1. ## General Solution

Find the general solution of:

du/dt = u^2-U

I found the integral of 1/u(u-1) du = int 1 dt to be -ln(u)+ln(u-1) = t using partial fractions so:

u = -1/(e^t-1) but u(0) should = 2 and it would leave -1/(1-1) in this case so undefined

Could anyone tell me where I've gone wrong, thanks

2. I found the integral of 1/u(u-1) du = int 1 dt to be -ln(u)+ln(u-1) = t using partial fractions so:
Well you forgot to add a constant.

$-ln(u)+ln(u-1)=t+C$

Then you get:

$u=\frac{1}{1-e^te^C}$

So just solve for C when u(0)=2.