growth and decay differential equations

[burt]

I am using modernstates.org to learn about this. The lecturer said that e^(kx+c) is equal to ce^kc. How can this be?

 

[Romsek]

\(\displaystyle \text{they probably said that $e^{kx+c} = C e^{kx}$, where $C = e^c$}\)

or because the constant can be anything they may have just said eh well call this one \(\displaystyle c\) as well.

 

[burt]

\(\displaystyle \text{they probably said that $e^{kx+c} = C e^{kc}$, where $C = e^c$}\)

So because c is a constant, e^c=c?

 

[topsquark]

\(\displaystyle \text{they probably said that $e^{kx+c} = C e^{kc}$, where $C = e^c$}\)

Danger Will Robinson! Danger!

Typo alert:
\(\displaystyle e^{kx+c} = C e^{kc}\)

Should be
\(\displaystyle e^{kx+c} = C e^{kx}\)

-Dan

 

[Romsek]

Danger Will Robinson! Danger!

Typo alert:
\(\displaystyle e^{kx+c} = C e^{kc}\)

So because c is a constant, e^c=c?

not exactly...

we know that the solution is \(\displaystyle e^{kx +c} = e^{kx}e^c\)

so let \(\displaystyle C = e^c\)
the solution becomes \(\displaystyle C e^{kx}\)

but you can say well \(\displaystyle C\) is just some constant, so I can forget about the old \(\displaystyle c\)
and just call this new constant \(\displaystyle c\) and it doesn't change the solution at all.

 

[burt]

not exactly...

we know that the solution is \(\displaystyle e^{kx +c} = e^{kx}e^c\)

so let \(\displaystyle C = e^c\)
the solution becomes \(\displaystyle C e^{kx}\)

but you can say well \(\displaystyle C\) is just some constant, so I can forget about the old \(\displaystyle c\)
and just call this new constant \(\displaystyle c\) and it doesn't change the solution at all.

I see - it's the difference between c and C.

 

[lookagain]

I am using modernstates.org to learn about this. The lecturer said that e^(kx+c) is equal to ce^kc. How can this be?

And you need grouping symbols around the exponent when you write it out horizontally. If you put all of
the above help and my suggestion together, you could have:

e^(kx+c) =

e^(c + kx) =

[e^c]*[e^(kx)]

Let C = e^c

So, [e^c]*[e^(kx)] =

Ce^(kx)