Understanding the Solution to dx/dt = x(t)(1-x(t)) Differential Equation

[Jamin2112]

Homework Statement

Where am I going wrong in solving this differential equation?

Homework Equations

x' = x(1-x) is what I'm after

The Attempt at a Solution

Separating variables: dx / [x(1-x)] = dt
Partial fraction decomposition on the left side yields: dx [1/x + 1/(1-x)] = dt.
Integrating both sides with respect to their variable: ln(x) + ln(1-x) = t + C
Simplifying the left side: ln(x(1-x)) = t + C
Raising both sides to the e power: x(1-x)=Ke^t

Nothing I can really do after this, though. I guess I just define it implicitly like that?

 

[Jamin2112]

Another question:

Any idea how I can get a formula for the x for which cos(x)=e^x?

 

[Mark44]

Homework Statement

Where am I going wrong in solving this differential equation?

Homework Equations

x' = x(1-x) is what I'm after

The Attempt at a Solution

Separating variables: dx / [x(1-x)] = dt
Partial fraction decomposition on the left side yields: dx [1/x + 1/(1-x)] = dt.
Integrating both sides with respect to their variable: ln(x) + ln(1-x) = t + C

Your mistake is just above.
$$ \int \frac{dx}{1 - x} \neq ln(1 - x)$$

Simplifying the left side: ln(x(1-x)) = t + C
Raising both sides to the e power: x(1-x)=Ke^t

Nothing I can really do after this, though. I guess I just define it implicitly like that?

 

[CompuChip]

Also, if you get x(1 - x) = f(t) you can definitely solve for x -- if you look more closely it's just a quadratic equation.
First correct the error pointed out by Mark though :)

As for your other question cos(x) = e^x does not have an algebraic solution.

 

[Jamin2112]

Also, if you get x(1 - x) = f(t) you can definitely solve for x -- if you look more closely it's just a quadratic equation.
First correct the error pointed out by Mark though :)

As for your other question cos(x) = e^x does not have an algebraic solution.

Ok. Because one of our homework questions was to solve x'(t) = e^t - cos(t). Someone asked the professor what to do about the fact that there are infinitely many fixed points, and he told us to just deal with x'(0)=0 but to explain how we know there are infinitely many other solutions.

 

[Jamin2112]

Your mistake is just above.
$$ \int \frac{dx}{1 - x} \neq ln(1 - x)$$

Ah, I see.

So then we have

x' = x(1-x)
---> dx / [x(1-x)] = dt
---> dx/x + dx/(1-x) = dt
---> ln(x) - ln(1-x) = t + c
---> ln[x/(1-x)] = t + c
---> x/(1-x) = Ce^t
---> (1-x)/x = Ke^-t
---> 1/x - 1 = Ke^-t
---> 1/x = 1 + Ke^-t
---> x(t) = 1/(1+Ke^-t)
---> x(0) = 1(1+K), K = x(0)-1
---> x(t) = 1/[1+(X(0)-1)e^-t]

 

[HallsofIvy]

One way to show it has more than one solution is to show a graph of $y= cos(x)- e^x$ and note that it crosses the x-axis multiple times.