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niah2411

  • 2 years ago

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

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  1. genius12
    • 2 years ago
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    are you supposed to find dy/dx?

  2. niah2411
    • 2 years ago
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    yup..

  3. niah2411
    • 2 years ago
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    im not confident of the answer in my book

  4. genius12
    • 2 years 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
    • 2 years ago
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    can you show solution??

  6. genius12
    • 2 years 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
    • 2 years ago
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    variable seperable method?

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

  9. niah2411
    • 2 years ago
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    yes

  10. niah2411
    • 2 years ago
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    maybe my book is wrong

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

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

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

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

  15. agent0smith
    • 2 years 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
    • 2 years ago
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    Can you stop tagging everyone. Please don't pay attention guys.

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

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

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

  20. niah2411
    • 2 years ago
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    oh..i see i understand that

  21. agent0smith
    • 2 years 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
    • 2 years ago
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    ohh can you give me your full solution so i can understand it really..pls?

  23. agent0smith
    • 2 years 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
    • 2 years ago
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    oh im sorry @agent0smith for the inconvenience

  25. genius12
    • 2 years 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
    • 2 years ago
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    Can you solve for 'y' now?

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

  28. agent0smith
    • 2 years 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
    • 2 years ago
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    @agent0smith I don't know what you're saying?

  30. agent0smith
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Yep, that's what i worked out. The book/yours is the same.

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