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Math Help - differential equation problem

  1. #1
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    differential equation problem

    solve the following differential equation dy/dx=x[(x^2 + 4)^(1/2)] with initial condition y(0)=1
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by myoplex11 View Post
    solve the following differential equation dy/dx=x[(x^2 + 4)^(1/2)] with initial condition y(0)=1
    consider the function:

    f(x) = (1/3)(x^2+4)^{3/2}

    df/dx = (1/3)(3/2) 2x (x^2+4)^{1/2} = x[(x^2 + 4)^(1/2)]

    So you can use this to integrate your DE you have:

    y = (1/3)(x^2+4)^{3/2} + C,

    when x=0 we require:

    1 = (1/3)(4)^{3/2} + C

    so:

    C = 1-8/3 = -5/3

    and so:

    y = (1/3)(x^2+4)^{3/2} - 5/3

    RonL
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  3. #3
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    Hello, myoplex11!

    Solve: .dy/dx .= .x(x + 4)^ .with initial condition y(0) = 1

    We have: .dy .= .x(x + 4)^ dx

    Integrate: .y .= .(1/3)(x + 4)^
    3/2 + C


    Initial condition: .y(0) = 1

    We have: .(1/3)(0 + 4)^
    3/2 + C .= .1 . . . . C = -5/3


    Therefore: .y .= .(1/3)(x + 4)^
    3/2 - (5/3)

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