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


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




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

The differential equation this relates to:


set $c=0$


set $y'=w$


solve this:



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

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

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

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Absolutely not!, in the first passage you get $J/K$, not $K$. – enzotib Sep 29 '12 at 18:15
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 $x_0$. – André Nicolas Sep 29 '12 at 18:16
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? – Korgan Rivera Sep 29 '12 at 18:19
If you want to rename $J/K$, you cannot use the same name of another already used variable. – enzotib Sep 29 '12 at 18:21
@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. – Graphth Sep 29 '12 at 18:23
up vote 3 down vote accepted

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. – Korgan Rivera Sep 29 '12 at 18:20
@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 $K, J, m$. – Graphth Sep 29 '12 at 18:20
Oh, I thought you were saying that J and K were functions. – Korgan Rivera Sep 29 '12 at 18:33
Thanks for the answer. What about renaming $-K\frac{a}{b}$ as just K? Can that work? – Korgan Rivera Sep 30 '12 at 5:15
@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. – Graphth Oct 1 '12 at 1:11

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