# Thread: Differential Linear Equation - formula from the world problem

1. ## Differential Linear Equation - formula from the world problem

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)
y= (C3)(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?

2. Originally Posted by someguy456345
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
Yes, very good.

dy/10y = dt/(400-6t) t<400/6

integrate each side respectively to get

ln|10y|/10 = ln|400-6t|/6 + C
Since ln|10y|/10= ln|y|/10+ ln10/10, you don't really need the "10" inside the logarithm. That could be incorporated into the constant, C. More important is the fact that the integral of dt/(400- 6t) is -ln|400- 6t|/6.

ln|10y|=ln|400-6t|^(5/3) + C
e^(ln|10y|) = e^(ln|(400-6t)^(5/3)| + C)
10y = (C2)(400-6t)^(5/3)
Since that should be $e^{-ln|(400-6t)^(5/3)|+ C$,
you should have $10y= (C2)(400- 6t)^{-5/3}$

y= (C3)(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?

3. The original equation is decreasing (and therefore has a negative sign) to begin with which i failed to include at first (it was actually the reason I was stuck on this problem for so long before going back over my work), so the integral of the RS cancelled that to become positive.

I understand what you mean by incorporating the 10 within the logarithm into the constant, thanks! I'll be sure to do that in the future.