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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

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  1. brinethery
    • 2 years ago
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    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
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    Proof that I actually tried hehe...

  3. Luigi0210
    • 2 years ago
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    @Jhannybean

  4. Jhannybean
    • 2 years ago
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    What if we expanded the RHS first?

  5. brinethery
    • 2 years ago
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    Try it... I am really unsure of what to do at this point.

  6. FutureMathProfessor
    • 2 years ago
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    Expand and use integrating factor

  7. Jhannybean
    • 2 years ago
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    Ah...

  8. brinethery
    • 2 years ago
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    Perhaps my math teacher led me astray. I'll give int factor a shot

  9. brinethery
    • 2 years ago
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    I can't do it. I get stuck.

  10. Jhannybean
    • 2 years ago
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    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
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    But will we ever be able to isolate y with this method? That's what I wonder.

  12. Jhannybean
    • 2 years ago
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    Yep.

  13. AccessDenied
    • 2 years ago
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    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
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    I personally don't like the "du/dt -1". But thats just my opinion.

  15. brinethery
    • 2 years ago
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    So integrating factor would be e^(-4t^2) for this, and then just solve?

  16. brinethery
    • 2 years ago
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    I think I am going to throw in the towel for tonight. Thank you all for your help anyway...

  17. zepdrix
    • 2 years ago
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    Try approaching the problem the way access suggested, it makes a lot more sense.

  18. zepdrix
    • 2 years ago
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    towel throwing time? :c aw

  19. brinethery
    • 2 years ago
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    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
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    \[\large u=t+4y\]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
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    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
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    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
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    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
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    That looks much better.

  25. brinethery
    • 2 years ago
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    Zepdrix was right.

  26. brinethery
    • 2 years ago
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    And accessdenied too :-). Thanks everyone!

  27. AccessDenied
    • 2 years ago
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    You are welcome! Glad it was sorted out. :P

  28. Jhannybean
    • 2 years ago
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    I like the way you worked ytour problem, @brinethery :)

  29. brinethery
    • 2 years ago
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    Thanks!

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