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Homogeneous equation, problem with algebra

  1. Sep 24, 2008 #1
    this is the given equation


    and here is all the work i've done so far:


    i moved v over and came up with this


    did a flip


    by partial fractions I got a=-3/2 and b=1/2

    so -3/2*ln|v-1|+1/2*ln|v+3|+c=ln|x|

    after making the -3/2 and 1/2 powers to ln and throwing e in




    that's the part I'm stuck at.
  2. jcsd
  3. Sep 25, 2008 #2


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    Hi chaotixmonjuish! :smile:

    hmm … are you staying up too late? :zzz:

    it's C((y+x)/x)-3/2((y+3x)/x)1/2=x , isn't it? :wink:
  4. Sep 25, 2008 #3
    The answer to this one is |y-x|=c|y+3x|^5

    i'm really concerned how this actually happens
  5. Sep 25, 2008 #4


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    that would be because …
    is wrong. :wink:
  6. Sep 25, 2008 #5
    Is the step completely wrong or are A and B wrong?
  7. Sep 25, 2008 #6


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    A and B :smile:
  8. Sep 25, 2008 #7
    I try to solve this DE using unconventional method using parameter t.
    dy/dx = dy/dt /dx/dt.
    Then dy/dt = 4y-3x and dx/dt=2x-y (Is this correct? )
    which is a system of DE with constant coefficients.
    X' = AX
    [tex] X=\left(\begin{array}{c}x\\y\end{array}\right) ,

    The eigenvalues of A are 1 and 5 with eigenvectors [tex] \left(\begin{array}{c}1\\1\end{array}\right) [/tex] and [tex] \left(\begin{array}{c}1\\-3\end{array}\right) [/tex] respectively.

    Hence the solution is
    x=c1et + c2e5t
    y=c1et - 3c2e5t

    which satisfies |y-x|=c|y+3x|^5.
  9. Sep 25, 2008 #8


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    It would have been better to tell us HOW you got that! I assume that, because this is a homogenous equation you let v= y/x so y= xv and y'= xv'+ v. (4y-3x)/(2x-y), dividing both numerator and denominator by x, is equal to [4(y/x)- 3]/[2- (y/x)]= (4v- 3)/(2- v).

    And v^2+ 2v- 3= (v- 1)(v+3). Now, let (2-v)/(-2+ 2v+ v^2)= A/(v-1)+ B/(v+3). Multiplying on both sides by (v-1)(v+3), 2- v= A(v+3)+ B(v-1). If v= 1, 2-1= 1= 4A. A= 1/4. If v= -3, 2-(-3)= 5= -4B and B= -5/4.


    Even with the correct coefficients, e^(ln(a)+ ln(b)) is NOT a+ b. ln(a)+ ln(b)= ln(ab) so e^(ln(a)+ ln(b))= e^(ln(ab))= ab.

    No, that's not the part you are stuck at!
  10. Sep 25, 2008 #9


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    oh come on … it was obvious that v was y/x :smile:

    frankly I'd rather solutions were set out as sparingly as possible, so that I can see what's going on …

    every line has to be checked anyway :rolleyes: … and by definition there's a mistake somewhere, or the OP wouldn't be asking …

    sometimes, too much detail is both off-putting and confusing! :confused:
    Hi matematikawan! :smile:

    ooh, I've never seen that before …

    looks good to me … you've defined a perfectly legitimate parameter t, and got a standard linear equation, leading to a parametric solution which is actually easier to graph than the book solution! yay! :biggrin:
  11. Sep 25, 2008 #10
    i got as my two coefficients -5/4 and 1/4, are those correct

    i'm still not sure how to move things around to as to get the book answer
  12. Sep 26, 2008 #11


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    Yes, -5/4 and 1/4.

    And that gives you C((y+x)/x)-5/4((y+3x)/x)1/4 = x , doesn't it?

    Then just raise both sides to the 4th power. :smile:
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