Undetermined Coefficients — Question 3

Question 3 – Solution

Strategy. Polynomial differentiation couples the sine and cosine components even when only one appears in the forcing.

Step 1: Complete the trial space. Use $$y_p=(at+b)\cos 2t+(ct+d)\sin 2t.$$ Frequency $2$ differs from the homogeneous frequency $1$, so no extra power of $t$ is needed for resonance. Nevertheless, differentiating the proposed cosine-only term twice produces $-4a\sin 2t$, which cannot be ignored.

Step 2: Match all terms. Applying $D^2+1$ to the complete trial gives $$(-3at-3b+4c)\cos 2t+(-3ct-3d-4a)\sin 2t.$$ Thus $a=-1/3$, $c=0$, $b=0$, $d=4/9$, and $$\boxed{y_p=-\tfrac 13t\cos 2t+\tfrac 49\sin 2t.}$$

Step 3: Fit the initial state. The particular data are $y_p(0)=0$ and $y_p'(0)=-1/3+8/9=5/9$. The homogeneous correction is therefore $-(5/9)\sin t$, giving $$\boxed{y=-\tfrac 13t\cos 2t+\tfrac 49\sin 2t-\tfrac 59\sin t.}$$ Its value and slope at zero vanish, and the correction leaves the verified forcing unchanged.

Step 4: Separate resonance from forcing growth. The forcing itself has a factor $t$ and is unbounded. At $t_n=2\pi n$, the solution is $y(t_n)=-t_n/3\to-\infty$. At $s_n=\pi/2+2\pi n$, it is $y(s_n)=s_n/3-5/9\to\infty$. Thus it is unbounded in both directions despite the absence of frequency resonance. Nonresonance alone is not a boundedness guarantee when the forcing amplitude grows.