2 (hopefully quick) Differential Equation Questions

Hi, I have a quick question to do with Differential equations.

So i have done the first part of my question no problems but for this next part it says:

Show x(t) is always non-decreasing for t=> 0, without solving the equation.

dx/dt = rx(1-x), where r is just a constant (rate of maximum growth in this case).

I get that it must have something to do with the derivative always being positive but i'm just not sure how I can show this when there are no t variables in the equation.

ALSO, I have another question with finding the general solution to this equation. I have gotten it as far as:

ln(x)-ln(1-x)=rt+c

ln(x/(1-x))=rt+c

x/(1-x)=e^(rt+c)

And I am positive it is correct to this point, however I just can't seem to finish it off and get it in the form x(t)= ......

Thanks for any help!!!

Re: 2 (hopefully quick) Differential Equation Questions

Quote:

Originally Posted by

**Cotty** Show x(t) is always non-decreasing for t=> 0, without solving the equation.

dx/dt = rx(1-x), where r is just a constant (rate of maximum growth in this case).

I get that it must have something to do with the derivative always being positive but i'm just not sure how I can show this when there are no t variables in the equation.

If it is non-decreasing that means it is always increasing. If dx/dy is always positive (neither 0 nor negative) what is the function doing?

-Dan

Re: 2 (hopefully quick) Differential Equation Questions

Quote:

Originally Posted by

**Cotty** ALSO, I have another question with finding the general solution to this equation. I have gotten it as far as:

ln(x)-ln(1-x)=rt+c

ln(x/(1-x))=rt+c

x/(1-x)=e^(rt+c)

Actually you are wrong on two counts.

First

$\displaystyle \frac{1}{x(1 - x)} = \frac{1}{x} + \frac{1}{1 - x}$

so you have a sign problem.

Also

$\displaystyle \int \frac{1}{x} = ln |x|$

So you have

$\displaystyle ln |x| + ln|1 - x| = rt + C$

$\displaystyle ln |x(1 - x)| = rt + C$

$\displaystyle | x(1 - x) | = e^{rt + C}$

or

$\displaystyle x(1 - x) = \pm e^{rt + C}$

Now multiply out and get everything on one side:

$\displaystyle x^2 - x \pm e^{rt + C} = 0$

Now use the quadratic formula. Go get 'em tiger.

One slight simplification here: If we define $\displaystyle \pm e^{rt + C} = \pm A e^{rt} \equiv A'e^{rt}$ for any A' not equal to 0 the equation is much simpler looking.

-Dan

Re: 2 (hopefully quick) Differential Equation Questions

Thanks for giving me a hand! So isn't the integral of 1/(1-x) = -ln(1-x), by using substitution? I think i might have arrived at the solution with it being:

x(t) = Ae^rt/ (1+Ae^(rt)) where A=e^c

And with your first answer, how can i show that the derivative is positive though if x is a function of t, but instead of it actually being in terms of t its just written as x(t), meaning that we don't actually know that when t is positive that x(t) is positive? Ie. if x(t)=1000-t, for all t<1000, won't the derivate be negative?

Also how do you write with the integrals and everything? People always answer my questions so neatly and mine is just a mess trying to write without the maths stuff!! haha

Re: 2 (hopefully quick) Differential Equation Questions

Quote:

Originally Posted by

**Cotty** Thanks for giving me a hand! So isn't the integral of 1/(1-x) = -ln(1-x), by using substitution? I think i might have arrived at the solution with it being:

x(t) = Ae^rt/ (1+Ae^(rt)) where A=e^c

And with your first answer, how can i show that the derivative is positive though if x is a function of t, but instead of it actually being in terms of t its just written as x(t), meaning that we don't actually know that when t is positive that x(t) is positive? Ie. if x(t)=1000-t, for all t<1000, won't the derivate be negative?

Also how do you write with the integrals and everything? People always answer my questions so neatly and mine is just a mess trying to write without the maths stuff!! haha

Ahem! Okay, I missed the extra sign. Good catch. But you still need the absolute value!

You proved from the original differential equation that the slope is always positive. The derivative of the explicit form will also show you the same.

The solution looks good. As for the LaTeX we have a subforum.

Re: 2 (hopefully quick) Differential Equation Questions

I will have a look at that, thank you very much for taking the time to help!