General Solution of a Doubly-Forced Oscillator (Resonant + Non-Resonant): 04-BS-1, December 2018, Q2

Question 2: General Solution of a Doubly-Forced Oscillator (Resonant + Non-Resonant) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $x''+4x=3\cos2t+4\cos3t$.

Find. The general solution $x(t)$.

Approach. The homogeneous natural frequency is $\omega_0=2$. The $3\cos2t$ term forces exactly at $\omega_0$ (resonance, needs a $t\sin\omega_0t$-type trial); the $4\cos3t$ term forces off-resonance and takes a plain $\cos3t$ trial. Superpose both particular solutions onto the homogeneous solution.

Homogeneous solution. $r^2+4=0\Rightarrow r=\pm2i$, so $x_h(t)=C_1\cos2t+C_2\sin2t$.
Resonant particular solution for $3\cos2t$. Since $\omega=2=\omega_0$, try $x_{p1}=t(A\cos2t+B\sin2t)$. Substituting and cancelling the (automatically vanishing) $t\cos2t,\,t\sin2t$ terms leaves $-4A\sin2t+4B\cos2t=3\cos2t$, giving $A=0,\ B=\tfrac34$: $$x_{p1}(t)=\tfrac34t\sin2t.$$
Non-resonant particular solution for $4\cos3t$. Try $x_{p2}=D\cos3t$: $-9D+4D=-5D=4\Rightarrow D=-\tfrac45$, so $$x_{p2}(t)=-\tfrac45\cos3t.$$
Superpose. $$x(t)=C_1\cos2t+C_2\sin2t+\tfrac34t\sin2t-\tfrac45\cos3t.$$

$$x(t)=\boxed{C_1\cos2t+C_2\sin2t+\tfrac34t\sin2t-\tfrac45\cos3t}$$

Quantity	Result
Natural frequency $\omega_0$	$2$
Resonant term ($3\cos2t$)	$\tfrac34t\sin2t$
Non-resonant term ($4\cos3t$)	$-\tfrac45\cos3t$
$x(t)$	$C_1\cos2t+C_2\sin2t+\tfrac34t\sin2t-\tfrac45\cos3t$

Concept & Theory

Topic. Superposition of resonant and non-resonant forcing in a linear oscillator.

Key relations. For $x''+\omega_0^2x=F\cos\omega_0t$ (resonance), $x_p=\dfrac{F}{2\omega_0}t\sin\omega_0t$. For $x''+\omega_0^2x=F\cos\omega t$ with $\omega\ne\omega_0$, $x_p=\dfrac{F}{\omega_0^2-\omega^2}\cos\omega t$.

Why this works. A forcing frequency equal to the natural frequency drives the amplitude linearly in $t$ because the plain $\cos\omega_0t$ trial already solves the homogeneous equation; multiplying by $t$ breaks the degeneracy exactly as it does for a repeated characteristic root. Superposition is valid because the ODE is linear, so each forcing term's particular solution can be found independently and added.

Common pitfall. Applying the resonant $t\sin\omega_0t$ trial to both forcing terms instead of only the one whose frequency matches $\omega_0$; sign errors when the non-resonant denominator $\omega_0^2-\omega^2$ is negative (here $4-9=-5$).

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 2 (§2.8 Modeling: Forced Oscillations, Resonance).

Question 2 of 8: General Solution of a Doubly-Forced Oscillator (Resonant + Non-Resonant)

Question 2: General Solution of a Doubly-Forced Oscillator (Resonant + Non-Resonant) (20 marks)