# Dx/dt = x(t)(1-x(t))

1. Jan 16, 2013

### Jamin2112

1. The problem statement, all variables and given/known data

Where am I going wrong in solving this differential equation?

2. Relevant equations

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

3. 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)=Ket

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

2. Jan 16, 2013

### Jamin2112

Another question:

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

3. Jan 17, 2013

### Staff: Mentor

$$\int \frac{dx}{1 - x} \neq ln(1 - x)$$

4. Jan 17, 2013

### 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) = ex does not have an algebraic solution.

5. Jan 17, 2013

### Jamin2112

Ok. Because one of our hw questions was to solve x'(t) = et - 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.

6. Jan 17, 2013

### Jamin2112

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) = Cet
---> (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]

7. Jan 17, 2013

### 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.