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niah2411

  • one year ago

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

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  1. genius12
    • one year ago
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    are you supposed to find dy/dx?

  2. niah2411
    • one year ago
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    yup..

  3. niah2411
    • one year ago
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    im not confident of the answer in my book

  4. genius12
    • one year ago
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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.

  5. niah2411
    • one year ago
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    can you show solution??

  6. genius12
    • one year ago
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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:\[\bf \therefore \ \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\]

  7. niah2411
    • one year ago
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    variable seperable method?

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

  9. niah2411
    • one year ago
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    yes

  10. niah2411
    • one year ago
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    maybe my book is wrong

  11. genius12
    • one year ago
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    add 2dy to both sides then divide both sides by 2dx.

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

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

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

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

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

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

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

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

  20. niah2411
    • one year ago
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    oh..i see i understand that

  21. agent0smith
    • one year ago
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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.

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

  23. agent0smith
    • one year ago
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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.

  24. niah2411
    • one year ago
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    oh im sorry @agent0smith for the inconvenience

  25. genius12
    • one year ago
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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: \[\bf -\frac{1}{2y^2}=\frac{ 2x^2\ln(x)-x^2 }{ 8 }+C\]

  26. genius12
    • one year ago
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    Can you solve for 'y' now?

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

  28. agent0smith
    • one year ago
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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.

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

  30. agent0smith
    • one year ago
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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.

  31. genius12
    • one year ago
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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

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

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