Undetermined Coefficients — Question 7

Question 7 – Solution

Strategy. Track the zero-frequency resonance and the oscillatory resonance separately.

Step 1: Calculate the resonant and nonresonant pieces. The characteristic polynomial is $r^2(r^2+1)$. Direct substitution gives $$L(x^2/2)=1,\qquad L(x\sin x)=-2\cos x,\qquad L(\cos 2x)=12\cos 2x.$$ Consequently, $$\boxed{y=\frac{a x^2}{2}-\frac b2x\sin x+\frac c{12}\cos 2x +k_0+k_1x+A\cos x+B\sin x.}$$ The factors $x^2$ and $x$ reflect the double root at zero and simple roots at $\pm i$, respectively.

Step 2: Classify boundedness exactly. If $a\ne 0$, division by $x^2$ gives a nonzero limit, so boundedness fails. Once $a=0$, a bounded solution would satisfy $$0=\lim_{x\to\infty}\frac{y(x)}x, \qquad \frac{y(x)}x=k_1-\frac b2\sin x+o(1).$$ Along sequences with sine equal to $1$ and $-1$, this requires both $k_1-b/2=0$ and $k_1+b/2=0$. Hence $$\boxed{\text{bounded on }[0,\infty)\iff a=b=k_1=0.}$$ These conditions suffice because all remaining terms are bounded.

Step 3: Check periodic compatibility independently. With $a=b=0$, choosing $k_1=0$ gives a $2\pi$-periodic solution for every $c$. Conversely, for a $2\pi$-periodic $C^4$ solution, integration gives $$\int_0^{2\pi}(y^{(4)}+y'')\,dx=0=2\pi a.$$ Integration by parts transfers the even derivatives onto $\cos x$ without boundary terms. Since $(D^4+D^2)\cos x=0$, orthogonality then gives $0=b\pi$. Thus $a=b=0$ is necessary and sufficient for existence of a $2\pi$-periodic response.

Step 4: Explain why zero mean is insufficient. The forcing $\cos x$ has zero average but produces the unavoidable term $-x\sin x/2$, so no solution is periodic. In contrast, $\cos 2x$ admits $y=\cos 2x/12$. A nonzero mean excites quadratic drift through the double zero root; a first harmonic excites linear oscillatory growth. Removing only the mean addresses just one of these two resonances.