# Application of Differential Equations.

• Nov 17th 2008, 08:07 PM
panda*
Application of Differential Equations.
A 25% nitric acid solution flows at a rate of 6 litres/min into a large tank. Initially, the tank holds 200 litres of 5% nitric acid solution. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 8 litres/min.

(a) If x(t) is the amount (in litres) of nitric acid in the tank after t minutes, prove that

$\frac{dx}{dt} + \frac{4x}{100-t} = \frac{3}{2}$

Okay .. I have been trying to do it, I'm close to the answer but .. not getting there. Please help me out :)?

I know that

$\frac{dx}{dt} =$ (Inflow Rate x Concentration) - (Outflow Rate) x $\frac{x}{v(t)}$

v(t) is the amount of solution in the tank at any minute which is defined as,

v(t) = 200 - (Inflow Rate - Outflow Rate)(t)
v(t) = 200 - 2t

so putting that back up to the dx/dt equation..

$\frac{dx}{dt} =$ (6 x 25%) - 8 x $\frac{x}{200 - 2t}$

Hmm, I don't know what went wrong.

Anyway, does it matter that they told us that there was 5% of the nitric acid in the solution? Or can I just ignore that?

Thank youu.
• Nov 17th 2008, 08:47 PM
sinewave85
Quote:

Originally Posted by panda*
Okay .. I have been trying to do it, I'm close to the answer but .. not getting there. Please help me out :)?

$\frac{dx}{dt} =$ (6 x 25%) - 8 x $\frac{x}{200 - 2t}$

Thank youu.

Yes, you are very close! In fact, you are only a few simple steps away from completing your proof. Watch this:

$\frac{dx}{dt} = (6)(0.25) - (8)\frac{x}{200 - 2t}$

$\frac{dx}{dt} = (6)(1/4) - \frac{8x}{200 - 2t}$

$\frac{dx}{dt} = \frac{6}{4} - \frac{8x}{200 - 2t}$

$\frac{dx}{dt} = \frac{3}{2} - \frac{4x}{100 - t}$

$\frac{dx}{dt} + \frac{4x}{100 - t} = \frac{3}{2}$

You had the answer, you just needed to put it in the same format as the equation provided.
• Nov 17th 2008, 10:54 PM
panda*
OH! My bad. I didn't notice it was 6/4, I immediately simplified it to 3/2!
Thank you so much! :)