (14 t)-a \sin (14 t)) = 40 \sin(14 t)$$
Equating terms, we get:
$$a = -\dfrac{10}{7}, b = 0$$
This gives a final solution of:
$$x(t) = c_1 \cos(14 t) + + c_2 \sin(14t) -\dfrac{10}{7} t \cos(14t)$$
