# Differential Equation Initial Value Problem

1. Feb 14, 2017

### Staff: Mentor

1. The problem statement, all variables and given/known data
A Solve the following initial value problem:

$\frac{dx}{dt}=-x(1-x)$
$x(0)=\frac{3}{2}$

B. At what finite time does $x→∞$

2. Relevant equations

3. The attempt at a solution

$\frac{dx}{dt}=x(x-1)$
$\frac{dx}{x(x-1)}=dt$
Partial fractions: $dx(\frac{-1}{x}-\frac{1}{x-1})=dt$
Integrating both sides: $ln|\frac{1}{x}|-ln|x-1|=t+c$
$ln|\frac{1}{x(x-1)}|=t+c$
e to the power of both sides and taking the constant $e^c$ as A: $\frac{1}{x(x-1)}=Ae^t$
Plugging in the initial value gives me $A=\frac{4}{3}$

My final equation is: $\frac{1}{x(x-1)} = \frac{4}{3}e^t$

What I don't understand is how $x$ is related to $t$ and how to figure out at what time x goes to infinity. Offhand I don't see any way for X to increase to infinity since t is an exponent of e, unless t goes to negative infinity.

2. Feb 14, 2017

### LCKurtz

Those fractions don't add up to what you started with.

You can always solve for $x$ in terms of $t$.

3. Feb 14, 2017

### cnh1995

Fix the partial fractions. A sign error can turn division into multiplication when you take logarithm.

4. Feb 14, 2017

### Staff: Mentor

Gah! That's no good! Looks like I dropped a negative sign.

New fractions: $\frac{1}{x-1}-\frac{1}{x}$
That means the problem becomes: $\frac{x-1}{x}=Ae^t$
$1-\frac{1}{x}=Ae^t$
$1-\frac{1}{\frac{3}{2}}=A$
$A=\frac{1}{3}$

That puts the final equation as: $x=\frac{3}{3-e^t}$
That means that x goes to infinity at t=3.

Indeed. It's just more difficult when your equation isn't correct in the first place.

5. Feb 14, 2017

### LCKurtz

Hopefully, that $t=3$ is just a typo.

6. Feb 15, 2017

### Staff: Mentor

I take it you mean that it should be $t→3$?

7. Feb 15, 2017

### LCKurtz

No. $t=3$ is not the correct answer.

8. Feb 15, 2017

### cnh1995

The entire denominator should tend to zero.

9. Feb 15, 2017

### Staff: Mentor

Christ, please tell me it's when $e^t→3$...

10. Feb 15, 2017

### Staff: Mentor

Indeed. I promise I'm not as stupid as I appear... maybe.

11. Feb 15, 2017

### cnh1995

Right.

12. Feb 15, 2017

### LCKurtz

What value of $t$ gives $e^t = 3$?

13. Feb 15, 2017

### Staff: Mentor

I believe it's $ln(3)$. But my basic math has been so bad for this question that I don't know if I'd double down on that bet.

14. Feb 15, 2017

### LCKurtz

$\ln 3$ is correct.

15. Feb 15, 2017

### cnh1995

Relax! That's just a silly mistake..No big deal!
Yes.

16. Feb 15, 2017

### Staff: Mentor

Relax? I don't see that equation on my formula sheets...