If y(t) is such that dy/dt = 13(t^3)(y^2), y(0) = -4, then compute y(1).

i know your suppose to switch it up such that dy/(y^2) = (t^3)dt ... and then you take the integral of both side and get (-)1/y = 1/4(t^4) + c.... and then its this part where i screw up on the correct algebra per say, need help on the last couple of steps