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Integration of Inverse of f(x)

  1. Jan 13, 2008 #1
    [SOLVED] Integration of Inverse of f(x)

    [tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy =

    [tex]\frac{1}{4y^{3}-18y^{2}+10y}[/tex](ln|[tex]{y^{4}-6y^{3}+5y^{2}}[/tex]|)


    is this correct way or doing it? thanks. i was a bit of confuse, if i should do it the above or partial fraction the denominator it? thanks
     
  2. jcsd
  3. Jan 13, 2008 #2

    mjsd

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    yes, factorise and then partial fractions is ok
     
  4. Jan 13, 2008 #3
    wait a minutes,i show my partial fraction. please point out where gone wrong ;) thanks
     
  5. Jan 13, 2008 #4

    mjsd

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    firstly you can further factorise y^2-6y+5
     
  6. Jan 13, 2008 #5
    [tex]\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex]

    =[tex]\frac{1}{y^{2}(y^{2}-6y+5)}[/tex]

    =[tex]\frac{1}{y^{2}(y-1)(y-5)}[/tex]

    So, by Partial fraction, is this correct?

    [tex]\frac{1}{y^{2}(y-1)(y-5)}[/tex]
    =[tex]\frac{A}{y}[/tex]+[tex]\frac{B}{y^{2}}[/tex]+[tex]\frac{C}{y-1}[/tex]+[tex]\frac{D}{y-5}[/tex]

    and then, by comparing coefficient of y^3, y^2, y.

    (A+C+D){y^{3}}=0
    (-6A+B-5C-D){y^{2}}=0
    (5A-6B)y=0
    answer is A=6/25, B=1/5, C=3/10. D=-87/50.

    but doubt my answer. anyone can point out?thanks
     
    Last edited: Jan 13, 2008
  7. Jan 13, 2008 #6
    i know, is just that i m slow using teX, please gimme a min.thanks
     
  8. Jan 13, 2008 #7

    mjsd

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    I can tell you the answer....
    :smile:

    no need use latex for such simple expression

    most likely you have treated the y^2 bit incorrectly
     
    Last edited: Jan 13, 2008
  9. Jan 13, 2008 #8
    after i substitute A,B,C,and D.

    i can get back my original 1/f(x)
     
  10. Jan 13, 2008 #9

    mjsd

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    looks ok so far
     
  11. Jan 13, 2008 #10
    lol, i uses teX as its more tidy, i go n check the y^2 bit now.thanks.
     
  12. Jan 13, 2008 #11

    mjsd

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    FYI i've got
    A=6/25
    B=1/5
    C=-1/4
    D=1/100
     
  13. Jan 13, 2008 #12
    if so, then
    [tex]\frac{1}{y^{2}(y-1)(y-5)}[/tex]=[tex]\frac{6}{25y}[/tex]+[tex]\frac{1}{5y^{2}}[/tex]+[tex]\frac{3}{10(y-1)}[/tex]-[tex]\frac{87}{50(y-5}[/tex]
     
    Last edited: Jan 13, 2008
  14. Jan 13, 2008 #13
    OOops,gimme a minute,i recheck :)
     
  15. Jan 13, 2008 #14
    you are right! :D :D :D thanks a lot.for pointing it out !:D
     
  16. Jan 13, 2008 #15
    [tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy =

    6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y). right?

    but this problem is actually make y the subject in this equation, how to go about it further?
    [tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy = [tex]\int\[/tex] dt


    the original question is [tex]\frac{dy}{dt}[/tex]=[tex]y^{4}-6y^{3}+5y^{2}[/tex] find y(t).


    if [tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy =

    6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y)
    then i m stuck to make y the subject. anyone to point out the mistake.thanks
     
    Last edited: Jan 13, 2008
  17. Jan 13, 2008 #16
    anyone, can show me the right way?
     
  18. Jan 13, 2008 #17
    another differential eq. question.

    y'=1+xy find y. how to go about it. i m clueless
    my method of multiply dx to (1+xy) doesn't work in this case.
     
  19. Jan 13, 2008 #18
    You can solve this by the solution for Linear Differential equation. The equation you have is:

    [tex]
    \frac{dy}{dx} = 1 + xy
    [/tex]

    which can be written as:

    [tex]
    \frac{dy}{dx} + y(-x) = 1
    [/tex]

    which is similar to:

    [tex]
    \frac{dy}{dx} + yP = Q
    [/tex]

    Here, [itex]P = (-x); Q = 1[/itex]

    So, you have:

    [tex]
    I.F = e^{\int (-x)dx}
    [/tex]
    [tex]
    I.F = e^{-\frac{x^2}{2}}
    [/tex]

    And hence, your solution is given by:

    [tex]
    ye^{\frac{-x^2}{2}} = \int (1)e^{\frac{-x^2}{2}}dx + C
    [/tex]

    But, it's gonna be a real pain with this:

    [tex]
    \int e^{\frac{-x^2}{2}}dx
    [/tex]

    a method to which i can't think of right now. Maybe there's some other way..
     
    Last edited: Jan 13, 2008
  20. Jan 13, 2008 #19

    Gib Z

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    That integral evaluates to the error function. No "method" for that one, you just have to know it. Haven't checked the rest of your post though, so not sure if the rest of your solution is right.
     
  21. Jan 14, 2008 #20
    great, this is something new to me, just started my linear algebra course.thanks.will study your solution
     
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