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Homework Help: Calc Final Help

  1. May 24, 2005 #1
    I am going to be gone all day tomorrow at a conference track meet and am unable to ask my teacher how to do integrating factors and differential operators. I leave tomorrow at 9:15 am and was hoping to have some examples to take with me to study.

    If someone could help me walk through a these it would be great

    integrating factor
    [tex] (x^4+2y)dx-xdy=0 [/tex] answer: [tex]2y=x^4+cx^2[/tex]
    I got it to
    [tex] \frac{dy}{dx}-\frac{2y}{x}=x^3 [/tex]

    [tex]e^\int{\frac{-2}{x}}=x^{-2} [/tex] = IF ... but what do you do then...

    [tex] x^{-2}\frac{dy}{dx}-x^{-2}\frac{2y}{x}=x [/tex]

    [tex] x^{-2}\frac{dy}{dx}-\frac{2y}{x^3}=x [/tex]

    so then you get yx^-2 but i don't know what to do from here

    more to come
    Last edited: May 24, 2005
  2. jcsd
  3. May 24, 2005 #2
    you get finally

    [tex]\frac{d}{dx}{(yx^{-2})} = x[/tex]

    Integrating both sides with respect to x gives,

    [tex]\int d(yx^{-2}) = x^{2} + C[/tex] **

    which gives

    [tex]yx^{-2} = x^{2} + C[/itex]

    Multiply both sides by [itex]x^2[/itex] and you're through.

    In step ** you have a total differential under the integral sign which is never a problem to integrate. If you have any more queries, please feel free to ask.

  4. May 27, 2005 #3
    Thanks for the help... thats really all i needed to know... was that last step
  5. May 27, 2005 #4
    Maybe I'm missing something here but on step ** did you not integrate x incorrectly?
  6. May 28, 2005 #5


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    yeah! [tex]\int x dx= \frac{1}{2}x^2+ C[/tex]!
    so [tex]yx^{-2}= \frac{1}{2}x^2+ C[/tex].
  7. Jun 8, 2005 #6
    Hi everyone

    Sorry for that mistake...must've forgotten to use my brain then ;-)

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