I believe A is correct. So, you just have a linear DE
y' + 0.055y = 550
which has the soution
y = c e^-0.55t + 10000
Since y(0) = 200,000
c = 190,000
Consider a lake of constant volume 12200 km^3, which at time t contains an amount y(t) tons of pollutant evenly distributed throughout the lake with a concentration y(t)/12200 tons/km^3.
assume that fresh water enters the lake at a rate of 67.1 km^3/yr, and that water leaves the lake at the same rate. suppose that pollutants are added directly to the lake at a constant rate of 550 tons/yr.
A) write a differential equation for y(t).
B) solve the differential equation for initial condition y(0)=200000 to get an expression for y(t). Use your solution to y(t) to describe in practical terms what happens to the amount of pollutants in the lake as t goes from 0 to infinity.
** i believe A is dy/dt = -67.1(y/12200) + 550 = -0.055y + 550
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