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Math Help - Solving Constant Differencial Equations

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    Solving Constant Differencial Equations

    Somebody asked me if I can teach how to solve equations:
    ay''+by' + cy = 0 on -\infty < x <\infty \ \ \ (1)
    Where a\not =0, b, c are any real numbers.

    It really is easy!

    Write the quadradic equation,
    ak^2 + bk + c = 0 \ \ \ (2)
    Since a\not = 0 this is a quadradic equation.

    Let \Delta = b^2 - 4ac, this is the discriminant.

    There are three possibilities:
    ------------------------------

    1) \Delta >0: In this case the equation (2) has two real solutions. Call them r,s. Then the solution to (1) is given by C_1e^{rx}+C_2e^{sx}. Where C_1,C_2 are any real numbers.

    2) \Delta = 0. In this case the equation (2) has one real solution. Call it t. Then the solution to (1) is given by C_1e^{tx}+C_2xe^{tx}.

    3) \Delta < 0. In this case the equation (2) has two complex solutions. Since a,b,c are real, the polynomial is a polynomial in real coefficients and hence the complex solutions come as complex conjuages. Thus let a\pm i\beta to the two solutions. Then C_1 e^{\alpha x}\sin \beta x+C_2 e^{\alpha x}\cos \beta x is the solution to (1).
    Last edited by ThePerfectHacker; June 15th 2007 at 10:31 AM.
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