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Thread: Operators

  1. #1
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    Operators

    Hello I really need help in understanding operators: the best way would be to show me how to solve:
    y''-y'-2y=e^(x)
    Thank you!
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  2. #2
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    Quote Originally Posted by deltafee View Post
    Hello I really need help in understanding operators: the best way would be to show me how to solve:


    Thank you!
    What exactly do you need to know. It can be rewritten as

    $\displaystyle D^2y-Dy-2y=e^{x}$ This can be factored as

    $\displaystyle (D^2-D-2)y=e^{x} \iff (D-2)(D+1)y=e^{x}$

    So the homogeneous system $\displaystyle (D-2)(D+1)y=0$ has solutions

    $\displaystyle e^{2x}$ because $\displaystyle (D-2)e^{2x}=0$ and

    $\displaystyle e^{-x}$ because $\displaystyle (D+1)e^{-x}=0$

    So it has the complimentary solution $\displaystyle y_c=c_1e^{2x}+c_2e^{-x}$

    Now we can solve the original

    $\displaystyle (D-2)(D+1)y=e^{x}$ by annihilating the right hand side with $\displaystyle (D-1)$ since $\displaystyle (D-1)e^{x}=0$


    $\displaystyle (D-1)(D-2)(D+1)y=(D-1)e^{x} \iff
    (D-1)(D-2)(D+1)y=0$ since that factor is not repeated the particular solution has the form

    $\displaystyle y_p=Ae^{x}$
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