differencial equation of flow

**Oranges&Lemons** · March 30th 2008, 05:54 AM

ok..so the question is...

"During a chemical reaction, a compound is being made from two other substances. At time t hours after the start of the reaction, x g of the compound has been produced. Assuming that x = 0 initially, and that

dx = 2(x-6)(x-3)
dt

show that it takes approximately 7 minutes to produce 2 g of the compund."

We have been started off with...

the integral of 1 over (x-6)(x-3) dx = the integral of 2 dt

i don't have any idea where this came from or why it is the first step...can anyone help?

**wingless** · March 30th 2008, 06:34 AM

You're doing it right. First we need an equation that shows how x is related to time.

Solve the differential equation. The question states: "Assuming that x = 0 initially".. so x(0)=0. You can find the integration constant from this initial condition. Now we have the equation. Use it to find the answer.

Remember: In the equation, the unit of t is hours as the question says: "At time t hours". Convert the units if needed.

**Soroban** · March 30th 2008, 06:56 AM

Hello, Oranges&Lemons!

Have you never seen Differential Equations or Integration . . . ever?

During a chemical reaction, a compound is being made from two other substances.
At time $t$ hours after the start of the reaction, $x$ grams of the compound has been produced.

Assuming that $x = 0$ initially, and that: . $\frac{dx}{dt} \:=\:2(x-6)(x-3)$
show that it takes approximately 7 minutes to produce 2 g of the compund.

We have: . $\frac{dx}{dt} \;=\;2(x-6)(x-3)$

Separate variables: . $\frac{dx}{(x-6)(x-3)} \;=\;2\,dt$

Use Partial Fractions on the left side, and we get: . $\frac{1}{3}\left(\frac{1}{x-6} - \frac{1}{x-3}\right)\,dx \;=\;2\,dt$

Integrate: . $\frac{1}{3}\bigg[\int\frac{dx}{x-6} - \int\frac{dx}{x-3}\bigg] \;=\;2\int dt$

. . $\frac{1}{3}\left[\ln(x-6) - \ln(x-3)\right] \;=\;2t + c \quad\Rightarrow\quad \ln\left(\frac{x-6}{x-3}\right) \;=\;6t + c<br />$

Exponentiate: . $\frac{x-6}{x-3} \;=\;e^{6t+c} \;=\;e^{6t}\cdot e^c \quad\Rightarrow\quad \frac{x-6}{x-3}\;=\;Ce^{6t}$

We are told that when $t=0,\:x=0$

We have: . $\frac{0-6}{0-3} \:=\:Ce^0\quad\Rightarrow\quad C \:=\:2$

. . Hence, the function is: . $\frac{x-5}{x-3} \;=\;2e^{6t}$

When $x = 2$ , find $t.$

We have: . $\frac{2-6}{2-3} \:=\:2e^{6t}\quad\Rightarrow\quad 2e^{6t} \:=\:4\quad\Rightarrow\quad e^{6t} \:=\:2$

. . $6t \:=\:\ln(2)\quad\Rightarrow\quad t \;=\;\frac{1}{6}\ln(2) \;=\;0.1155\text{ hours} \;=\;6.9315\text{ minutes}<br />$

Therefore: . $t \;\approx\;7\text{ minutes}$

**Oranges&Lemons** · March 30th 2008, 07:29 AM

yes, i have seen differentiation and integration before...lots of times..i just couldnt figure out how to do this particular question.

thanks for your help.

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