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Find a synchronous solution of the form \( \mathrm{A} \cos \Omega \mathrm{t}+\mathrm{B} \sin \Omega \mathrm{t} \) to the given forced oscillator equation using the method of insertion, collecting terms, and matching coefficients to solve for A and B . \( \mathrm{y}^{\prime \prime}+2 \mathrm{y}^{\prime}+4 \mathrm{y}=6 \sin 3 \mathrm{t}, \Omega=3 \) A solution is \( \mathrm{y}(\mathrm{t})=\square \)

Ask by Cook Rogers. in the United States
Mar 17,2025

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The solution is \( y(t) = -\frac{36}{61} \cos(3t) - \frac{30}{61} \sin(3t) \).

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Beyond the Answer

To solve the forced oscillator equation \( y'' + 2y' + 4y = 6 \sin(3t) \), we start by making an ansatz for the particular solution of the form \( y_p(t) = A \cos(3t) + B \sin(3t) \). First, we calculate the first and second derivatives of \( y_p \): 1. \( y_p' = -3A \sin(3t) + 3B \cos(3t) \) 2. \( y_p'' = -9A \cos(3t) - 9B \sin(3t) \) Now we substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the original forced oscillator equation: \[ (-9A \cos(3t) - 9B \sin(3t)) + 2(-3A \sin(3t) + 3B \cos(3t)) + 4(A \cos(3t) + B \sin(3t)) = 6 \sin(3t) \] This simplifies to: \[ (-9A + 6B + 4A) \cos(3t) + (-9B - 6A + 4B) \sin(3t) = 6 \sin(3t) \] Combining like terms: \[ (-5A + 6B) \cos(3t) + (-5B - 6A) \sin(3t) = 0 \cos(3t) + 6 \sin(3t) \] Now we can match the coefficients for \( \cos(3t) \) and \( \sin(3t) \): 1. For the \( \cos(3t) \) term: \( -5A + 6B = 0 \) (1) 2. For the \( \sin(3t) \) term: \( -5B - 6A = 6 \) (2) From equation (1), we can express \( A \) in terms of \( B \): \[ A = \frac{6}{5}B \] Now substituting \( A \) in equation (2): \[ -5B - 6\left(\frac{6}{5}B\right) = 6 \] \[ -5B - \frac{36}{5}B = 6 \] \[ -\left(5 + \frac{36}{5}\right)B = 6 \] \[ -\frac{61}{5}B = 6 \] \[ B = -\frac{30}{61} \] Now substituting \( B \) back to find \( A \): \[ A = \frac{6}{5}\left(-\frac{30}{61}\right) = -\frac{36}{61} \] Thus, our solution for the particular solution \( y_p(t) \) is: \[ y_p(t) = -\frac{36}{61} \cos(3t) - \frac{30}{61} \sin(3t) \] Therefore, the final solution for the original forced oscillator equation is: \[ y(t) = y_h(t) + y_p(t) = C_1 e^{-t} \cos(\sqrt{3} t) + C_2 e^{-t} \sin(\sqrt{3} t) - \frac{36}{61} \cos(3t) - \frac{30}{61} \sin(3t) \] In the answer box, the specific phrase to fill will be: \[ y(t) = C_1 e^{-t} \cos(\sqrt{3} t) + C_2 e^{-t} \sin(\sqrt{3} t) - \frac{36}{61} \cos(3t) - \frac{30}{61} \sin(3t) \]

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