# how to solve a particular question with a differential equation

A tank contains 100 gal of brine in which 40 lb of salt are dissolved. It is desired to reduce the concentration of salt to 0.1 pounds per gal by pouring in pure water at the rate of 5 gal per minute and allowing the mixture (which is kept uniform by stirring) to flow out at the same rate. How long will this take?

• So what progress have you made? Have you set up a differential equation? What is the difficulty? – almagest Jan 1 at 16:56
• $dy/dt$= salt at which it enters the tank-rate at which salt leaves the tank. Since its pure water entering the tank thus salt entering is zero. $dy/dx=0-(y/100) *5$. the differential equation which we get is $Log y=-t/20+c$. at $y=40$ and $t=0$ we could get the value of $c=Log40$ – haider Jan 1 at 17:01

Let $$s(t)$$ lbs be the total quantity of salt in the tank. So $$s(0)=40$$. We want to find $$T$$ so that $$s(T)=10$$.

In time $$\delta t$$ min we have $$5\ \delta t$$ of fresh water entering the tank and the same amount of salty water leaving the tank, taking with it $$0.05s\delta t$$ of salt.

So $$s(t+\delta t)=s(t)-0.05s\ \delta t$$ and hence $$s'(t)=-0.05t$$. Integrating we get $$s(t)=s(0)e^{-0.05t}=40e^{-0.05t}$$.

So $$e^{0.05T}=40/10=4$$, giving $$T=20\ln 4$$ min $$\approx 27.7$$ min.

It looks like you've basically solved the problem. When the concentration is 0.1, this means the amount of salt in the tank is 10 lbs, so you can solve for the time when $$\log{y} = \log{10}$$ using your equation.

I hope this helps.

• I am not able to understand how come will be the amount of salt 10 lbs when the concentration is 0.1. Is it like 01 in 100 liters of brine that will give me 10 lbs of salt – haider Jan 1 at 17:52
• That is the idea, although the problem is in lbs and gallons, so it would 10 lbs in 100 gallons of brine = 10/100=0.1 The original mixture was 40 lbs of salt in 100 gallons of water to produce 100 gallons of brine at 0.4 concentration. "Brine" is just the name for salt mixed with water. I hope this helps. – ad2004 Jan 1 at 17:55