ok, so im reading my lecture notes again and here goes..
Solve the IVP dx/dt = 2x^2 with condition x(5) = -3
Solution: x(t) = -1/c + 2t
At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
so c = -29/3
now this is where i dont understand! (how did they get 6t ?? )
and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)