Undetermined Coefficients — Question 2

Question 2 – Solution

Strategy. Shift the triple root to zero, where the coefficient calculation becomes polynomial differentiation.

Step 1: Remove the homogeneous overlap. The characteristic root $1$ has multiplicity three. Every proposed term $e^x(Ax^2+Bx+C)$ is already homogeneous and gives zero under $(D-1)^3$. The corrected trial is $$y_p=x^3e^x(Ax^2+Bx+C).$$

Step 2: Solve the shifted coefficient equations. Since $(D-1)(e^xv)=e^xv'$, the equation becomes $v'''=x^2+1$. For $v_p=Ax^5+Bx^4+Cx^3$, matching gives $$60Ax^2+24Bx+6C=x^2+1, \qquad A=\frac 1{60},\quad B=0,\quad C=\frac 16.$$ The complete solution is consequently $$\boxed{y=e^x\left(c_0+c_1x+c_2x^2+\frac{x^3}{6}+\frac{x^5}{60}\right).}$$

Step 3: Impose and verify the initial data. With $v=e^{-x}y$, the initial values are $$v(0)=y(0)=0,\quad v'(0)=y'(0)-y(0)=0,\quad v''(0)=y''(0)-2y'(0)+y(0)=0.$$ Thus $c_0=c_1=c_2=0$. The polynomial $v=x^3/6+x^5/60$ and its first two derivatives vanish at zero, so the corresponding initial data for $y=e^xv$ vanish as well. Its third derivative is $v'''=1+x^2$, and the shift identity verifies the forcing exactly.

Step 4: Identify unavoidable resonant growth. For every choice of $c_0,c_1,c_2$, $$\boxed{\lim_{x\to\infty}\frac{e^{-x}y(x)}{x^5}=\frac 1{60}.}$$ The lower-degree terms disappear after division by $x^5$. Hence $y(x)$ is eventually positive and asymptotic to $e^xx^5/60$, so it is unbounded. Homogeneous terms have polynomial degree at most two after removing $e^x$ and cannot cancel the forced degree-five term.