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Math Help - Solving diffrential equations

  1. #1
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    Solving diffrential equations

    Well i couldn't go to school today for various reasons.
    So i'm trying to teach myself some stuff and am having trouble understanding the method.

    Find a differential equation satisfied by y = Ax^2 + Bx^0.5 for ALL values of A and B
    The method goes like this :
    if
    y = Ax^2 + Bx^0.5
    then
    \frac{dy}{dx} = 2Ax + \frac{1}{2}Bx^0.5
    and \frac{d^2y}{dx^2} = 2A - \frac{1}{4}Bx^0.5

    Ok, i get that and i also get you can use elimination to then solve this...but i don't get what they've done...

    By eliminating B from the first and second equation...they get
    y - 2x\frac{dy}{dx} = A(x^2 - 4x^2) + B(x^0.5 - x^0.5)
    This is where im confused...where the hell do they even get 2x\frac{dy}{dx}
    from?

    Note: Its meant to be x to the power of a half...but latex doesnt like it lol.

    Thank-you
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
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    France
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    Quote Originally Posted by AshleyT View Post
    Well i couldn't go to school today for various reasons.
    So i'm trying to teach myself some stuff and am having trouble understanding the method.

    The method goes like this :
    if
    y = Ax^2 + Bx^0.5
    then
    \frac{dy}{dx} = 2Ax + \frac{1}{2}Bx^0.5
    and \frac{d^2y}{dx^2} = 2A - \frac{1}{4}Bx^0.5

    Ok, i get that and i also get you can use elimination to then solve this...but i don't get what they've done...

    By eliminating B from the first and second equation...they get
    y - 2x\frac{dy}{dx} = A(x^2 - 4x^2) + B(x^0.5 - x^0.5)
    This is where im confused...where the hell do they even get 2x\frac{dy}{dx}
    from?

    Note: Its meant to be x to the power of a half...but latex doesnt like it lol.

    Thank-you
    Hi,
    Actually there is a little mistake which answers your question
    y = Ax^2 + Bx^{\frac{1}{2}}
    then
    \frac{dy}{dx} = 2Ax + \frac{1}{2}Bx^{-\frac{1}{2}}
    and \frac{d^2y}{dx^2} = 2A - \frac{1}{4}Bx^{-\frac{3}{2}}

    Then
    2x \frac{dy}{dx} = 4Ax^2 + Bx^{\frac{1}{2}}
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