Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

brinethery

  • 2 years ago

Solve the separable differential equation for y by making the substitution u=t+4y dy/dt = (t+4y)^2 Oh yeah, and y(0) = 7

  • This Question is Closed
  1. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    This is what I thought the answer is, but it's wrong. If it helps at all, I believe this is a Bernoulli equation. :-/

  2. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Proof that I actually tried hehe...

  3. Luigi0210
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @Jhannybean

  4. Jhannybean
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What if we expanded the RHS first?

  5. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Try it... I am really unsure of what to do at this point.

  6. FutureMathProfessor
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Expand and use integrating factor

  7. Jhannybean
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah...

  8. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Perhaps my math teacher led me astray. I'll give int factor a shot

  9. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I can't do it. I get stuck.

  10. Jhannybean
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Lets see...Integrating factor is \(\large e^{\int p(t)y}\)\[\large \cfrac{dy}{dt} = t^2 +8yt+16y^2\]So you'll have \[\large \cfrac{dy}{dt} -8yt -16y^2 = t^2\]

  11. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But will we ever be able to isolate y with this method? That's what I wonder.

  12. Jhannybean
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yep.

  13. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    When you substitute: u=t+4y Could we not just take the derivative of both sides with respect to t: du/dt = 1 + 4 dy/dt and then solve for dy/dt in terms of du/dt to substitute as well? We then have: dy/dt = (t + 4y)^2 <==> 1/4(du/dt - 1) = u^2 1/4 (du/dt - 1) = u^2 du/dt - 1 = 4u^2 du/dt = 4u^2 + 1 Because is that not in some sense a separable DE?

  14. Jhannybean
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I personally don't like the "du/dt -1". But thats just my opinion.

  15. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So integrating factor would be e^(-4t^2) for this, and then just solve?

  16. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think I am going to throw in the towel for tonight. Thank you all for your help anyway...

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

    Try approaching the problem the way access suggested, it makes a lot more sense.

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

    towel throwing time? :c aw

  19. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes :-). I slept 2 hours last night (just couldn't fall asleep). My brain's not working right. Thank you all again, I always appreciate math help.

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

    \[\large u=t+4y\]Taking the derivative of both sides WRT y you get,\[\large \frac{du}{dy}=\frac{dt}{dy}+4\]"Dividing" through by dt/dy gives us,\[\large \frac{du}{dy}\cdot\frac{dy}{dt}=\color{orangered}{1}+4\frac{dy}{dt}\]OOoo Ok I think I see the mistake she made. Forgot the 1 up there, \[\large \frac{du}{dt}=\color{orangered}{1}+4\frac{dy}{dt}\]Which simplifies to,\[\large \frac{du}{dt}=1+4u^2\]

  21. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think the d/dy (t) simply was taken to be 0, more or less like a partial derivative. Since the work only shows the du/dy = 4 Otherwise, it seems the work would have likely been clean. :)

  22. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I'll tell you what all: I'll ask my teacher in the morning how to approach this thing and I'll post the full solution tomorrow. Thank you all for your help.

  23. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That sounds like a fair deal. I am thinking this way looks most promising right now... \( \displaystyle \frac{du}{dt} = 1 + 4u^2 \) \( \displaystyle \frac{1}{1 + 4u^2} \; du = 1 \; dt \) \( \displaystyle \int \frac{1}{1 + 4u^2} \; du = \int 1 \; dt \) \( \displaystyle \frac{1}{2} \tan^{-1} \left(2u\right) = t + C \) \(u = t + 4y \) \( \displaystyle \frac{1}{2} \tan^{-1} \left( 2(t + 4y)\right) = t + C \) I think that looks fairly reasonable. :P

  24. Jhannybean
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That looks much better.

  25. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Zepdrix was right.

  26. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    And accessdenied too :-). Thanks everyone!

  27. AccessDenied
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You are welcome! Glad it was sorted out. :P

  28. Jhannybean
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I like the way you worked ytour problem, @brinethery :)

  29. brinethery
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks!

  30. 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.