calculus pure time differential equations?

Let N be the amount of the "something" that's increasing at the specified rate. Then:

dN(t)/dt = 1000/(2+3t)^(3/2)

This is a separable differential equation:

dN(t) = 1000/(2+3t)^(3/2) dt

Integrate it:

N(t) - c = -2000/(3*sqrt(2+3t))

Where c is a constant of integration that must be determined from the initial condition. Now use that condition to solve for c (i.e., plug in the value N(0) = 1000, and solve for c):

1000 + 2000/(3*sqrt(2)) = c

c = 1000*(1+(sqrt(2))/3)

N(t) = 1000*(1+(sqrt(2))/3) -2000/(3*sqrt(2+3t))

As t - infinity, the second term on the right hand side goes to zero, so N(t) is bounded, and approaches the limit, 1000*(1+(sqrt(2))/3).