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Initial value problem

  1. Dec 19, 2004 #1

    kreil

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    Gold Member

    Rewrite the differential equation [tex]\frac{dy}{dx}=x{\sqrt{y}}[/tex] in the form y=f(x) given the initial condition f(3)=25.

    I am new to integration so I am unsure about my work on this problem.

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

    [tex]dy=(dx)(x)(\sqrt{y})[/tex]

    [tex]\frac{dy}{\sqrt{y}}=(dx)(x)[/tex]

    [tex]\int{\frac{dy}{\sqrt{y}}}=\int{(x)(dx)}[/tex]

    [tex]2y^{\frac{1}{2}}=\frac{1}{2}x^2+ C[/tex]

    [tex]10=\frac{9}{2}+C[/tex]

    [tex]C=\frac{11}{2}[/tex]

    [tex]2y^{\frac{1}{2}}=\frac{1}{2}x^2+\frac{11}{2}[/tex]

    [tex]y=(\frac{1}{4}x^2+\frac{11}{4})^2[/tex]

    If I did it correctly, is there an easier way to do it? If I messed up, where?

    Thanks
     
  2. jcsd
  3. Dec 19, 2004 #2
    Looks right to me.

    It doesn't seem like that much work to me.
     
  4. Dec 19, 2004 #3

    kreil

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    Gold Member

    No, it's not a lot of work, I just thought that there might be others ways to arrive at the same answer.

    Thanks
     
  5. Dec 19, 2004 #4

    dextercioby

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    Homework Helper

    Nope,there's no simpler way to integrate that diff.eq. than the variable separation method.
    Nice work!!

    Daniel.
     
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