The general solution to the forced undamped harmonic oscillator equation $\frac{d^2 y}{dt^2} + 4y = 8 \cos(2t)$ exhibiting pure resonance is given by:

(A) $y(t) = c_1 \cos(2t) + c_2 \sin(2t) + 2t \sin(2t)$

(B) $y(t) = c_1 \cos(2t) + c_2 \sin(2t) + 2 \cos(2t)$

(C) $y(t) = c_1 \cos(2t) + c_2 \sin(2t) - 2t \cos(2t)$

(D) $y(t) = c_1 e^{2t} + c_2 e^{-2t} + 2t \sin(2t)$

Answer Key:

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### 1. Strategy & Theoretical Principles For $y'' + \omega_0^2 y = F_0 \cos(\omega t)$ with driving frequency $\omega = \omega_0 = 2$ (resonance): 1. Homogeneous solution: $y_c(t) = c_1 \cos(2t) + c_2 \sin(2t)$. 2. Because $\cos(2t)$ is in the null space of the differential operator, the particular solution must be multiplied by $t$: $$y_p(t) = t [A \cos(2t) + B \sin(2t)]$$ ### 2. Analytical Formulation & Calculation Compute derivatives of $y_p$: $$y_p' = A \cos(2t) + B \sin(2t) + t [-2A \sin(2t) + 2B \cos(2t)]$$ $$y_p'' = -4A \sin(2t) + 4B \cos(2t) - 4t [A \cos(2t) + B \sin(2t)]$$ Substitute into $y_p'' + 4y_p$: $$y_p'' + 4y_p = -4A \sin(2t) + 4B \cos(2t) = 8 \cos(2t)$$ Equating coefficients: - $-4A = 0 \implies A = 0$ - $4B = 8 \implies B = 2$ Therefore, the particular solution is: $$y_p(t) = 2t \sin(2t)$$ Total general solution: $$y(t) = c_1 \cos(2t) + c_2 \sin(2t) + 2t \sin(2t)$$ ### 3. Distractor Forensics - **Option A:** Correct secular resonance term $2t \sin(2t)$. - **Option B:** Fails because $(D^2+4)\cos(2t) = 0$. - **Option C:** Sign error on $B$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** In resonance, the particular solution amplitude grows linearly with time ($t$). - **Standard Reference:** Kreyszig, E., *Advanced Engineering Mathematics*, Chapter 2.

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