Question

How do you reconcile where d & c are constants?

y[x] = d E^( r x - c r)

Reconcile with the formula
y[x] = k E^(r x)

Answer 1

I cant seem to work out how you would get c out of the exponent?

Answer 2

its easy if c is defined to something, but if it's just c.. ??

Answer 3

c and d are constants... well using product rule that d is going to go bye bye if you leave the second term alone and deal with the first

Answer 4

or do you leave c in the exponent?

Answer 5

is r the variable in this case?

Answer 6

but does taking the derivative mean to 'reconcile'?

Answer 7

ah, the actual question went..

When you start with
y'[x] = r y[x] ,
give it one data point, and solve for y[x] , you always get something of the form
y[x] = k e^(r x)
where k is a constant determined by the data point.
Try it with the data point y[c] = d where c and d are constants:

Answer 8

This gives you
y[x] = d e^((-r) c + r x) .
How do you reconcile this output with the formula
y[x] = k e^(r x)
above?

Answer 9

It looks like d and c need to combine somehow to form k ?

Answer 10

no.. and ugh sorry for the wait my batteries died on my wireless mouse

Answer 11

y'[x]/y[x] = r
is there a formula for just k?

Answer 12

y[0] ?

Answer 13

no???

Answer 14

I get y[0] = d E^-(cr)

Answer 15

so that would be k, right?

Answer 16

y[x] = d e^(-cr + r x) .
y[x] = d e^(-cr) E^(r x) .

y[x] = d/e^(cr) E^(r x) .
k = d/e^(cr)
r = r

Answer 17

let me know if Im off mark.. thanx.. hitting the sack

Answer 18

I would analyze it this way
\[y'[x] = r y[x] \\ \frac{dy}{dx} = r y \\ \frac{dy}{y}= r dx \\ \int \frac{dy}{y}= \int r\ dx\]
now integrate, and explicitly note the constant of integration:
\[ \ln y +c_1= rx + c_2 \\ \ln y = r x +(c_2-c_1) \\ \ln y = r x + c\]
I put in a constant on both sides, but as you can see, because they are arbitrary,
you can combine them into a single arbitrary constant.
Most people would just write the constant on one side, like this
\[ \int \frac{dy}{y}= \int r\ dx \\ \ln y = r x + c \]
now make each side the exponent of "e"
\[ y = e^{rx+c}\\ y= e^c\ e^{rx} \]
"c" is an arbitrary number, so that means e^c is also some number. Call it k. Thus
\[ \ y= k\ e^{rx} \]

Answer 19

that was awesome @phi
I think it was due to @hughfuve 's notation on the equation that got me really confused so I couldn't go on further, but when you've changed it, I can understand that we need to use separation of variables

Answer 20

wow that's awesome hey.
separation of variables and
integration

Answer 21

to finish up the details
data point y[c] = d where c and d are constants:
this means when x=c, y = d, i.e. point (c,d) is a solution to the equation
\[ y = k\ e^{rx} \]
plug in x=c , y=d and solve for k:
\[ d= k\ e^{rc} \\ k = d\ e^{-rc} \]

notice this matches with
***
How do you reconcile where d & c are constants?
y[x] = d E^( r x - c r)
Reconcile with the formula y[x] = k E^(r x)
***
k= d E^(cr)
(some constant determined by the fixed constants c , d and r)