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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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    Absolutely not!, in the first passage you get $J/K$, not $K$.2012-09-29
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    The first step should presumably read $\frac{y}{K}=e^{mx}+\frac{J}{K}$. The $J$ is still around, and won't go away. A usual way to get rid of $J$ is if you know $y$ at some specific $x_0$.2012-09-29
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    Yeah but $\frac{J}{K}=K$, since they're both unknowns. If If I have J+K, I can just rename the conglomerate unknown either J or K. Can't I do this with multiplication too?2012-09-29
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    If you want to rename $J/K$, you cannot use the same name of another already used variable.2012-09-29
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    @KorganRivera In some occasions, you could replace $J/K$ with $K$, though you should be clear you are doing this by writing it out. HOWEVER, in this case that is not okay because you still have another $K$ in the equation and thus your two $K$'s are different. You can not all of the sudden declare that they are the same.2012-09-29
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    Can you explain what you mean by wanting to get rid of $J$? Do you just want to have two unknowns? Then you'll need some sort of information like the value of $y$ at some value $x$ (or a couple such points).2012-09-29
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    I thought that if two unknown values combine, it doesn't matter what name you give it, it's still unknown. So, given that both Ks are unknown, it can't be proven that they're not the same thing.2012-09-29
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    @Graphth When you have a second-order differential equation, and the coefficient of $y''$ is zero, you can solve and get something of the form of my first equation. But I always end up with J and that complicates things.2012-09-29
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    How about you give us an example of such a second-order differential equation you are starting with.2012-09-29
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    @Graphth Well I suppose so. I'll append it to the question.2012-09-29
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    Okay, so now that you have added to your question, the new question is, why do you think you can just set $c = 0$? By the way you wrote it, $c$ is some fixed constant, possibly 0 but possibly not 0. Then, you set it equal to 0.2012-09-29
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    What?? Is it against the law?2012-09-29
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    @KorganRivera Yes, and you are under arrest.2012-09-29
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    Well I'm wasting my time here, forget I asked.2012-09-29
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    You're wasting our time too, don't forget that. Either $c$ is some unknown fixed constant, or it's 0. If it's 0, then there's no reason to write it in the first place. If it's not assumed to be 0, then you're solving a different differential equation when you all of the sudden decide you want it to be 0. So, you can set $c = 0$ if you want, but you're not solving the original differential equation you wrote. You're just solving $ay'' + by' = 0$. That's all I'm trying to make clear here.2012-09-29
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    Why would you use the same variable name for two different variables? Why?2012-09-29
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    @SiliconCelery Really? Ok listen. Here's the deal here. Unknown constants are neat. If I have two unknown constants A and B, I can say that A+B = A, if I want to. That way, I'm only using one symbol. I could call it Z, I could call it Jennifer if I wanted to. In any case, it's an unknown value. So long as I have some symbol in there that represents 'some unknown value'. It's similar to $\infty+\infty=\infty$.2012-09-29
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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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    They're not functions. Like I said, they're unknown constants.2012-09-29
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    @KorganRivera $K$ and $J$ are unknown constants, $y = Ke^{mx} + J$ is a function, where you input a value of $x$ and get out a value of $y$. The function depends on the fixed values of the unknown constants $K, J, m$.2012-09-29
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    Oh, I thought you were saying that J and K were functions.2012-09-29
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    Thanks for the answer. What about renaming $-K\frac{a}{b}$ as just K? Can that work?2012-09-30
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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