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anonymous

  • 5 years ago

Can someone help me solve this Differential equation? dy+lnxy dx=(4x+lny)dx

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  1. anonymous
    • 5 years ago
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    I think this is where I start? \[dy/dx+\ln xy=4x+\ln y\]

  2. anonymous
    • 5 years ago
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    You just have to use log laws to open the product of xy in the LHS log into a sum. That is,\[\frac{dy}{dx}+\ln xy = 4x + \ln y \rightarrow \frac{dy}{dx}+\ln x + \ln y = 4x + \ln y\]The logs of y cancel and you're left with\[\frac{dy}{dx}=4x-\ln x \rightarrow y=2x^2-x \ln x +x+c\]

  3. anonymous
    • 5 years ago
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    \[=2x^2+x(1-\ln x)+c\]

  4. anonymous
    • 5 years ago
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    Do you know how to integrate ln(x)?

  5. anonymous
    • 5 years ago
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    Yes, I'm not too familiar with the laws of logs thats where I needed help. Does dy just = y?

  6. anonymous
    • 5 years ago
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    Oh, are you asking if dy becomes y as in \[\frac{dy}{dx}=4x- \ln x \rightarrow \int\limits_{}{}dy=\int\limits_{}{}4x-\ln x dx \rightarrow \]\[y=2x^2-x \ln x + x+c\]?

  7. anonymous
    • 5 years ago
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    If so, yes.

  8. anonymous
    • 5 years ago
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    Right... and isn't \[\int\limits_{}^{} lnx dx = x-\ln(x)-x\] ? You have +x

  9. anonymous
    • 5 years ago
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    No...\[\int\limits_{}{}\ln x dx=x \ln x -x+c\]I have a + because I skipped a step: when you sub. the result of the integral in, you'll have\[2x^2-(x \ln x -x) +c\]which is equal to \[2x^2-x \ln x + x +c\]

  10. anonymous
    • 5 years ago
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    You can integrate ln(x) using integration by parts, taking u=ln(x), dv=dx and going from there.

  11. anonymous
    • 5 years ago
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    Ohh I see I forgot to distribute the - back into it.

  12. anonymous
    • 5 years ago
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    Yep

  13. anonymous
    • 5 years ago
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    You can always check your solution by subbing it into the differential equation.

  14. anonymous
    • 5 years ago
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    Ok and for the final anwer you just moved the x on the back, to the front and factored a little right?

  15. anonymous
    • 5 years ago
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    Yeah, for the final answer (which technically you don't need to do since the first answers the question), I've just taken out the common factor of x.

  16. anonymous
    • 5 years ago
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    Great. Thanks for your help :)

  17. anonymous
    • 5 years ago
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    No probs.

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