Integration of Inverse of f(x)

  1. [SOLVED] Integration of Inverse of f(x)

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


    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. mjsd

    mjsd 858
    Homework Helper

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

    mjsd 858
    Homework Helper

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



    So, by Partial fraction, is this correct?


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

    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. i know, is just that i m slow using teX, please gimme a min.thanks
  8. mjsd

    mjsd 858
    Homework Helper

    I can tell you the answer....

    no need use latex for such simple expression

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

    i can get back my original 1/f(x)
  10. mjsd

    mjsd 858
    Homework Helper

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

    mjsd 858
    Homework Helper

    FYI i've got
  13. if so, then
    Last edited: Jan 13, 2008
  14. OOops,gimme a minute,i recheck :)
  15. you are right! :D :D :D thanks a lot.for pointing it out !:D
  16. [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 =

    then i m stuck to make y the subject. anyone to point out the mistake.thanks
    Last edited: Jan 13, 2008
  17. anyone, can show me the right way?
  18. 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. You can solve this by the solution for Linear Differential equation. The equation you have is:

    \frac{dy}{dx} = 1 + xy

    which can be written as:

    \frac{dy}{dx} + y(-x) = 1

    which is similar to:

    \frac{dy}{dx} + yP = Q

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

    So, you have:

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

    And hence, your solution is given by:

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

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

    \int e^{\frac{-x^2}{2}}dx

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

    Gib Z 3,347
    Homework Helper

    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. great, this is something new to me, just started my linear algebra course.thanks.will study your solution
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