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RolyPoly

  • 2 years ago

\[ (2y^2 - 6xy)dx + (3xy-4x^2)dy =0\] Find an integrating factor of the form \(x^ny^m\) and solve the equation...

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  1. abb0t
    • 2 years ago
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    1st check to make sure that they are exact.

  2. RolyPoly
    • 2 years ago
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    \(M = 2y^2 - 6xy \) \(N= 3xy-4x^2\) \(\frac{\partial M}{\partial y}=4y - 6x\) \(\frac{\partial N}{\partial x}=3y-8x\) \[\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}=\frac{3y-8x-(4y - 6x)}{2y(y-3x)}\]Doesn't go right..

  3. abb0t
    • 2 years ago
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    \[\frac{ ∂M }{ ∂x }=\frac{ ∂N }{ ∂y }\]

  4. abb0t
    • 2 years ago
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    check your partial derivatives. they should match up

  5. Outkast3r09
    • 2 years ago
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    either that or your teacher is horrible and wants you to solve this another way ll

  6. UnkleRhaukus
    • 2 years ago
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    integrating factor!

  7. RolyPoly
    • 2 years ago
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    But I couldn't find a suitable integrating factor... :'(

  8. Outkast3r09
    • 2 years ago
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    ah yes this is not an exact solution one

  9. abb0t
    • 2 years ago
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    UnkleR is right. Find the integrating factor to make it exact.

  10. RolyPoly
    • 2 years ago
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    That's why I tried that partial.. - partial.. / ... to find one..

  11. Outkast3r09
    • 2 years ago
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    is that a hint... try getting one in such that it looks like above

  12. abb0t
    • 2 years ago
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    You were on the right path.

  13. abb0t
    • 2 years ago
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    \[\frac{ M_y-N_x }{ -M }\]

  14. abb0t
    • 2 years ago
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    That's your integrating factor.

  15. RolyPoly
    • 2 years ago
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    The way I learn to get an integrating factor is by showing \(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) or \(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) equals to a function of x/y .. But for this (and the next question), I got troubles..

  16. UnkleRhaukus
    • 2 years ago
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    \[\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}\] \[R=R(x,y)=x^ny^m\] \[\frac{\partial R(x,y)M}{\partial y}= \frac{\partial R(x,y)N}{\partial x}\] \[R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}\]

  17. RolyPoly
    • 2 years ago
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    How... does... that... work...?

  18. UnkleRhaukus
    • 2 years ago
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    which bit

  19. RolyPoly
    • 2 years ago
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    \[\frac{\partial R(x,y)M}{\partial y}= \frac{\partial R(x,y)N}{\partial x}\]And \[R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}\]

  20. UnkleRhaukus
    • 2 years ago
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    well we want a integrating factor that will make the partial derivatives equal (which will makes the equation exact) applying the product rule for derivatives on both sides we get that last line i have used mixed notation for derivatives for some reason \[Z_w\leftrightarrow\frac{\partial Z}{\partial w}\](these mean the same thing, just written differently )

  21. UnkleRhaukus
    • 2 years ago
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    so you know \[R,M,N\]substitute these in

  22. UnkleRhaukus
    • 2 years ago
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    take the partial derivatives

  23. RolyPoly
    • 2 years ago
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    Is R the integrating factor?

  24. UnkleRhaukus
    • 2 years ago
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    yes

  25. RolyPoly
    • 2 years ago
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    So, I should multiply the equation by R, right?

  26. RolyPoly
    • 2 years ago
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    \[R=x^ny^m\]\[M=2y^2-6xy\]\[N=3xy-4x^2\] \[\frac{\partial R}{\partial y} M + \frac{\partial M}{\partial y} R= \frac{\partial R}{\partial x} N+ \frac{\partial N}{\partial x}R\] \[LS\]\[=mx^ny^{m-1}(2y^2-6xy) + (4y-6x)x^ny^m\]\[=2mx^ny^{m+1}-6mx^{n+1}y^m + 4y^{m+1}x^n – 6x^{n+1}y^m\]\[(2m+4)x^ny^{m+1}-(6m+1)x^{n+1}y^m\] \[RS\]\[=nx^{n-1}y^m(3xy-4x^2) + (3y-8x)x^ny^m\]\[=3nx^ny^{m+1}-4nx^{n+1}y^m+3x^ny^{m+1} – 8x^{n+1}y^m\]\[(3n+3)x^ny^{m+1} – (4n+8)x^{n+1}y^m\] So, \[(2m+4)x^ny^{m+1}-(6m+1)x^{n+1}y^m = (3n+3)x^ny^{m+1} – (4n+8)x^{n+1}y^m\] 2m+4 = 3n+3 6m+1 = 4n+8 m=5/2 , n=2 But that is not right :\

  27. UnkleRhaukus
    • 2 years ago
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    hmm, i cannot see any error in your working , how do you know its not right?

  28. RolyPoly
    • 2 years ago
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    Because the book whispered me the answer :(

  29. UnkleRhaukus
    • 2 years ago
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    \[\tiny \text{can you whisper it to me too please , }\]

  30. RolyPoly
    • 2 years ago
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    \[\tiny \color{white}{\mu =xy} \]

  31. UnkleRhaukus
    • 2 years ago
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    i see

  32. hartnn
    • 2 years ago
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    \(\huge \color{red}{\mu =xy}\) now, i see also.

