## anonymous one year ago 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)

1. anonymous

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

2. anonymous

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

3. UsukiDoll

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

4. anonymous

or do you leave c in the exponent?

5. UsukiDoll

is r the variable in this case?

6. anonymous

but does taking the derivative mean to 'reconcile'?

7. anonymous

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:

8. anonymous

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?

9. anonymous

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

10. UsukiDoll

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

11. anonymous

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

12. anonymous

y[0] ?

13. UsukiDoll

no???

14. anonymous

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

15. anonymous

so that would be k, right?

16. anonymous

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

17. anonymous

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

18. phi

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 $\ y= k\ e^{rx}$

19. UsukiDoll

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

20. anonymous

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

21. phi

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)