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Homework Help: Help solving a first order DE

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve the following DE:
    (6x[tex]^{4}+x^{3}[/tex]+10x[tex]^{2}[/tex]+2x-4)yy' = (y[tex]^{2}[/tex]+y+2)(12x[tex]^{4}[/tex]+2x[tex]^{3}[/tex]+20x[tex]^{2}[/tex]+6x-7)

    2. Relevant equations
    All we really know technique wise is seperation of variables then integrate. Basically this is the first week of DE class.

    3. The attempt at a solution
    I divided both sides by the left parantheses part and did division. Put the y's on the left and I am stuck at integration.

    [tex]\int\frac{y}{y^2+y+2}[/tex]dy =2[tex]\int\frac{2x+1}{6x^4+x^3+10x^2+2x-4}[/tex]dx

    There is where I left off. I am having difficulty with both integrals, if they are even possible. Did I miss something in an early step? Not sure where to proceed or if I am heading in the right direction. Help would be great.
  2. jcsd
  3. Jan 19, 2009 #2


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    That's a pretty masochistic problem. One place you did go wrong is that the ratio of x polynomials is 2+(2x+1)/(6*x^4+x^3+10*x^2+2*x-4). The 2 doesn't come out as a multiplicative factor. Now you just have a couple of moderately nasty (but doable) integrals. For the y integral you need to start by completing the square in the denominator, do a ordinary substitution and then a trig substitution. The x integral is a partial fractions project. You started it out right by doing the division.
  4. Jan 19, 2009 #3


    Staff: Mentor

    The right side should be
    [tex]\int (2 + \frac{2x + 1}{6x^4 + x^3 + 10x^2 + 2x - 4}) dx[/tex]

    not 2 time the integral that you show.

    For the integral on the left, you could rewrite the numerator as y + 1/2 - 1/2, and then split it into two integrals, with y + 1/2 in the numerator of the first, and -1/2 in the numerator of the second. After integration, these would be K*ln(y^2 + y + 2) and arctan(something).

    For the integral on the right, you'll get 2x + something, where the "something" is going to take some work. You should start by factoring the 6x^4 + ... expression in the denominator, if possible. If you can get this expression factored, you can use partial fractions to break up (2x + 1)/(6x^4 + ... - 4) into the sum of easier to work with rational functions.
  5. Jan 19, 2009 #4
    When you the complete the square do you add 1/(1/4) for completing the square in the bottom as (y+1/2)^2 so add 4 to the right?
  6. Jan 19, 2009 #5


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    Wouldn't that mean you subtract 1/4 from the 2? y^2+y+2=(y+1/2)^2+7/4, isn't that it?
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