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Given

$y=Ke^{mx}+J$

where K and J are both unknown constants, can I do this:

$\frac{y}{K}=e^{mx}+\frac{J}{K}$

Since J and K are both unknowns, I can just combine them and call the combined unknown K.

$\frac{J}{K}=K$

$\frac{y}{K}=e^{mx}+K$

$y=K(e^{mx}+1)$

I'm really trying to get rid of J altogether.


The differential equation this relates to:

$ay''+by'+cy=0$

set $c=0$

$ay''+by'=0$

set $y'=w$

$aw'+bw=0$

solve this:

$e^{\frac{b}{a}x}w=K$

$w=Ke^{\frac{-b}{a}x}$

$y=K\int e^{\frac{-b}{a}x}$

$y=-K \frac{a}{b}e^{\frac{-b}{a}x} + J$

$y=Ke^{\frac{-b}{a}x} + J$

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    @KorganRivera Except, you're doing something different that's not actually allowed. This isn't two constants of integration that you just add together. One is in front of $e^{mx}$ and one is out by itself. It's not at all the same situation as you just described. And, it has nothing to do with $\infty + \infty = \infty$.2012-09-29

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No, your starting function is $y = Ke^{mx} + J$ and your last is $y = Ke^{mx} + K$. Unless $K = J$, these are clearly different functions.

When you divided both sides by $K$, you should get

$\frac{y}{K} = e^{mx} + \frac{J}{K}.$

Korgan, since you still don't understand, here are all the details. Let's start at the equation just above. If we want to replace $\frac{J}{K}$ with some other constant, let's use a different letter so it's less confusing. So, let $I = \frac{J}{K}$. This gives

$\frac{y}{K} = e^{mx} + I$

but we still have $K$ in the denominator, and never have we found any justification to know that $K = I$ so we can't just replace $K$ with $I$. After all, $I = \frac{J}{K}$. We never said anything about $I = K$. If you want to solve for $K$ to get rid of it, you have to solve $I = \frac{J}{K}$ for $K$ which gives $K = \frac{J}{I}$. So, our new equation reads

$\frac{y}{\frac{J}{I}} = e^{mx} + I$

Now, if you really want to call the new constant $K$, now would be a less confusing time to change it to $K$, so we would have

$\frac{y}{\frac{J}{K}} = e^{mx} + K$

Multiplying both sides by $\frac{J}{K}$ to clear the denominator gives

$y = \frac{J}{K} e^{mx} + J$

Notice, it STILL does not give

$y = K e^{mx} + K$

This would only happen in the specific case where $\frac{J}{K} = K$ and $J = K$, which would lead to $J = K = \pm 1$. But, we don't know $J$ or $K$ so we can't assume their values to make things simpler. They're unknown, yet fixed. They're not waiting around for us to simply pick their value.

And, also notice that where we end up is actually more complicated than where we started.

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    @KorganRivera As I showed in this example above, you can do such a thing, but you have to be very careful about the other variables that are around and make sure you account for all of them. So, instead, there would be much less chance for confusion if you renamed it using a new variable name.2012-10-01