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Quick question about Differential Equations.

  1. Aug 3, 2011 #1
    I was working on a simple differential problem which caused me some confusion

    original eq =

    dy/dx = -x/2y

    which can easily be altered into

    y dy = (-x/2) dx

    This finally transforms into (after antiderivatives are determined)

    (y^2/2) = -(x^2/4) + c

    While I can see that this logically is equivallent to the problem above I just don't see how the answer was derived. The x side of the equation makes perfect sense to me, but what I can't tell is how the y was derived to be (y^2/2). In my mind it should have just been y...

    That aside I can see that the transformation is logical, but I can't figure out how it was derived.

    Thanks,
    Ozone.
     
  2. jcsd
  3. Aug 3, 2011 #2

    HallsofIvy

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    I don't see how you could say that! You say you see how [itex](x/2) dx[/itex] gives [itex]x^2/4+ c[/itex], I presume by recognizing that
    [tex]\frac{1}{2}\int xdx= \frac{1}{2}\left(\frac{1}{2}x^2\right)+ c= \frac{x^2}{4}+ c[/tex]

    Welll, the left side is not different:
    [tex]\int y dy= \frac{1}{2}y^2+ c[/tex]

    [tex]\int y dy= \frac{1}{2}\int x dx[/tex]
    [tex]\frac{1}{2} y^2+ c_1= \frac{1}{4} x^2+ c_2[/tex]

    I have written the two constants differently because there is no reason to think they have to be the same. Now subtract [itex]c_1[/itex] from both sides:
    [tex]\frac{1}{2}y^2= \frac{1}{4}x^2+ (c_2- c_1)= \frac{1}{4}x^2+ c[/tex]
    where "c" is just [itex]c_2- c_1[/itex].
     
  4. Aug 4, 2011 #3
    Thanks hallsofivy this is exactly what I was looking for. Also I will be sure to use the forums coding next time :). (I responded to your other post after I made this post).
     
  5. Aug 4, 2011 #4
    Set F(x,y)≡-x/2y then 0=dF=(∂F/∂x)dx+(∂F/∂y)dy and, -(∂F/∂x)/(∂F/∂y)=dy/dx. Reversing the steps should answer your question.
     
  6. Aug 6, 2011 #5

    I was not able to correct my rough work above, so i shall do it now. You (`ozone`) say that, "I can't figure out how it was DERIVED".

    The `derivation` follows from simple multivariable calculus. Given dy/dx=-x/2y, rewrite it as 2ydy+xdx=0. Next, introduce F∈C¹(ℝ²) and set F(x,y)=constant (level curves), so that 0=dF=∂F/∂ydy+∂F/∂xdx.

    Identify ∂F/∂y=2y (*), and ∂F/∂x=x (**). From (*), F(x,y)=y²+c₁(x) and from (**), F(x,y)=½x²+c₂(y). The obvious choice is, F(x,y)=½x²+y²+constant.
     
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