Differential Equation Initial Value Problem

[Drakkith]

Homework Statement

A Solve the following initial value problem:

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

B. At what finite time does $x→∞$

Homework Equations

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.

 

[LCKurtz]

Homework Statement

A Solve the following initial value problem:

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

B. At what finite time does $x→∞$

Homework Equations

The Attempt at a Solution

$\frac{dx}{dt}=x(x-1)$
$\frac{dx}{x(x-1)}=dt$
Partial fractions: $dx(\frac{-1}{x}-\frac{1}{x-1})=dt$

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

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.

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

 

[cnh1995]

What I don't understand is how xxx is related to ttt 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.

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

 

[Drakkith]

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

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.

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

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

 

[LCKurtz]

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.

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

 

[Drakkith]

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

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

 

[LCKurtz]

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

 

[cnh1995]

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

The entire denominator should tend to zero.

 

[Drakkith]

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

 

[Drakkith]

The entire denominator should tend to zero.

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

 

[cnh1995]

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

Right.

 

[LCKurtz]

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

 

[Drakkith]

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

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.

 

[LCKurtz]

$\ln 3$ is correct.

 

[cnh1995]

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

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

I believe it's ln(3)ln(3)ln(3).

Yes.

 

[Drakkith]

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

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