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Homework Help: DE with initial conditions

  1. Jan 29, 2006 #1

    Pengwuino

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    Gold Member

    Ok I have a problem here. I have gotten to the unfortunate point where I feel like I'm nearly done, Mathematica has given me a solution that is correct… but of course, I can't figure out how I was suppose to get from point A to B. I currently have:

    [tex]2x - 6y\sqrt {x^2 + 1} \frac{{dy}}{{dx}} = 0,y(0) = 3[/tex]

    So I did this…

    [tex]\begin{array}{l}
    2xdx = 6y(x^2 + 1)^{1/2} \frac{{dy}}{{dx}} \\
    \int {2x(x^2 + 1)^{ - 1/2} dx = } \int {6ydy} \\
    u = x^2 + 1 \\
    dx = \frac{{du}}{{2x}} \\
    2u^{1/2} = 3y^2 + c \\
    2(x^2 + 1)^{1/2} = 3y^2 + c \\
    \end{array}[/tex]

    Kinda not sure where to go from here…. Or if I did that right in the first place

    The correct answer according to Mathematica is...

    [tex]\frac{{\sqrt {{\rm 25 + 2}\sqrt {{\rm 1 + x}^{\rm 2} } } }}{{\sqrt 3 }}[/tex]

    So yup, put x=0 and 3 pops out.
     
    Last edited: Jan 29, 2006
  2. jcsd
  3. Jan 29, 2006 #2
    It looks right, now use your initial condition to solve for the constant of integration.
     
  4. Jan 29, 2006 #3

    Pengwuino

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    Gold Member

    oh my god, sorry guys, i guess i had one of those "forgot basic algebra" moments. I tried to square root the left side of the equation before bringing c to the left...
     
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