Author Topic: dcc + differential equations = atrocious (Read 10747 times) Tweet Share

Mao · « **Reply #45 on:** July 12, 2008, 11:27:08 pm »

Quote from: phagist_ on July 12, 2008, 09:32:48 pm

(sorry if there are mistakes am hungover and tired)

hehe

dcc · « **Reply #46 on:** July 14, 2008, 09:51:56 pm »

A way of solving the OP:

Let Q be the amount of salt in grams in the tank at time t.

$Q(0) = 30\; g$ .

$Q_{in} = 9\; L\; \mbox{min}^{-1}$

$V(t) = 250 - 4t$

$Q_{out} = \dfrac{13Q}{250 - 4t}$

Firstly: $\dfrac{dQ}{dt} = 9 - \dfrac{13Q}{250 - 4t}$

$\implies Q' + \dfrac{13}{250 - 4t}Q = 9$

Multiply through by:

$\left(250 - 4t\right)^{-\frac{13}{4}}$

and we get:

$\implies \left(250 - 4t\right)^{-\frac{13}{4}} Q' + 13\left(250 - 4t\right)^{-\frac{17}{4}}Q = 9 \left(250 - 4t\right)^{-\frac{13}{4}}$

$\implies \left[\left(250 - 4t\right)^{-\frac{13}{4}}Q \right]' = 9 \left(250 - 4t\right)^{-\frac{13}{4}}$

Therefore by the fundamental theorem of calculus:

$(250 - 4t)^{-\frac{13}{4}} \cdot Q = \int \dfrac{9}{\left(250 - 4t\right)^\frac{13}{4}}\; dt = \left(250 - 4t\right)^{-\frac{9}{4}} + C$

$Q = \left(250 - 4t\right)^{\frac{13}{4}} \left( \dfrac{1}{\left(250 - 4t\right)^{\frac{9}{4}}} + C \right)$

$Q = \left(250 - 4t\right) + C\left(250 - 4t\right)^\frac{13}{4}$

Using the initial condition Q(0) = 30g

$30 = 250 + C \cdot 250^\frac{13}{4} \implies C = \dfrac{30 - 250}{250^\frac{13}{4}}$

$\therefore \; Q = (250 - 4t) - \dfrac{220}{250^\frac{13}{4}} \cdot (250 - 4t)^\frac{13}{4}$

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Author Topic: dcc + differential equations = atrocious (Read 10747 times) Tweet Share

Mao

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dcc

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