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Prove a given equation is a solution of a given differential equation

Let L be defined by

    \[ L(y) = y'' + ay' + by, \]

for a, b \in \mathbb{R} constants.

Let c \in \mathbb{R} and b \neq \omega^2. Using the previous exercise prove that the differential equation

    \[ L(y) = c \cos (\omega x) \]

has a particular solution given by

    \[ y = A \cos (\omega x - \alpha), \]

where

    \[ A = \frac{c}{\sqrt{(b-\omega^2)^2 + a^2 \omega^2}} \quad \text{and} \quad \tan \alpha = \frac{a \omega}{b - \omega^2}. \]


Proof. Incomplete.

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