1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook