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Math Help - Experts in differential equations, plz solve this.....

  1. #1
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    Experts in differential equations, plz solve this.....

    1)
    If { 1(x), 2(x) } is one set of two linearly independent solution on a ≤ x ≤ b and { ψ1(x), ψ2(x) } be another set of two linearly independent solution, then show that there exists a constant c ≠ 0 such that
    W[ ψ1(x), ψ2(x) ] = c[ 1, 2 ](x) for all a ≤ x ≤ b.
    where W[] is wronskian.

    2)
    If one of the solution of the linear equation
    (1-x) y'' + xy' - y=0,
    is y1(x)=x; find the second solution y2(x) such that y1(x) and y2(x) are linearly independent on some interval L.
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  2. #2
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    Quote Originally Posted by akshatha View Post
    1)
    If { 1(x), 2(x) } is one set of two linearly independent solution on a ≤ x ≤ b and { ψ1(x), ψ2(x) } be another set of two linearly independent solution, then show that there exists a constant c ≠ 0 such that
    W[ ψ1(x), ψ2(x) ] = c[ 1, 2 ](x) for all a ≤ x ≤ b.
    where W[] is wronskian.

    2)
    If one of the solution of the linear equation
    (1-x) y'' + xy' - y=0,
    is y1(x)=x; find the second solution y2(x) such that y1(x) and y2(x) are linearly independent on some interval L.
    Here's 2) Let y = x u so y' = x u' + u and y'' = x u'' + 2u' so your ODE becomes

    (1-x)(xu'' + 2u') + x(xu'+u) - xu = 0 or x(1-x)u'' +(x^2-2x+2)u' = 0.

    If you let v = u' your ODE then becomes (1-x)(xu'' + 2u') + x(xu'+u) - xu = 0 or x(1-x)v' +(x^2-2x+2)v = 0. Separating gives

    \frac{dv}{v} = - \frac{x^2 -2x + 2}{x(1-x)}dx

    and upon integrating

    \ln v = x + \ln |x-1| - 2 \ln |x| or v = \frac{(x-1)e^x}{x^2}

    and since u' = v then

    u' = \frac{(x-1)e^x}{x^2}

    Integrating again gives

    u = \frac{e^x}{x}

    and so your second solution is y = xu = e^x and the general solution y = c_1 x + c_2 e^x
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  3. #3
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    Here's an answer for 1). Let us assume that your functions, \phi_1 , \phi_2 satisfy some ODE

    a y'' + by' + cy = 0,\;\;\; a \ne 0

    So a \phi_1'' + b\phi_1' + c\phi_1 = 0, and a \phi_2'' + b\phi_2' + c\phi_2 = 0. Then multiplying the first by \phi_2 and the second by \phi_1 and subtracting gives

    a\left(\phi_2 \phi_1'' - \phi_1 \phi_2'' \right) + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0 or a\left(\phi_2 \phi_1' - \phi_1 \phi_2' \right)' + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0.

    If W_{\phi} = W(\phi_1,\phi_2) is the usual Wronskian, then the above equation becomes

    a W_{\phi} ' + b W_{\phi} = 0

    Similarly, for \psi_1\; \text{and}\, \psi_2

    a W_{\psi} ' + b W_{\psi} = 0

    Thus,

    \frac{W_{\phi} ' }{W_{\phi} } = - \frac{b}{a} = \frac{W_{\psi} ' }{W_{\psi} } or \frac{W_{\phi} ' }{W_{\phi} } = \frac{W_{\psi} ' }{W_{\psi} }

    from which we obtain  W_{\phi} = c W_{\psi}.
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  4. #4
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    Quote Originally Posted by danny arrigo View Post
    Here's an answer for 1). Let us assume that your functions, \phi_1 , \phi_2 satisfy some ODE

    a y'' + by' + cy = 0,\;\;\; a \ne 0

    So a \phi_1'' + b\phi_1' + c\phi_1 = 0, and a \phi_2'' + b\phi_2' + c\phi_2 = 0. Then multiplying the first by \phi_2 and the second by \phi_1 and subtracting gives

    a\left(\phi_2 \phi_1'' - \phi_1 \phi_2'' \right) + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0 or a\left(\phi_2 \phi_1' - \phi_1 \phi_2' \right)' + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0.

    If W_{\phi} = W(\phi_1,\phi_2) is the usual Wronskian, then the above equation becomes

    a W_{\phi} ' + b W_{\phi} = 0

    Similarly, for \psi_1\; \text{and}\, \psi_2

    a W_{\psi} ' + b W_{\psi} = 0

    Thus,

    \frac{W_{\phi} ' }{W_{\phi} } = - \frac{b}{a} = \frac{W_{\psi} ' }{W_{\psi} } or \frac{W_{\phi} ' }{W_{\phi} } = \frac{W_{\psi} ' }{W_{\psi} }

    from which we obtain  W_{\phi} = c W_{\psi}.
    thanks
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  5. #5
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    Exclamation

    Quote Originally Posted by akshatha View Post
    1)
    If { 1(x), 2(x) } is one set of two linearly independent solution on a ≤ x ≤ b and { ψ1(x), ψ2(x) } be another set of two linearly independent solution, then show that there exists a constant c ≠ 0 such that
    W[ ψ1(x), ψ2(x) ] = c[ 1, 2 ](x) for all a ≤ x ≤ b.
    where W[] is wronskian.

    2)
    If one of the solution of the linear equation
    (1-x) y'' + xy' - y=0,
    is y1(x)=x; find the second solution y2(x) such that y1(x) and y2(x) are linearly independent on some interval L.
    This is a well-known fact (Abel's identity) Abel's identity - Wikipedia, the free encyclopedia
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