If dx/dt=kx, and if x=2 when t=0 and x=6 when t=1, then k=

[karush]

If $\frac{dx}{dt}=kx$, and if $x=2$ when $t=0$ and $x=6$ when $t=1$,
then $k=$

or rewriten as

$\frac{dx}{kx} = dt$

I saw this posted on another forum but didn't understand how this was integrated.. the ans appears to \(\displaystyle ln 3\)

 

[MarkFL]

We have the separable ODE:

\(\displaystyle \frac{dx}{dt}=kx\)

\(\displaystyle \frac{1}{x}\,dx=k\,dt\)

\(\displaystyle \ln|x|=kt+C\)

Now, we know the point:

\(\displaystyle (t,x)=(0,2)\)

Hence:

\(\displaystyle C=\ln(2)\)

Thus:

\(\displaystyle \ln|x|=kt+\ln(2)\)

We also know the point:

\(\displaystyle (t,x)=(1,6)\)

Hence:

\(\displaystyle k=\ln(6)-\ln(2)=\ln(3)\)