# Math Help - Another differential equation

1. ## Another differential equation

Hello

We got a container holding 100 liters of saltwater (concentration: 1%).

We pour 5 liters/min saltwater (2% conc.) and remove 5 liters/min of the liquid in the tank.

The concentration of salt is the same in all of the container.

When is the concentration 1.5%?
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Ok I have nearly solved this problem, some details that don't fit in though.

My (near) solution:

Some trivial facts:
$y(0) = 1$
$y(infinity) = 2$

where y(t) is the amount of salt at a specific time.

$y' = \frac{1}{10} - \frac{5y(t)}{100}$

We get this equation:

$y' + \frac{5}{100}y = \frac{1}{10}$

Solve it...

$y(t)\cdot e^{\frac{5x}{100}} = \frac{1}{10} \cdot \frac{100}{5}e^{\frac{5x}{100}} + C$

$y(t) = \frac{1}{2} + Ce^{-\frac{x}{20}}$

Well I think that I got it wrong somewhere, the above equation doesn't seem to fit.

2. Originally Posted by Tau
Hello

We got a container holding 100 liters of saltwater (concentration: 1%).

We pour 5 liters/min saltwater (2% conc.) and remove 5 liters/min of the liquid in the tank.

The concentration of salt is the same in all of the container.

When is the concentration 1.5%?
----

Ok I have nearly solved this problem, some details that don't fit in though.

My (near) solution:

Some trivial facts:
$y(0) = 1$
$y(infinity) = 2$

where y(t) is the amount of salt at a specific time.

$y' = \frac{1}{10} - \frac{5y(t)}{100}$

We get this equation:

$y' + \frac{5}{100}y = \frac{1}{10}$

Solve it...

$y(t)\cdot e^{\frac{5x}{100}} = \frac{1}{10} \cdot \frac{100}{5}e^{\frac{5x}{100}} + C$

$y(t) = \frac{1}{2} + Ce^{-\frac{x}{20}}$

Well I think that I got it wrong somewhere, the above equation doesn't seem to fit.

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