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I'm trying to model the following:

Assume that paraffin and diesel are mixed together in a tank with a capacity of 1000 litres. Pure diesel is pumped into the tank at a rate of 20 litres per day while pure paraffin is pumped into the tank at the rate of 60 litres per day. The mixture in the tank is always kept well mixed together, and is pumped out at a rate of 80 litres per day. Let X denote the amount of diesel in the tank on day t measured in litres.

I did not consider the part in bold when modelling the equation because the volume coming in is equal to volume coming out. My differential equation modelling the above is:

$\frac{dx}{dt} = 20 - \frac{80x}{1000} $

$\frac{dx}{dt} = 20 - {0.08x} $

Now suppose I want to solve for a particular solution such that:

If x(0) = 500, solve the differential equation for X.

My general solution is:

$\int{\frac{1}{20-0.08x}dx} = \int t $

${\frac{-1}{0.08}ln|20-0.08x|} = t + c$

${e^{ln|20-0.08x|}} = e^{-0.08(t + c)}$

$20-0.08x = e^{-0.08(t + c)}$

My attempt of particular solution is:

$20-0.08(500) = e^{-0.08((0) + c)}$

$-20 = e^{-0.08c}$

At this point in time there is no value of c such that this equation is valid (because we cannot take a ln of a negative number) and therefore there is no particular solution. Where did I go wrong? What is even more absurd is that if I remodel the equations so that X is the paraffin and not the diesel (which is another way to model it), I get a particular solution.

Dean P
  • 548

1 Answers1

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I am a bit new here, so I might not be able to depict everything properly as I am not familiar with MathJax.

The problem is when you're integrating 1/x, you're simply putting it as ln(x) whereas it should be ln|x|. If you correct that you'll be able to get a particular solution for your equation. You can read more about it here

What is the integral of 1/x?

Sauhard Sharma
  • 1,174