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    Let L be defined by L(y)=y\prime\prime+ay\prime+by for a,b \in \mathbb{R}

    Let w \in \mathbb{R} and A \in \mathbb{C}. If either b \neq w^2 pr aw \neq 0 prove that the differential equation L(y)=Ae^{iwx} has a solution y=Be^{iwx}. Express the value of B in terms of a,b,A and w.

    This one is pretty straight forward I think, here is my attempt.

    Let y=Be^{iwx} with either b \neq w^2 or aw \neq 0. This gives me L(y)=-Bw^2e^{iwx}+aBiwe^{iwx}+bBe^{iwx  }


    B=\frac{A}{b-w^2+iaw} if b-w^2+iaw \neq 0 and since b-w^2 \neq 0 or aw \neq 0 means b-w^2_iaw \neq 0 therefore I have proven L(y)=Ae^{iex} has a solution y=Be^{iex}

    I'm pretty sure this is correct, can someone check it for me? Thank you.

    Looks good to me.

    You're given L(y), and you're given y, so it's just a case of plugging them in to the equation and rearranging to find B, then checking the fraction's denominator can't come to zero.

    All good!
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Updated: February 4, 2017
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