Linear Homogeneous Differential Equations — Question 1

Question 1 – Solution

Strategy. Separate exponential growth, polynomial drift, decay and oscillation before imposing boundedness.

Step 1: Read all multiplicities. The roots are $0$ (twice), $1$, $-2$, and $-1\pm 2i$. Their multiplicities total six, so a real fundamental family gives $$\boxed{y=c_0+c_1x+Ae^x+Be^{-2x}+e^{-x}(C\cos 2x+S\sin 2x).}$$ The six constants are independent. In particular, the double zero root requires both $1$ and $x$.

Step 2: Remove forward growth. If $A\ne 0$, the $e^x$ term dominates as $x\to+\infty$. Once $A=0$, the remaining exponential terms decay, so boundedness also requires $c_1=0$. Conversely those two conditions suffice: $$\boxed{\text{bounded on }[0,\infty)\iff A=c_1=0.}$$ This is a four-dimensional subspace, with free constants $c_0,B,C,S$.

Step 3: Distinguish a limit from zero. Every forward-bounded solution tends to $c_0$. Thus the decaying subspace has $A=c_1=c_0=0$ and dimension three. For example, $1+e^{-2x}$ is bounded and nonconstant on $[0,\infty)$ but tends to $1$.

Step 4: Impose the backward restriction. Start with a forward-bounded solution. Multiplication by $e^{2x}$ and passage to $x\to-\infty$ shows $B=0$, since bounded $y$ would give a zero limit. If $(C,S)\ne(0,0)$, write $C\cos 2x+S\sin 2x=R\cos(2x-\phi)$ with $R>0$. Along a sequence tending to $-\infty$ at cosine maxima, $Re^{-x}$ is unbounded. Hence $C=S=0$ as well, leaving $$\boxed{y(x)=c_0\quad\text{as the entire-line bounded family}.}$$ Different growth rates and the explicit oscillatory subsequence rule out hidden cancellation.