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Homework Help: Little help with differential equations

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{dy}{dx} = \sqrt{xy^3} , y(0) = 4[/tex]

    3. The attempt at a solution


    [tex]\frac{dy}{dx} = x^{\frac{1}{2}}y^{\frac{3}{2}}

    \Rightarrow \int y^{-\frac{3}{2}}dy = \int x^{\frac{1}{2}}dx
    \Rightarrow -2y^{-\frac{1}{2}} = \frac{2}{3}x^{\frac{3}{2}} + C
    \Rightarrow y^{-\frac{1}{2}} = -\left(\frac{1}{3}x^{\frac{3}{2}} + \frac{1}{2}C\right)
    \Rightarrow y^{\frac{1}{2}} = -\left(\frac{1}{\frac{1}{3}x^{\frac{3}{2}} + \frac{1}{2}C}\right)
    \Rightarrow y = \frac{1}{(\frac{1}{3}x^{\frac{3}{2}} + \frac{1}{2}C)^2}

    Then Putting in the conditions mentioned above:

    4 = \frac{1}{(0 + \frac{1}{2}C)^2}
    \Rightarrow \frac{1}{4}C^2 = \frac{1}{4}
    \Rightarrow C = 1

    Yet my book shows [tex]C = \frac{3}{2}[/tex]
  2. jcsd
  3. Feb 18, 2009 #2


    Staff: Mentor

    Your work looks fine to me except near the end. C can also equal -1.
    Is the book's answer y = 1/[1/3*x^(3/2) + 1/2 * 3/4]^2? (I.e., same as yours but with C = 3/2 rather than C = 1 as you have it.)

    I don't see why your general solution wouldn't be a solution to the DE, and I don't see any problems with your use of the initial condition to find the particular solution. I would want to verify that the book's solution solves the DE and initial condition. After all, math text occasionally have typos.
  4. Feb 18, 2009 #3
    you mean there can be more then one particular solution?
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