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UnkleRhaukus

  • 5 years ago

\[{dy \over dx}={4y^2 \over x^2}+5{y \over x}+1\]

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  1. TuringTest
    • 5 years ago
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    bookmark*

  2. UnkleRhaukus
    • 5 years ago
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    i think it is a Homogeneous Equation requiring a change of variables

  3. UnkleRhaukus
    • 5 years ago
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    (a linear shift

  4. TuringTest
    • 5 years ago
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    I saw your work on it yesterday and saw no mistakes, so I'm curious as well

  5. Akshay_Budhkar
    • 5 years ago
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    i believe we have to solve using y=mx where m= y/x

  6. TuringTest
    • 5 years ago
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    that's what rhaukus did yesterday, but he got the wrong answer... ideas imperialist?

  7. Akshay_Budhkar
    • 5 years ago
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    dy=mdx + xdm

  8. anonymous
    • 5 years ago
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    I'm trying a couple of things, I'll let you know if anything works.

  9. UnkleRhaukus
    • 5 years ago
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    thanks

  10. Akshay_Budhkar
    • 5 years ago
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    m + x dm/dx = 4m^2 + 5 m +1

  11. Akshay_Budhkar
    • 5 years ago
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    x dm /dx= 4m^2+4m+1

  12. Akshay_Budhkar
    • 5 years ago
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    dx/x=dm/4m^2+4m+1

  13. Akshay_Budhkar
    • 5 years ago
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    um then integrate :P

  14. Akshay_Budhkar
    • 5 years ago
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    u can do from here?

  15. UnkleRhaukus
    • 5 years ago
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    ill give it a go ,

  16. anonymous
    • 5 years ago
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    Akshay is right, that's how you do it

  17. anonymous
    • 5 years ago
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    is y = -1/2 x a solution?

  18. anonymous
    • 5 years ago
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    a particular* solution.

  19. UnkleRhaukus
    • 5 years ago
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    \[ln(x)={1 \over 4x+2}+c\]

  20. anonymous
    • 5 years ago
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    \[\ln(x)=\frac{-1}{4v+2}+C\] before you get too far without the negative sign and with x instead of v

  21. anonymous
    • 5 years ago
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    yeah, \[y=-\frac{1}{2}x\]is one solution. just let y = mx like akshay said, and when you plug that in you get an equation for m.

  22. anonymous
    • 5 years ago
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    Sorry, m instead of v in what I said above, I always use v's for these types of questions!

  23. Akshay_Budhkar
    • 5 years ago
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    lol i prefer m :P

  24. UnkleRhaukus
    • 5 years ago
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    i meant \[lnx=-1/(4m+2) +c\]

  25. anonymous
    • 5 years ago
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    Me too! I prefer 'V'

  26. watchmath
    • 5 years ago
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    \(y'/x=4u^2/x+5u/x+1\) where \(u=y/x\) \(y'/x-u/x=4u^2/x+4u/x+1/x\) \(\frac{xy'-y}{x^2}=4u^2/x+4u/x+1/x\) \(du/dx=4u^2/x+4u/x+1/x\) \(du/(4u^2+4u+1)=dx/x\) \(du/(2u+1)^2=dx/x\) \(-\frac{1}{2}(2u+1)^{-1}=\ln|x|+C\) \(-\frac{1}{2}(\frac{2y}{x}+1)^{-1}=\ln|x|+C\) Let see ...

  27. Akshay_Budhkar
    • 5 years ago
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    just re-substitute the m now to get your equation.. i think we are done, aren't we?

  28. UnkleRhaukus
    • 5 years ago
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    \[{-1 \over 2}(2{y \over x}+1)^{-1}=ln|x| +C\] ... \[x+2(2y+x)ln|kx|=0\] where \[(e^C=k)\]

  29. Akshay_Budhkar
    • 5 years ago
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    i think that your answer is correct :D

  30. TuringTest
    • 5 years ago
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    I thought you said it had an arctan in it Unk? what did you do wrong last time?

  31. Akshay_Budhkar
    • 5 years ago
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    did u try wolfphram @turing?

  32. TuringTest
    • 5 years ago
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    I'm going by unk's last post, no haven't looked at wolf looking at this http://openstudy.com/users/unklerhaukus#/updates/4f14e000e4b0b9109c92149b

  33. watchmath
    • 5 years ago
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    \(-1/(\ln|x|+C) =4(y/x)+2\) \(-x/(\ln |x|+C)=4y+2x\) \(y=1/4( -x/(\ln|x|+c) -2x)\)

  34. Akshay_Budhkar
    • 5 years ago
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    @turing just check with wolfphram.. it is possible to have two different ways of represnting the same function

  35. TuringTest
    • 5 years ago
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    I just assumed they were too different, but I suppose you are right.

  36. Akshay_Budhkar
    • 5 years ago
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    not a Guru at using wolhphram but i assume this is right http://www.wolframalpha.com/input/?i=dy%2Fdx%3D%284y%5E2%2Fx%5E2%29+%2B+5y%2Fx+%2B+1

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