This is a first order differential equation, but it's taught in calc 2. If I'm in the wrong forum, please move me!

It is in verbose detail because I'm having trouble with both problems ( 2 problems, 2 hours... ) and hope that it's a similar issue so I can fix both.

"A tank contains 100 gal of fresh water. A solution containing 1 lb/gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal/min and the mixture is pumped out of the tank at the rate of 3 gal/min. Find the maximum amount of fertilizer in the tank and the time required to reach the maximum."

The book's given answer is "y(27.8)=14.8 lb, t = 27.8 min" verbatim.

So, I start out by writing dy/dt = 1 gal/min * 1 lb/gal - 3 gal/min * Y/(100+(1-3)*t) lb/gal.

I simplified that to dy/dt = 1 lb/min - 3*y/(100-2t) lb/min.

Then I chucked my units of measurement and put it into standard form.

dy/dt + 3Y/(100-2t)=1

I used the integrating factor e^integral(3/(100-2t) = e^(3*ln(100-2t) =This gave me d((100-2t)^-3*y)=(100-2t)^-3.

(100-2t)^-3

I integrated, got (100-2t)^-3*y=1/4 * (100-2t)^-2 + C

I multiplied across by (100-2t)^3 and ended up with y=1/4 * (100-2t) + C.

y(0)=100, which means 100=100/4+C, or C=75.

And now I'm completely cornered. Taking the derivative of what I have gets me -1/2, so I can't find out when the slope is 0 and the tank is full, nor go any further with the problem. Where did I go wrong in it?