
Differential Equation
A tank contains http://webwork.math.uwyo.edu/webwork...ba0c4005e1.png kg of salt and http://webwork.math.uwyo.edu/webwork...0f746f9fb1.png L of water. A solution of a concentration http://webwork.math.uwyo.edu/webwork...cbec96e741.png kg of salt per liter enters a tank at the rate http://webwork.math.uwyo.edu/webwork...fc95b6a081.png L/min. The solution is mixed and drains from the tank at the same rate.
Find the amount of salt in the tank after http://webwork.math.uwyo.edu/webwork...e94e38c1d1.png hours.
I got something like 36 kg and my equation was
A= (320e^(8t/1000)320)/8
but i dont think this is correct

Let C be the concentration of salt in the tank.
Each minute, 0.04 kg/L * 8L = 0.32 kilos of salt goes in and C kg/L * 8 L = 8C goes out.
$\displaystyle \frac{dC}{dt} = 0.32  C*8 $
$\displaystyle \frac{dC}{0.32  C*8} = dt $
$\displaystyle ln(0.32  C*8) = t + K$
$\displaystyle ln(0.32  C*8) = t + K$
$\displaystyle ln(0.32  C*8) = t + K$
$\displaystyle 0.32  C*8 = e^{t + K}$
$\displaystyle C = 0.04  Ke^{t}$
Find K with initial condition C(0) = 0.08
$\displaystyle C = 0.04 + 0.04e^{t}$
After 5 hours, C = 0.4 + almost nothing so Q = C*1000 = 40kg