Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

rosho

dy/dx=2y+6, y(1)=6 (calculus II) the answer is y(x)=9e^(2x-2)-3

  • one year ago
  • one year ago

  • This Question is Closed
  1. Best_Mathematician
    Best Response
    You've already chosen the best response.
    Medals 0

    explain what u need to find

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

    \[\large \frac{dy}{dx}=2y+6\] So this appears to be separable :) Divide each side by (2y+6), and move the dx to the other side, think of this step as multiplication. \[\large \frac{dy}{2y+6}=dx\] Understand that part? From here we can integrate both sides.

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

    yes, got that part!

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

    \[\large \int\limits \frac{dy}{2y+6}=\int\limits dx\]Integrating the right side is pretty straight forward,\[\large \int\limits\limits \frac{dy}{2y+6}=x+c\] Do you understand how to do the left side? You can apply a `u sub` if it will help.

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

    go ahead ur doing great @zepdrix

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

    all I need is the steps, i understand most of it.

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

    Just in case you're having trouble with the left side, a u substitution would look like this, \[\large u=2y+6 \qquad \qquad \qquad \frac{1}{2}du=dy\] Which changes our integral to,\[\large \frac{1}{2}\int\limits\frac{du}{u}=x+c\] Giving us,\[\large \frac{1}{2}\ln|2y+6|=x+c\]

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

    I already got that far. Now to put in the form y=9e^(2x-2)-3.

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

    Multiply both sides by 2,\[\large \ln|2y+6|=2x+c\] We distributed the 2 to each term, but we simply `absorbed` the other 2 into the C, since it's an arbitrary value. Now exponentiate both sides, `rewrte both sides as exponents with a base of e`.\[\huge e^{\ln|2y+6|}=e^{2x+c}\]

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

    The exponential and the logarithm are `inverse` operations of one another, so on the left they will simply "undo" one another.\[\huge 2y+6=e^{2x+c}\]

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

    oh, i get it!

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

    Understand how to solve for C from here? :)

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

    i forgot that c was a constant.

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

    and could be treated like that. now it's easy i think.

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

    but what about the 9?

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

    Are you not coming up with a 9? :o

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

    y=e^(2x+c)/2-3

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

    There's what I've got so far.

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

    Ok cool. You need to solve for C at some point :) That's where your 9 will come from.

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

    Don't fully comprehend, could you please help a little more.

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

    At the start they gave us a coordinate pair that we can NOW use to solve for C.\[\large y(1)=6 \qquad \qquad (1,6)\] Plug the 1 in for X, the 6 in for Y, and solve for C! :) I have a feeling this will be a tad easier to work with if we fiddle with the C first. Using a rule of exponents we can write the C like this. \[\huge 2y+6=e^{2x}e^c\] But now recognize, that e^c is just another arbitrary constant value.\[\huge 2y+6=Ce^{2x}\]Then subtract, divide,...\[\huge y=Ce^{2x}-3\]Again the 2 got absorbed into the C. This isn't totally necessary. I just think the problem will be a lot easier to work with from this form.

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

    \[\huge (\color{royalblue}{1},\color{orangered}{6}) \qquad \rightarrow \qquad \color{orangered}{y}=Ce^{2\color{royalblue}{x}}-3\]Understand how to plug these values in? :)

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

    I finally got it!!! Thanks a lot!

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

    yay! \c:/

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

    try solving mine plz

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

    The silly bread problem? c:

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

    ya

    • one year ago
    • Attachments:

See more questions >>>

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.