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Math Help - Second Order Differential Equation

  1. #1
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    Second Order Differential Equation

    Hi,

    I have a second order differential equation as shown below. I have no idea how to start and solve the problem. Appreciate if anyone could help me on this.

    <br />
(A - B/n)Q - A(C-n)dQ/dn + d2Q/dn2 = 0<br />

    Thank you very much.
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  2. #2
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    Quote Originally Posted by jen81 View Post
    Hi,

    I have a second order differential equation as shown below. I have no idea how to start and solve the problem. Appreciate if anyone could help me on this.

    <br />
(A - B/n)Q - A(C-n)dQ/dn + d2Q/dn2 = 0<br />

    Thank you very much.
    \frac{d^2Q}{dn^2} - A(C - n)\frac{dQ}{dn} + \left(A - \frac{B}{n}\right)Q = 0.


    I'm assuming that this is a second order linear CONSTANT COEFFICIENT ODE, so write the characteristic equation:

    m^2 - A(C - n)m + \left(A - \frac{B}{n}\right)Q = 0.


    Solve this Quadratic equation for m.

    You will end up with either:

    1. Two real distinct solutions

    2. Two real repeated solutions

    3. Complex conjugates.


    The solution to the ODE depends on the solutions of the characteristic equation.
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  3. #3
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    Hi,

    Thanks ProveIt.
    The solution to the equation above if assuming B = 0 is
    <br />
C1\exp[-A/2(n-C)^2]\int \exp[A/2(n'-C)^2]dn');   C1 = constant<br />

    I tried the method as suggested but I couldn't get to this solution.
    Thank you very much.
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  4. #4
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    The differentiation is with respect to n, albeit non-standard notation, so it's a DE with variable coefficients. And if I let B=0 and just change n to x to make it more standard looking, it's then:

    Q''-A(C-x)Q'+AQ=0

    So other than power series, we could get it into normal form letting:

    Q=v\,\text{exp}\left\{1/2\int A(C-x) dx\right\}

    which then gets it to:

    v''+Iv=0

    with: I=1/2A+1/4 A^2(C-x)^2

    but that's as far as I can take it.
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