Just solving 2 simultaneous differential equations - am i being thick!?

Sorry, i meant sin(2alpha)/2. Will have a go now..

I'm not convinced of some of your manipulation; it seems you've written $\sin \alpha \cos \alpha = \sin \frac{\alpha}{2}$.

But aside from that, find the complementary function, the extra term on the RHS is just a constant, so look for a PI of the form x = K for some constant K.

Yes. After the correction integrate the equation twice with respect to x.

You can't just do that though can you. x is a function of t..

sorry i meant integrate x with respect to t.

that's like saying what's the integral of y with respect to x and not knowing what the function y is!

I'm not convinced of some of your manipulation; it seems you've written $\sin \alpha \cos \alpha = \sin \frac{\alpha}{2}$.

But aside from that, find the complementary function, the extra term on the RHS is just a constant, so look for a PI of the form x = K for some constant K.