Solving an Initial Value Problem for x(t=ln2): Step-by-Step Solution

[Drakkith]

Homework Statement

Solve the initial value problem:
$\frac{dx}{dt} = x(2-x)$, $x(0) = 1$
for $x(t=ln2)$.

Homework Equations

The Attempt at a Solution

I moved the right side to the left and multiplied both sides by dt to get:
$\frac{dx}{x(2-x)} = dt$

Integrating gave me:
$\frac{ln|x|}{2} - \frac{ln|x-2|}{2} = t+C$

Then:
$ln|x| - ln|x-2| = 2t + 2C$
$ln|\frac{x}{x-2}| = 2t + 2C$
$\frac{x}{x-2} = e^{2t}e^{2c}$
$\frac{1}{1-\frac{2}{x}} = Ke^{2t}$

Manipulating for a while, I end up with:
$x=\frac{2}{Ke^{2t}}$

Since $x(0) = 1$, I set the left side to 1 and solve for K, winding up with $k=3$.
However, when trying to solve for $x(ln2)$ I end up with $\frac{2}{11}$, which isn't one of my possible answers.

Does my process look even remotely correct?

 

[LCKurtz]

Homework Statement

Solve the initial value problem:
$\frac{dx}{dt} = x(2-x)$, $x(0) = 1$
for $x(t=ln2)$.

Homework Equations

The Attempt at a Solution

I moved the right side to the left and multiplied both sides by dt to get:
$\frac{dx}{x(2-x)} = dt$

Integrating gave me:
$\frac{ln|x|}{2} - \frac{ln|x-2|}{2} = t+C$

Then:
$ln|x| - ln|x-2| = 2t + 2C$
$ln|\frac{x}{x-2}| = 2t + 2C$
$\frac{x}{x-2} = e^{2t}e^{2c}$
$\frac{1}{1-\frac{2}{x}} = Ke^{2t}$

So you could say $\frac x {x-2} = Ke^{2t}$ Put $t=0,~x=1$ there to get $\frac x {x-2} = -e^{2t}$. Now put $t=\ln 2$ in that, then solve for $x$. See if that fixes it.

 

[Ray Vickson]

Homework Statement

****************.

$ln|x| - ln|x-2| = 2t + 2C$
$ln|\frac{x}{x-2}| = 2t + 2C$
$\frac{x}{x-2} = e^{2t}e^{2c}$
$\frac{1}{1-\frac{2}{x}} = Ke^{2t}$

For $t$ near 0 you have $x$ near 1, and so $x/(x-2) = e^{2C} e^{2t}$ is impossible unless you let $C$ be a complex number. However, $\ln(|x|/|x-2|) = \ln(x/(2-x))$, and so having $x/(2-x) = e^{2C} e^{2t}$ is OK. Of course, when you re-wrote $e^{2C}$ as $K$ and then forgot the origin of $K$, you were then able to have a valid formula!

 

[LCKurtz]

$ln|\frac{x}{x-2}| = 2t + 2C$
$\frac{x}{x-2} = e^{2t}e^{2c}$
$\frac{1}{1-\frac{2}{x}} = Ke^{2t}$

For $t$ near 0 you have $x$ near 1, and so $x/(x-2) = e^{2C} e^{2t}$ is impossible unless you let $C$ be a complex number. However, $\ln(|x|/|x-2|) = \ln(x/(2-x))$, and so having $x/(2-x) = e^{2C} e^{2t}$ is OK. Of course, when you re-wrote $e^{2C}$ as $K$ and then forgot the origin of $K$, you were then able to have a valid formula!

@Drakkith: Ray is quite correct and I didn't notice you had been sloppy with the absolute value signs. A better way for you to have written it would be like this:$$
ln\left | \frac{x}{x-2}\right | = 2t + 2C$$ $$
\left|\frac{x}{x-2}\right | =e^{2t + 2C}=e^{2C}e^{2t}$$ $$
\frac{x}{x-2} =\pm e^{2C}e^{2t} = Ke^{2t}$$

 

[Drakkith]

So you could say $\frac x {x-2} = Ke^{2t}$ Put $t=0,~x=1$ there to get $\frac x {x-2} = -e^{2t}$. Now put $t=\ln 2$ in that, then solve for $x$. See if that fixes it.

Yes, that seemed to work. I got x=8/5, which is one of the possible answers. Thanks guys!