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Thread: Initial Value problem

  1. #1
    Member Altair's Avatar
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    Initial Value problem

    The question asks to "solve" the following differential equation.

    $\displaystyle D^2(D-3)y = 6; Y(1) = 1, y'(1) = -1, y''(1)=-2$

    We have not been told yet how to use the Method Of Undetermined Coefficients or by variable coefficients.

    We have, however, solved similar questions with partial fractions.

    I therefore tried the question in the same way.

    $\displaystyle D^2(D-3)y = 6e^{0x}$

    $\displaystyle y = 6\frac{e^{0x}}{(D^2)(D-3)}$

    $\displaystyle \frac{1}{(D^2)(D-3)} = \frac{A}{D} + \frac{B}{D^2} + \frac{C}{D-3}$

    After solving I get,

    $\displaystyle A = -\frac{2}{3}$

    $\displaystyle B = -\frac{1}{3}$

    $\displaystyle c = -\frac{1}{9}$

    Then after solving (as D is the differential operator) I get the following.

    $\displaystyle (\frac{-2x}{3} -\frac{-x^2}{3} - \frac{1}{27}) * 6$

    Where as the answer is simply $\displaystyle -x^2 + x + 1$
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  2. #2
    Member Altair's Avatar
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    Can't do it by Binomial expansion as well.
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