  33. UnkleRhaukus
    • 2 years ago
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    well we were close \[x^2y^{5/2}(2y^2 - 6xy)\text dx + x^2y^{5/2}(3xy-4x^2)\text dy =0\] \[(2x^2y^{9/2} - 6x^3y^{7/2})\text dx +(3x^3y^{7/2}-4x^4y^{5/2})\text dy =0\] \[\frac{\partial MR}{\partial y}=9x^2y^{7/2}-21x^3y^{5/2}\] \[\frac{\partial NR}{\partial x}=9x^2y^{7/2}-16x^3y^{5/2}\]

  34. UnkleRhaukus
    • 2 years ago
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    \[21\sim16\]

  35. RolyPoly
    • 2 years ago
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    I.... don't .... understand.... :(

  36. UnkleRhaukus
    • 2 years ago
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    \[R=\mu\]

  37. RolyPoly
    • 2 years ago
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    Hmm... I should do it all over again?!

  38. RolyPoly
    • 2 years ago
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    xy vs x^2 y^(5/2) They look too different to me!

  39. hartnn
    • 2 years ago
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    life would be so much easier if we could use substitution...and not IF.

  40. RolyPoly
    • 2 years ago
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    Then, this, again, proves that life is not easy :(

  41. UnkleRhaukus
    • 2 years ago
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    well i used 6 bits of paper , and got the same answer we had before, bother

  42. RolyPoly
    • 2 years ago
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    6... bits?! of paper?! :(

  43. hartnn
    • 2 years ago
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    \(LS=(2m+4)x^ny^{m+1}-(6m+6)x^{n+1}y^m\)

  44. hartnn
    • 2 years ago
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    checking RS

  45. hartnn
    • 2 years ago
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    2m+4 = 3n+3 6m+6 = 4n+8 m=1 n=1

  46. hartnn
    • 2 years ago
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    just one silly mistale!

  47. hartnn
    • 2 years ago
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    *mistake

  48. UnkleRhaukus
    • 2 years ago
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    oh

  49. UnkleRhaukus
    • 2 years ago
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    i see it now , such a tiny tiny mistale

  50. hartnn
    • 2 years ago
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    thank you so much @UnkleRhaukus , i learned a new method today...R=x^m y^n

  51. UnkleRhaukus
    • 2 years ago
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    OK, so now we have the integrating factor \[R(x,y)=\mu(x,y)=xy\], lets integrate!

  52. hartnn
    • 2 years ago
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    actually , i wanted to see the look at @RolyPoly face, when he/she finds out how \(\tiny tiny\) the error was :P

  53. UnkleRhaukus
    • 2 years ago
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    \[\int \mu N\text dx\]\[=\int (2xy^3−6x^2y^2)\text dx\]\[=x^2y^3-2x^3y^2+g(y)\] \[\int\mu M\text dy\]\[=\int(3x^2y^2−4x^3y)\text dy\]\[=x^2y^3-2x^3y^2+h(x)\] \[\implies g(y)=h(x)=0\] \[f(x,y)=x^2y^3-2x^3y^2=c\]

  54. UnkleRhaukus
    • 2 years ago
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    or the other way \[f(x,y)=\int \mu N\text dx=x^2y^3-2x^3y^2+g(y)\] \[\frac{f(x,y)}{\text dy}=3x^2y^2-4x^3y+g'(y)=\mu M=3x^2y^2−4x^3y\] \[\implies g'(y)=0\]\[g(y)=c_1\] \[f(x,y)=x^2y^3-2x^3y^2+c_1=0\]

  55. UnkleRhaukus
    • 2 years ago
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    \[c_1=-c\]

  56. UnkleRhaukus
    • 2 years ago
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    woo!

  57. UnkleRhaukus
    • 2 years ago
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    wasn't that fun?

  58. hartnn
    • 2 years ago
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    \[ \begin{array}l\color{red}{\text{y}}\color{orange}{\text{e}}\color{#e6e600}{\text{s}}\color{green}{\text{,}}\color{blue}{\text{ }}\color{purple}{\text{o}}\color{purple}{\text{f}}\color{red}{\text{c}}\color{orange}{\text{o}}\color{#e6e600}{\text{u}}\color{green}{\text{r}}\color{blue}{\text{s}}\color{purple}{\text{e}}\color{purple}{\text{ }}\color{red}{\text{:}}\color{orange}{\text{)}}\color{#e6e600}{\text{}}\end{array} \]

  59. RolyPoly
    • 2 years ago
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    Mummy!!!! I want to die :'(

  60. hartnn
    • 2 years ago
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    lol :P

  61. hartnn
    • 2 years ago
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    plz don't die.

  62. RolyPoly
    • 2 years ago
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    She! It's a she!

  63. RolyPoly
    • 2 years ago
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    Thanks for rescue!!!! I'm sorry for my silly mistake!!

  64. hartnn
    • 2 years ago
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    \[ \begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#e6e600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array} \]

  65. UnkleRhaukus
    • 2 years ago
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    *mistale

  66. RolyPoly
    • 2 years ago
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    Hmmm.. So must I use \[\frac{\partial R}{\partial y} M + \frac{\partial M}{\partial y} R= \frac{\partial R}{\partial x} N+ \frac{\partial N}{\partial x}R\] to find R?! You know it's a pain :(

  67. hartnn
    • 2 years ago
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    yes, it is pain, but there is different kind of pleasure when we arrive at correct answer after all that!

  68. RolyPoly
    • 2 years ago
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    I must take this pain then.. Thanks again for all of your help!! Much appreciated!! (Btw, who wants the medal?)

  69. UnkleRhaukus
    • 2 years ago
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    if you were wondering what the solution 'looks' like ,...

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  70. UnkleRhaukus
    • 2 years ago
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    solutions*

  71. RolyPoly
    • 2 years ago
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    Ugly :( How do you get the plot?

  72. UnkleRhaukus
    • 2 years ago
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    i used a graphing program ,

  73. RolyPoly
    • 2 years ago
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    That is..?

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is replying to Can someone tell me what button the professor is hitting...

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