Linear Homogeneous Differential Equations — Question 4

Question 4 – Solution

Strategy. Read the highest polynomial degree at each distinct exponential frequency, including complex conjugates.

Step 1: Find sufficient root multiplicities. The shift identity $P(D)(e^{\lambda x}q)=e^{\lambda x}P(D+\lambda)q$ shows that a quadratic polynomial multiplying $e^{2x}$ is killed by $(D-2)^3$. Also $$x\sin 3x=\frac{x}{2i}(e^{3ix}-e^{-3ix}),$$ so each of $\pm 3i$ requires multiplicity two. A sufficient real monic polynomial is $$\boxed{P(r)=(r-2)^3(r^2+9)^2,\qquad \deg P=7.}$$

Step 2: Prove minimality, not just sufficiency. Suppose a polynomial $Q$ has a root of multiplicity $m$ at $\lambda$; write $Q(r)=(r-\lambda)^mR(r)$ with $R(\lambda)\ne 0$. Applied to $e^{\lambda x}q(x)$, the polynomial $R(D+\lambda)q$ retains the degree of $q$ because its leading coefficient is multiplied by $R(\lambda)$. Thus $D^m$ kills it exactly when $m>\deg q$.

Exponential-polynomial modes with distinct complex exponents are linearly independent on $\mathbb R$; this is the generalized characteristic-root independence theorem. Therefore the contributions at $2,3i,-3i$ cannot cancel each other. Every annihilating $Q$ must have multiplicities at least $3,2,2$, so must be divisible by $P$ and have degree at least seven.

Step 3: Recover the entire solution space. The real general solution is $$\boxed{y=e^{2x}(a_0+a_1x+a_2x^2)+(b_0+b_1x)\cos 3x +(c_0+c_1x)\sin 3x.}$$ There are seven free real constants. A monic degree-seven annihilator divisible by the monic degree-seven $P$ must equal $P$, proving uniqueness at minimal order.

Step 4: Interpret the forced companion modes. Both $xe^{2x}$ and $\cos 3x$ belong to this solution space. The two written summands package several repeated-root modes: their number is not the differential order. The lower bound concerns nonzero constant-coefficient operators only; it does not assert a minimum order among variable-coefficient equations.