This is taken directly from my textbook in section 9.5: Linear differential equations
A tank with a capacity of 400L is full of a mixture of water and chlorine with a concentration of 0.05g of chlorine per litre. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 4L/min. The mixture is kept stirred and is pumped out at a rate of 10L/min. Find the amount of chlorine in the tank as a function of time.
Note: I tried very hard but couldn't get anything more than a separable differential equation, this makes me sure my attempt is wrong as the section is linear differential equations but here it goes.
What do we know?
20g of chlorine to start --> y(0) = 20
Volume at time t is (400L - 6L/min x tmin)
the concentration at time t is y(t)/(400-6t) kg/L
The rate of chlorine departure y(grams)/(400-6t)(litres) * 10 (L/min) = 10y / (400-6t) kg/min of departure
this would make me think that dy/dt = -10y/(400-6t) --> separable equation
dy/10y = dt/(400-6t) t<400/6
integrate each side respectively to get
ln|10y|/10 = ln|400-6t|/6 + C
ln|10y|=ln|400-6t|^(5/3) + C
e^(ln|10y|) = e^(ln|(400-6t)^(5/3)| + C)
10y = (C2)(400-6t)^(5/3)
Sub t=0, y =20
20 = (C4)(400)^(5/3)
C3 = 0.000921
y = 0.000921(400-6t)^(5/3)
lim as t-->400/6 is 0
What do you guys think?