Undetermined Coefficients — Question 4

Question 4

On $\mathbb R$, consider $$y^{(4)}-y=6x^2+8e^{-x}+10\cos x.$$ Its three forcing terms require different coefficient calculations.

Tasks

1. Factor the characteristic polynomial, write the full real homogeneous solution, and select a separate particular-solution trial for each forcing term.

2. Determine all coefficients and assemble one particular solution.

3. Verify each forcing contribution directly using $L=D^4-1$, rather than relying only on the trial-selection rule. Write the complete general solution.

4. Decide whether any solution is bounded on $[0,\infty)$. Prove your answer even when homogeneous constants are chosen to cancel all removable growing terms.

Question 4 – Solution

Strategy. Use linearity to separate polynomial forcing from two distinct resonant frequencies.

Step 1: Identify roots and trials. The characteristic roots are $1,-1,i,-i$, all simple, so $$y_h=Ae^x+Be^{-x}+C\cos x+D\sin x.$$ Use $ax^2+bx+c$ for $6x^2$, $kxe^{-x}$ for $8e^{-x}$, and $x(m\cos x+n\sin x)$ for $10\cos x$. Only the last two blocks are resonant.

Step 2: Compute the coefficients. The fourth derivative of a quadratic is zero, giving $a=-6,b=c=0$. For a simple root $\lambda$ of $P(r)=r^4-1$, the shift identity gives $P(D)(xe^{\lambda x})=P'(\lambda)e^{\lambda x}$. Since $P'(-1)=-4$, we obtain $k=-2$.

For the trigonometric terms, $L(x\cos x)=4\sin x$ and $L(x\sin x)=-4\cos x$. Therefore $m=0,n=-5/2$, and $$\boxed{y_p=-6x^2-2xe^{-x}-\frac 52x\sin x.}$$

Step 3: Verify all three residuals. Direct differentiation gives $$L(-6x^2)=6x^2,\qquad D^4(xe^{-x})=e^{-x}(x-4),$$ $$D^4(x\sin x)=x\sin x-4\cos x.$$ Thus the second and third terms produce $8e^{-x}$ and $10\cos x$ exactly. The full solution is $\boxed{y=y_h+y_p}$; no homogeneous constants are determined without additional data.

Step 4: Rule out bounded solutions. If $A\ne 0$, the $Ae^x$ term dominates every other term, so boundedness is impossible. If $A=0$, division by $x^2$ gives $$\lim_{x\to\infty}\frac{y(x)}{x^2}=-6,$$ because the remaining oscillatory growth is only $O(x)$ and the other exponential terms decay. Hence even after removing the positive exponential, quadratic forcing remains. No solution is bounded on the positive half-line.