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Homework Help: Can someone please help me factorise a fraction (u-2)(-u^2 +3u +2)

  1. Aug 6, 2008 #1
    1. The problem statement, all variables and given/known data
    I am working on a topic called differential equations, and Im stuck on some working out.

    2. Relevant equations[/b]

    Can someone please help me to factorise the following fraction (u-2)(-u^2+ 3u+ 2)
    I have to integrate it, and I know I can without factorising, but it is so messy, my professor said to factorise the fraction so it looks a bit like a1/(bu+ c)+ a2/(du+ e)

    Well thats what the professor said, I have no idea what it means... I can factorise as far as
    = u/(-u^2 +3u +2) - 2/(-u^2 +3u +2)

    And I know that -u^2 +3u +2 = (-u + 1)(u-2) +4

    But not sure how to put it all together

    Here is the working out for the whole question just in case you're interested
    y' &= \frac{y+2x}{y-2x} \\
    Let y &= ux \\
    \text{Then we have} y' &= \frac{ux + 2x}{ux - 2x} \\
    &= \frac{u+2}{u-2} \\
    \text{Now } y' &= \frac{dy}{dx} = \frac{d(ux)}{dx} = x \frac{dy}{dx} + u \\
    \text{So we can say that} x \frac{du}{dx} + u& = \frac{u+2}{u-2} \\
    x \frac{du}{dx} &= \frac{u+2}{u-2} - u \\
    x \frac{du}{dx} &= \frac{u+2}{u-2} - \frac{u^2-2u}{u-2} \\
    x \frac{du}{dx} &= \frac{u+2- u^2+2u}{u-2} \\
    x \frac{du}{dx} &= \frac{-u^2+3u+2}{u-2} \\

    Then I have to integrate both sides of this equation...which is what I think I need to factorise in order to make it nice and neat

    \frac{u-2}{-u^2 + 3u + 2}du &= \frac{1}{x} dx
    Last edited: Aug 6, 2008
  2. jcsd
  3. Aug 6, 2008 #2


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    That can't be factored using integer coefficients but you know that if an expression factors as (x-a)(x-b) then a and b must be roots of (x-a)(x-b)= 0. Use the quadratic formula to find the roots of x2- 3x- 2= 0.
  4. Aug 6, 2008 #3


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    Try completing the square of [tex]-u^2+3u+2[/tex]. Then use a simple algebraic identity to decompose it into partial fractions.
  5. Aug 7, 2008 #4
    I think i've gotten a little closer, The question I'm trying to do is to solve a d.e. using change of variables, so once I get this part worked out I should be able to get it

    So this is how far I've gone
    Since [tex] u^2- 3 u -2= 0 [/tex]
    has two roots
    [tex] u_1 = (3+ 17^{0.5})/2, u_2= (3- 17^{0.5})/2 [/tex]


    [tex] u^2- 3 u -2= (u - (3+ 17^{0.5})/2 ) (u- (3- 17^{0.5})/2 ) [/tex]

    Originally I had
    [tex] \frac{u-2}{-u+3u+2}du = \frac{1}{2}dx [/tex] - I'll call this equation (2)
    So now I've got the LHS as

    [tex] S = -u /((u - (3+ 17^{0.5})/2 ) +2 /(u- (3- 17^{0.5})/2 ) [/tex]
    I changed the u-2 to -u+2 since I made a similar change on the denominator in order to solve it...

    Anyhow, I integrated that and ended up with something really messy invloving logs

    [tex] -2(u + 17^{0.5} ln(u-17^{0.5}-3)+3ln(u-17^{0.5}-3)-17^{0.5}-3+4ln(u+17^{0.5}-3) = ln(x) + C [/tex]

    (I integrated both sides of equation (2) above)

    Well the prob is now, I'm probably wrong anyway, but the next step is to solve for u, and then I have to sub back in the original change of varaible which was y=xu and end up with an expression for y in terms of x and C... before I go on, can anyone tell me if I should bother working through those logs, or is it wrong? Thanks :)
  6. Aug 8, 2008 #5


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    Dearly Missed

    Let us take this as a GENERAL case, shal we?
    We are to decompose:
    where the unindexed u is our variable, the indexed u's known numbers, and A and B constants to be determined.
    We therefore must have:
    [tex]u-u_{0}=A(u-u_{2})+B(u-u_{1})[/tex], or by comparing coefficients, we get the system of equations:
    whereby we arrive at the solutions:

    This is much simpler than using specific numbers!
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