Original post by DohaerisI don't mind helping you out. From what I can see, you used partial fractions, but there's no need for that (at least in the way I solved it.) I'll try and take it step-by-step.
First, they said M is the mass of the fuel that was burned, from this we know that M is a constant.
Next, we're told that the sum of the mass of waste produced and the mass of unburned fuel remaining is always constant, this'll come into play later on.
We're given that X is the mass of waste produced.
Then we're told that the rate of increase of X, dX/dt, is k times that the mass of fuel remaining.
The mass of fuel remaining will have to be M-X, where M is the original mass of unburned fuel, because we know that the sum of waste produced and unburned fuel remaining is constant. So if you take 'Y' as the unburned fuel remaining, then Y+X=M, so Y=M-X
From this, we can deduce the equation to be dX/dt = k*(M-X), and you can carry on the to solve the differential equation as you would any other, though you have to remember that M and k are constants.
Do you get it?