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Math Help - Inverse operators

  1. #1
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    Inverse operators

    Hey I need to solve this problem by means of inverse operators:
    y''-3y'+2y=x
    Thank you!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by deltafee View Post
    Hey I need to solve this problem by means of inverse operators:

    y''+3y'+2y=x
    Thank you!
    Just FYI your questions are very vague.
    I think this is what you mean...

    \displaystyle y''+3y'+2y=x \iff (D^2+3D+2)y=x
    Solving for y gives

    \displaystyle y=\frac{1}{(D+1)(D+2)}[x]

    Since x is a polynomial we can express each operator as a geometric series, but 1st partial fractions gives

    \displaystyle \frac{1}{(D+1)(D+2)}=\frac{1}{D+1}+\frac{-1}{D+2}
    Then
    \displaystyle \frac{1}{1+D}=\sum_{n=0}^{\infty}(-1)^nD^n
    and
    \displaystyle \frac{1}{2+D}=\frac{1}{2}\frac{1}{1+\frac{D}{2}}=\  sum_{n=0}^{\infty}(-1)^n\frac{D^n}{2^n}

    Combing the sums gives

    \displaystyle \frac{1}{(D+1)(D+2)}=\sum_{n=0}^{\infty} \left( \frac{2^n+1}{2^n}\right)(-1)^nD^n

    Note that since the right hand side is only a degree 1 polynomial we only need two terms from the series

    \displaystyle 1-\frac{3}{2}D

    \displaystyle \left( 1-\frac{3}{2}D\right)x=x-\frac{3}{2}

    So this gives \displaystyle y_p=x-\frac{3}{2}

    You still need to use the eigenfunctions to find the complimentary solution.
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  3. #3
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    Very elegant!
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