A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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)

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or do you leave c in the exponent?

  5. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    is r the variable in this case?

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but does taking the derivative mean to 'reconcile'?

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    y[0] ?

  13. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    no???

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so that would be k, right?

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  18. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  21. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 5

    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)

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.