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Differential Equation

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{dy}{dx}=\frac{x+3y}{3x+y}[/tex]


    2. Relevant equations

    M(x,y)dx+N(x,y)dy=0

    Exact equation (I think)

    3. The attempt at a solution

    Ok, so I try to put it in the form of the of an exact equation, like above, but I end up with

    -(x+3y)dx+(3x+y)dy=0

    This results in the partial of M with respect to y being -3 and the Partial of N with respect to x being 3, which is not equal!

    I tried investigating to see if there is an integrating factor but [tex]\frac{M_{y}-N_{x}}{N}[/tex] and [tex]\frac{N_{x}-M_{y}}{M}[/tex] are not functions of just one variable. What am I doing wrong? Different method? Algebra mistake?

    Thanks
     
  2. jcsd
  3. Sep 14, 2010 #2

    rock.freak667

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    Homework Helper

    Try using a substitution of y=vx.
     
  4. Sep 15, 2010 #3
    ok, so

    [tex]y=vx[/tex]

    and

    [tex]dy=vdx+xdv[/tex]



    [tex](-x-3vx)dx+(3x+vx)(vdx+xdv)=0[/tex] So far so good...



    [tex]-xdx+xv^{2}dx+3x^{2}dv+vx^{2}dv=0[/tex] Collect like terms...



    [tex]x(v^{2}-1)dx+x^{2}(3+v)dv=0[/tex] Factor out like terms...



    [tex]\frac{1}{x}dx=\frac{-(3+v)}{(v^{2}-1)}dv[/tex] Seperate variables...is this right? I tend to make a lot of simple errors, no matter how many times I check my work...

    Thanks!
     
    Last edited: Sep 15, 2010
  5. Sep 15, 2010 #4

    rock.freak667

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    Homework Helper

    That should be correct. Now just split the function in 'v' into partial fractions and integrate.
     
  6. Sep 15, 2010 #5
    Ok so...

    [tex]\frac{1}{x}=\frac{1}{v+1}-\frac{2}{v-1}[/tex]

    [tex]\int\frac{1}{x}dx=\int(\frac{1}{v+1}-\frac{2}{v-1})dv[/tex]

    [tex]ln(x)=ln(v+1)-2ln(v-1)+ln(c)[/tex] Ok I don't really know what to do next, heres a try....

    [tex]ln(x)=ln(c\frac{v+1}{(v-1)^{2}})[/tex]

    Is this right? I'm not very good with natural logs, hopefully I didn't make a mistake...

    Next step would be exponentiate both sides? (Assuming what I just did was right)
     
  7. Sep 15, 2010 #6

    rock.freak667

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    Homework Helper

    Well you don't really have to exponentiate both sides since if lnx=lny then x=y.

    So essentially, you can just remove the 'ln' and then replace 'v'.
     
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