Linear Homogeneous Differential Equations — Question 3

Question 3 – Solution

Strategy. Retain the polynomial factors attached to repeated conjugate roots and test growth at suitable phases.

Step 1: Build the real basis. The polynomial is $(r^2+4)^2$, with roots $\pm 2i$, each twice. Thus $$y=(A+Bx)\cos 2x+(C+Dx)\sin 2x.$$ Omitting the factors $x\cos 2x$ and $x\sin 2x$ would lose half the solution space.

Step 2: Fit the initial vector. At zero the four derivatives are $$A,\qquad B+2C,\qquad -4A+4D,\qquad -12B-8C.$$ The data give $A=D=0$, $B=-1$ and $C=1/2$, hence $$\boxed{y=\tfrac 12\sin 2x-x\cos 2x.}$$ Indeed $y'=2x\sin 2x$, $y''=2\sin 2x+4x\cos 2x$, and $y'''=8\cos 2x-8x\sin 2x$, verifying all four data. Each basis term satisfies the factored equation.

Step 3: Exhibit growing extrema. For $x_n=n\pi/2$, we have $$y'(x_n)=0,\qquad y(x_n)=(-1)^{n+1}\frac{n\pi}{2},\qquad y''(x_n)=4x_n(-1)^n\ne 0.$$ These are alternating strict extrema with unbounded magnitude. In particular, purely imaginary roots alone do not imply boundedness when they are repeated.

Step 4: Classify the bounded family. If $(B,D)\ne(0,0)$, choose a sequence $x\to\infty$ on which $B\cos 2x+D\sin 2x=\sqrt{B^2+D^2}>0$. The linear-in-$x$ part grows, whereas $A\cos 2x+C\sin 2x$ is bounded. Therefore boundedness is equivalent to $B=D=0$, a two-dimensional subspace. A nonzero member of that subspace attains its positive amplitude infinitely often, so it cannot tend to zero. An unbounded member cannot tend to zero either.