Question 1 of 8: Resonant Initial Value Problem for a Forced Harmonic Oscillator

Question 1: Resonant Initial Value Problem for a Forced Harmonic Oscillator (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. Linear, constant-coefficient, nonhomogeneous ODE $y''+4y=6\cos(2t)$ with $y(0)=1,\ y'(0)=0$.

Find. The unique solution $y(t)$.

Approach. Solve the homogeneous equation (undamped oscillator, natural frequency $\omega=2$), notice that the forcing frequency coincides exactly with $\omega$ — pure resonance — so the particular-solution trial needs an extra factor of $t$, then fix the two constants from the initial conditions.

1. Homogeneous solution. Characteristic equation $r^2+4=0\Rightarrow r=\pm2i$, so $$y_h(t)=C_1\cos2t+C_2\sin2t.$$
2. Particular solution — resonance case. The forcing $6\cos2t$ has the same frequency as the homogeneous oscillation, so the bare trial $A\cos2t+B\sin2t$ would solve the homogeneous equation and cannot work. Try instead $y_p=t(A\cos2t+B\sin2t)$. Substituting and matching coefficients (or using the standard resonance formula $y_p=\dfrac{F}{2\omega}t\sin\omega t$ for forcing $F\cos\omega t$ with $F=6,\ \omega=2$), $$y_p(t)=\frac{6}{2(2)}t\sin2t=\frac32t\sin2t.$$
3. Apply the initial conditions. General solution $y(t)=C_1\cos2t+C_2\sin2t+\tfrac32t\sin2t$. $y(0)=C_1=1$. Differentiating, $y'(t)=-2C_1\sin2t+2C_2\cos2t+\tfrac32\sin2t+3t\cos2t$, so $y'(0)=2C_2=0\Rightarrow C_2=0$.

$$y(t)=\boxed{\cos2t+\tfrac32t\sin2t}$$