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anonymous

  • 5 years ago

Find the solution of the following differential equation that passes through (-1,-1). You may use separation of variables. dy/dt = [ 2t(y+1) ] / [y] This is what I have after separation of variables and integration of both sides: y - ln(y+1) = t^2 +c Plugging in (-1,-1) would not work as ln(0) does not exist. Is there another way to solve this problem? Thanks in advance!

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  1. myininaya
    • 5 years ago
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    this is weird my solution also has ln(y+1)

  2. myininaya
    • 5 years ago
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    it seems like they would give points for which the function is defined

  3. anonymous
    • 5 years ago
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    my solution doesn't.

  4. myininaya
    • 5 years ago
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    what did you get?

  5. myininaya
    • 5 years ago
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    it has to be that it is undefined. i feel confident in my answer

  6. myininaya
    • 5 years ago
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    let me ask loiskin though one sec

  7. myininaya
    • 5 years ago
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    oh he left nvm

  8. anonymous
    • 5 years ago
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    hint: it is NON-LINEAR ordinary diffeq

  9. myininaya
    • 5 years ago
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    okay we have y dy/(y+1)=2t dt. Ingratiating both sides gives us y+1-ln(y+1)=t^2/2+C

  10. myininaya
    • 5 years ago
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    for the y part, I used a substitution y+1=u so dy=du and we we can find y=u-1 so we have int([u-1]/u, u)=u-lnu= y+1-ln(y+1)

  11. myininaya
    • 5 years ago
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    therefore, the function does not exist at (-1,-1)

  12. myininaya
    • 5 years ago
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    oops the t side is both to have t^2+C, that was a type-0

  13. myininaya
    • 5 years ago
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    suppose to not both to lol stupid type-0s

  14. anonymous
    • 5 years ago
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    nope. you didn't finish the problem.

  15. myininaya
    • 5 years ago
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    ln(y+1) does not exist for y=-1

  16. myininaya
    • 5 years ago
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    (-1,-1) is not in the domain or range

  17. anonymous
    • 5 years ago
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    keep going.

  18. myininaya
    • 5 years ago
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    just tell me what you are thinking. I'm not the one who needs this problem done. lol

  19. myininaya
    • 5 years ago
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    that guy left along time ago it looks like

  20. myininaya
    • 5 years ago
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    did you want me to say there isn't a constant that exist with the initial condition y(-1)=-1. Is this what you wanted me to say?

  21. anonymous
    • 5 years ago
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    I got that answer, but I'm looking at something.

  22. anonymous
    • 5 years ago
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    Hmm, I have to say, given that this is a first order differential equation, and you've found a solution, and solutions are unique, if the solution satisfies the d.e., then that's it! Couldn't it be the case you have a typo. for the BC's?

  23. myininaya
    • 5 years ago
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    he already left. I just wanted to make sure I said what I needed to about the solution since quantish felt I was wrong about the solution or something. I'm not really sure what quantish thinks. Thanks for looking lokisan.

  24. anonymous
    • 5 years ago
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    Do you have maple, mathmatica, or matlab?

  25. myininaya
    • 5 years ago
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    i have maple

  26. anonymous
    • 5 years ago
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    I've run a few different things in Wolfram Alpha, and I'm pretty sure no solutions exist.

  27. anonymous
    • 5 years ago
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    so wolframalpha said no solution exist?

  28. anonymous
    • 5 years ago
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    Wolfram Alpha just confirmed what I worked out above. I tried to get it to repeat the process and solve given the initial condition and it reports "No solutions exist". http://www.wolframalpha.com/input/?i=-1+%3D+-W%28-e^%28-x-1-1%29%29-1+

  29. anonymous
    • 5 years ago
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    There's something wrong with your question, myininaya.

  30. myininaya
    • 5 years ago
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    it isn't mine lol

  31. myininaya
    • 5 years ago
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    i was just worried about it because it makes me worry when people disagree with me

  32. anonymous
    • 5 years ago
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    Oh, well, I think you have the weight of evidence on your side ;)

  33. myininaya
    • 5 years ago
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    ok thanks for supporting my solution.

  34. anonymous
    • 5 years ago
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    No problems.

  35. anonymous
    • 5 years ago
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    I played around with that too and kept getting similar answers. I'll have to ask my professor. :/

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