OK I really don't understand how to use this method of numerical integration.

I have the coupled differential equations:

dx/dt = f(I, I') = dI/dt

df/dt = g(I, I') = -(R/L)f(I, I') - I/(LC)

Now I have a source that says to solve the formulae

dI/dt = I’ and

dI’/dt = f(I’, I) = -(R/L)I’ – I/(LC)

You use tn+1 = tn + h

In+1 = In + (K1 + 2k2 + 2k3 + k4)/6

I’n+1 = I’n + (l1 + 2l2 + 2l3 + l4)/6

where k1 = h x I’

l1 = h x f(I’, I)

k2 = h x (I’ + l1/2)

l2 = h x f(I’ + l1/2, I + k1/2)

k3 = h x (I’ + l2/2)

l3 = h x f(I’ + l2/2, I + k2/2)

k4 = h x (I’ + l3)

l4 = h x f(I’ + l3, I + k3)

However I cannot see how these derive from the normal runge kutta formulae for a single differential equation, nor can I see how to apply this to my problem. Finally, once I have solved the initial problem I have to solve

d2I/dt2 + (R/L)dI/dt + I/(LC) = (1/L)dV/dt

Where V = V0cos(wt)

The LHS is the part I need to solve for the first part using runge kutta, having split it into the coupled equations shown at the top.

Plese, please, please help!