# Fint r and k in $\frac{dy}{dt} = ry(K-y)$

$$\frac{dy}{dt} = ry(K-y) \Leftrightarrow \frac{1}{k}\ln \frac{y}{y-k} = rt +c$$

And now I'm stuck. I know that $y(0) = 10^4$ and $y(1) = 2 \times 10^4$ and a very big $y(t)$ gives $10^5$.

I need to find out how to find C and then solve for K and r. I've tried a few ways but usually the logaritm gives me a long and complicated value for C.

You are probably told that for very large times $t$, $y(t)\sim10^5$. Since $y(t)\to k$ when $t\to\infty$, this means that you know $k$. Additionally, $$\frac{1}{k}\ln\left|\frac{y(0)}{k-y(0)}\right| = c,\qquad\frac{1}{k}\ln \left|\frac{y(1)}{k-y(1)}\right| = r +c.$$ Two equations for two more unknowns $r$ and $c$... we cool.

• Where do you get $y(t)\rightarrow k$ when $t\rightarrow \inf$ from? – iveqy May 19 '14 at 18:44
• From the fact that $y'(t)\gt0$ when $y(t)\lt k$ and $y'(t)\lt0$ when $y(t)\gt k$ hence $y(t)$ increases to $k$ when $y(0)\lt k$ and $y(t)$ decreases to $k$ when $y(0)\gt k$ (this is also called the phase diagram of the equation). – Did May 19 '14 at 18:50

At $t=0$,$$\ln\left|{\frac{10^4}{10^4-k}}\right|=kc$$ At $t=1$,$$\ln{\left|\frac{2\cdot10^4}{2\cdot10^4-k}\right|}=kr+kc$$
Subtracting, $$\ln{\frac{10^4}{10^4-k}\cdot\frac{2\cdot10^4-k}{2\cdot10^4}}=-kr$$
$$\implies\ln{\frac{2\cdot10^4-2k}{2\cdot10^4-k}}=kr$$
When $t\rightarrow\infty, \ln{\frac{y}{y-k}}\rightarrow\infty\implies y\rightarrow k\implies k=\boxed{10^5}$

$$\ln{\frac{2-2\cdot10}{1-10}}=\ln2=r\cdot10^5$$
$\implies r=\boxed{\frac{\ln2}{10^5}}$

• Nice additions to your post. I wonder what they remind me. – Did May 19 '14 at 18:52
• @did Are you referring to the added modulus in the natural logarithms? I added them because when I substituted the value of $k$, a black hole appeared. – Shaurya Gupta May 19 '14 at 18:54