Mixing problem, brine solution

A 400-gal tank initially contains 100 gal of brine containing 50 lb of salt. Brine containing 1 lb of salt per gallon enters the tank at the rate of 4 gal/min, and the well mixed brine in the tank flows out at the rate of 3 gal/min. How much salt will the tank contain after 100 minutes?

So, I get the integrating factor but then I'm lost.

I(t)=(t+100)^3

Thanks for any help.

Re: Mixing problem, brine solution

Quote:

Originally Posted by

**Lamont** A 400-gal tank initially contains 100 gal of brine containing 50 lb of salt. Brine containing 1 lb of salt per gallon enters the tank at the rate of 4 gal/min, and the well mixed brine in the tank flows out at the rate of 3 gal/min. How much salt will the tank contain after 100 minutes?

So, I get the integrating factor but then I'm lost.

I(t)=(t+100)^3

Thanks for any help.

What is your Differential Equation?

Re: Mixing problem, brine solution

I tried setting this up but I'm not sure if it's right.

dy/dt = 4-3y/(100+t); y(0)=50

Re: Mixing problem, brine solution

Quote:

Originally Posted by

**Lamont** I tried setting this up but I'm not sure if it's right.

dy/dt = 4-3y/(100+t); y(0)=50

That's right.

$\displaystyle \frac{dy}{dt}+\frac{3y}{100+t}=4$

Integrating factor $\displaystyle =\exp\left(\int \frac{3}{100+t}dt\right)=\exp[3\ln (100+t)]=(100+t)^3$

$\displaystyle (100+t)^3\frac{dy}{dt}+3y(100+t)^2=4(100+t)^3$

$\displaystyle \frac{d}{dt}[(100+t)^3y]=4(100+t)^3$

Can you proceed?

Re: Mixing problem, brine solution

Yeah, I think so. Thanks!