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Math Help - Partial Diff (2)

  1. #1
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    Partial Diff (2)

    If V =x f(u) and u = y/x,

    determine the value of n such that

    x^2(∂^2V / ∂x^2) + nxy( ∂^2V / ∂x∂y) + y^2 ( ∂^2V / ∂y^2) = 0


    Need help to check my working..

    ∂V/∂x = xf'(u) + f(u)

    ∂^2V/∂x^2 = xf"(u) + f'(u) +f'(u)

    For u = y/x,
    ∂u/∂x = -y/x^2
    ∂^2u/∂x^2 = 2y/x^3

    Sub into ∂V/∂x and ∂^2V/∂x^2,

    I get ∂V/∂x= -y/x + y/x =0
    ∂^2V/∂x^2= x(2y/x^3) - (2y/x^2) =0

    ∂V/∂y = xf'(u)
    ∂^2V/∂y^2 = xf"(u)

    ∂u/∂y = 1/x
    ∂^2u/∂x^2 = 0

    Sub into ∂V/∂y and ∂^2V/∂y^2,

    ∂V/∂y = xf'(u) = 1
    ∂^2V/∂y^2 = xf"(u) = 0


    And,
    ∂^2V / ∂x∂y= (∂/ ∂x)(∂V/∂y)<----------(Do i need to find ∂^2u / ∂x∂y?)
    = (∂/ ∂x)xf'(u)<-----------( i am using ∂u/∂y for this function)
    = x(0) + 1/x = 1/x

    Sub into given eqn,

    x^2(0) + nxy(1/x) + y^2(0) = 0
    ny = 0
    n=0

    Is this correct?
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  2. #2
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    Errors

    Right away I can see a glaring error:

    Need help to check my working..

    ∂V/∂x = xf'(u) + f(u)
    You forgot that u is a function of x and y. This should be
    \frac{\partial V}{\partial x}=xf'(u)\frac{\partial u}{\partial x}+f(u)
    \frac{\partial V}{\partial x}=xf'(u)\cdot\left(-\frac{y}{x^2}\right)+ f(u)
    \frac{\partial V}{\partial x}=f(u)-\frac{yf'(u)}{x}

    Your second derivative will also need adjusted accordingly.

    Shortly thereafter, you substitute \frac{\partial u}{\partial x} in as f'(u). This is simply wrong. f(u) is some unknown (arbitrary) function, and thus f'(u) is simply the derivative of that function, and cannot be substituted with anything else. (Try picturing the sample case f(u)=\sin{u}. Then f'(u)=\cos{u}, which you can see is most definitely not \frac{\partial u}{\partial x}=-\frac{y}{x^2}

    --Kevin C.
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  3. #3
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    Ok i understand what you are saying

    but that means the final answer will be in terms of derivatives of f(u)?

    or how should i go about solving this question?

    Thks
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  4. #4
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    You will be able to combine like terms with regards to f(u), f'(u), and f''(u); and, in fact, some of them will cancel, but mostly, you will get an equation involving one or more of those, but there will be one (and only one) value of n for which that equation will be true for all f(u).

    --Kevin C.
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  5. #5
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    ∂V/∂x = f(u) - yf'(u)/x

    ∂^2V/∂x^2 = f"(u)y/x^3

    ∂V/∂y = f'(u)
    ∂^2V/∂y^2 = f"(u)/x

    ∂^2V / ∂x∂y= (∂/ ∂x)(∂V/∂y) = f'(u)(-y/x^2)

    Sub into given eqn and factorise,

    f"(u)(y/x) (1 - ny+y) = 0

    therefore

    1-ny+y = 0
    n=(y+1)/y

    Is this the final answer?
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  6. #6
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    No

    No.

    You were correct on \frac{\partial V}{\partial x}=f(u)-\frac{y}{x}f'(u), but not \frac{\partial^2 V}{\partial x^2}.
    \frac{\partial^2 V}{\partial x^2}=\frac{\partial}{\partial x}\frac{\partial V}{\partial x}
    \,\,=\frac{\partial}{\partial x}\left(f(u)-\frac{y}{x}f'(u)\right)
    \,\,=f'(u)\frac{\partial u}{\partial x}+\frac{y}{x^2}f'(u)-\frac{y}{x}f''(u)\frac{\partial u}{\partial x}
    \,\,=-\frac{y}{x^2}f'(u)+\frac{y}{x^2}f'(u)-\left(\frac{y}{x}\right)\left(-\frac{y}{x^2}\right)f''(u)
    \,\,=\frac{y^2}{x^3}f''(u).

    Next,
    \frac{\partial^2 V}{\partial x\partial y}=\frac{\partial}{\partial x}\frac{\partial V}{\partial y}
    \,\,=\frac{\partial}{\partial x}f'(u)
    \,\,=f''(u)\frac{\partial u}{\partial x}
    \,\,=-\frac{y}{x^2}f''(u)

    Thus:
    x^2\frac{\partial^2 V}{\partial x^2}+nxy\frac{\partial^2 V}{\partial x\partial y}+y^2\frac{\partial^2 V}{\partial y^2}=0
    becomes:
    x^2\cdot\frac{y^2}{x^3}f''(u)+nxy\cdot\left(-\frac{y}{x^2}f''(u)\right)+y^2\cdot\frac{1}{x}f''(  u)=0
    \frac{y^2}{x}f''(u)-n\frac{y^2}{x}f''(u)+\frac{y^2}{x}f''(u)=0
    (2-n)\frac{y^2}{x}f''(u)=0
    And the answer should be obvious from that.

    --Kevin C.
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  7. #7
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    now i get it, thanks
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