Undetermined Coefficients — Question 3

Question 3 – Solution

Strategy. Account for the double conjugate roots before imposing the initial state.

Step 1: Use the full resonance multiplicity. The roots $\pm i$ both have multiplicity two. The functions $\cos x,\sin x,x\cos x,x\sin x$ are all homogeneous, so an $x$-weighted trial still gives zero. Use $y_p=x^2(A\cos x+B\sin x)$.

Step 2: Match the forcing directly. A first application gives $(D^2+1)(x^2\cos x)=2\cos x-4x\sin x$, and a second gives $-8\cos x$. Likewise, $(D^2+1)^2(x^2\sin x)=-8\sin x$. Thus $A=-1/8$, $B=0$, and $$y=-\frac{x^2\cos x}{8}+(a+bx)\cos x+(c+dx)\sin x.$$

Step 3: Recover the initially resting solution. The particular term has initial vector $(0,0,-1/4,0)$. The homogeneous term has vector $(a,b+c,-a+2d,-3b-c)$. Setting their sum to zero gives $a=b=c=0,d=1/8$, hence $$\boxed{y=\frac{x\sin x-x^2\cos x}{8}.}$$ The two numerator terms cancel to order two: $x\sin x-x^2\cos x=x^4/3+O(x^6)$ near zero. Thus $y$ and its first three derivatives vanish there. The term $x\sin x/8$ is homogeneous, while the already checked particular term produces $\cos x$.

Step 4: Separate quadratic forcing from linear freedom. At $x=2\pi n$, $$\boxed{y(2\pi n)=-\frac{(2\pi n)^2}{8}.}$$ This proves unbounded quadratic growth in magnitude along that sequence. Every homogeneous term is $O(x)$ on the positive half-line, so no correction can cancel these quadratic values. The figure shows the response and the valid envelopes $\pm(x^2+x)/8$.