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niah2411

differential equation of xy^3lnxdx-2dy=0

  • 8 months ago
  • 8 months ago

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  1. genius12
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    are you supposed to find dy/dx?

    • 8 months ago
  2. niah2411
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    yup..

    • 8 months ago
  3. niah2411
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    im not confident of the answer in my book

    • 8 months ago
  4. genius12
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    Just re-arrange. Take the -2dy to the other side, divide both sides by dx, and then divide both sides by 2.

    • 8 months ago
  5. niah2411
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    can you show solution??

    • 8 months ago
  6. genius12
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    First add 2dy to both sides:\[\bf xy^3\ln(x)dx-2dy=0 \implies xy^3\ln(x)dx=2dy\]Now divide both sides by dx:\[\bf xy^3\ln(x)=2\frac{dy}{dx}\]Now divide both sides by 2:\[\bf \therefore \ \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\]

    • 8 months ago
  7. niah2411
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    variable seperable method?

    • 8 months ago
  8. genius12
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    @niah2411 Do you see what I did?

    • 8 months ago
  9. niah2411
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    yes

    • 8 months ago
  10. niah2411
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    maybe my book is wrong

    • 8 months ago
  11. genius12
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    add 2dy to both sides then divide both sides by 2dx.

    • 8 months ago
  12. niah2411
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    our methods are the same and also the answers, the book answer is maybe wrong

    • 8 months ago
  13. genius12
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    May be. It often happens that the publishers make a mistake in the solutions.

    • 8 months ago
  14. niah2411
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    the book answer is x^2(lnx^2-1)+4y^-2=c

    • 8 months ago
  15. agent0smith
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    That looks like you're supposed to find the solution to the equation, then, not just find dy/dx

    • 8 months ago
  16. genius12
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    Can you stop tagging everyone. Please don't pay attention guys.

    • 8 months ago
  17. niah2411
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    @agent0smith can you give me a solution if you have one?

    • 8 months ago
  18. agent0smith
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    Sorry, don't really remember much of this part of calculus :(

    • 8 months ago
  19. niah2411
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    @genius12 can you give me a solution using variable separable method?

    • 8 months ago
  20. niah2411
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    oh..i see i understand that

    • 8 months ago
  21. agent0smith
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    Oh, yes you can use separation of variables: \[\Large \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\] multiply both sides by dx, divide both sides by y^3\[\Large \int\limits \frac{ x\ln(x) }{ 2 } dx= \int\limits \frac{dy}{y^3}\]Now you can integrate.

    • 8 months ago
  22. niah2411
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    ohh can you give me your full solution so i can understand it really..pls?

    • 8 months ago
  23. agent0smith
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    Make it a little simpler to integrate:\[\Large \frac{ 1 }{ 2 } \int\limits\limits x \ln(x) dx= \int\limits\limits y^{-3} dy\] On the left it looks like you'll need to use integration by parts. Sorry, I'm too tired to go through it all now, someone else can take over.

    • 8 months ago
  24. niah2411
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    oh im sorry @agent0smith for the inconvenience

    • 8 months ago
  25. genius12
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    @niah2411 I'll take over from where @agent0smith left off. To continue, we will use integration by parts for the "xln(x)" integral: \[\bf \frac{ 1 }{ 2 }\int\limits_{}^{}xln(x) \ dx \implies \frac{ 1 }{ 2 } \left( \frac{x^2\ln(x)}{2}-\int\limits_{}^{}\frac{ x }{ 2} \ dx \right)=\frac{ 1 }{ 2 } \left( \frac{x^2\ln(x)}{2}-\frac{ x^2 }{ 4} \right)\]\[\bf =\frac{ 2x^2\ln(x)-x^2 }{ 8 }+C\]Now the right-side would just be:\[\bf \int\limits_{}^{}y^{-3} \ dy=-\frac{y^{-2}}{2}\]Now we equate the results: \[\bf -\frac{1}{2y^2}=\frac{ 2x^2\ln(x)-x^2 }{ 8 }+C\]

    • 8 months ago
  26. genius12
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    Can you solve for 'y' now?

    • 8 months ago
  27. genius12
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    @agent0smith You still here? Help me awaken @niah2411 ? lol

    • 8 months ago
  28. agent0smith
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    Hehe yeah I'm still here. After saying i was too tired, i did the int by parts on paper (i thought it would take longer than it did, or i would've just typed it up). Looks like he didn't need to find y anyway, this was the answer: x^2(lnx^2-1)+4y^-2=c which you can get by moving terms around and multiplying by 8.

    • 8 months ago
  29. genius12
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    @agent0smith I don't know what you're saying?

    • 8 months ago
  30. agent0smith
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    This was the answer that he gave, from the book: x^2(lnx^2-1)+4y^-2=c Which you can get from your result.

    • 8 months ago
  31. genius12
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    According to my result, which is correct, the answer should be:\[\bf x^2(2\ln(x)-1)+4y^{-2}=C\]So yes, the book answer is correct and I believe he misinterpreted the meaning of "solution" or whatever. They were just looking for separation of variables and then simply bring over the x's and y's' on one side and equating it to the constant. @agent0smith

    • 8 months ago
  32. agent0smith
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    Yep, that's what i worked out. The book/yours is the same.

    • 8 months ago
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