anonymous
  • anonymous
Can someone help me solve this Differential equation? dy+lnxy dx=(4x+lny)dx
Mathematics
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anonymous
  • anonymous
Can someone help me solve this Differential equation? dy+lnxy dx=(4x+lny)dx
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I think this is where I start? \[dy/dx+\ln xy=4x+\ln y\]
anonymous
  • anonymous
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\[\frac{dy}{dx}=4x-\ln x \rightarrow y=2x^2-x \ln x +x+c\]
anonymous
  • anonymous
\[=2x^2+x(1-\ln x)+c\]

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anonymous
  • anonymous
Do you know how to integrate ln(x)?
anonymous
  • anonymous
Yes, I'm not too familiar with the laws of logs thats where I needed help. Does dy just = y?
anonymous
  • anonymous
Oh, are you asking if dy becomes y as in \[\frac{dy}{dx}=4x- \ln x \rightarrow \int\limits_{}{}dy=\int\limits_{}{}4x-\ln x dx \rightarrow \]\[y=2x^2-x \ln x + x+c\]?
anonymous
  • anonymous
If so, yes.
anonymous
  • anonymous
Right... and isn't \[\int\limits_{}^{} lnx dx = x-\ln(x)-x\] ? You have +x
anonymous
  • anonymous
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 \[2x^2-x \ln x + x +c\]
anonymous
  • anonymous
You can integrate ln(x) using integration by parts, taking u=ln(x), dv=dx and going from there.
anonymous
  • anonymous
Ohh I see I forgot to distribute the - back into it.
anonymous
  • anonymous
Yep
anonymous
  • anonymous
You can always check your solution by subbing it into the differential equation.
anonymous
  • anonymous
Ok and for the final anwer you just moved the x on the back, to the front and factored a little right?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Great. Thanks for your help :)
anonymous
  • anonymous
No probs.

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