Linear Homogeneous Differential Equations — Question 5

Question 5 – Solution

Strategy. Use polynomial interpolation at the roots to extract growing modes directly from the initial data.

Step 1: Construct mode selectors. The polynomial factors as $(r-1)(r+1)(r-2)(r+2)$. Thus $y$ is a linear combination of $e^{\lambda x}$ for $\lambda=-2,-1,1,2$. Since $Q(D)e^{\mu x}=Q(\mu)e^{\mu x}$ and $\Pi_\lambda(\mu)$ is $1$ if $\mu=\lambda$ and $0$ otherwise, $$\Pi_\lambda(D)y=A_\lambda e^{\lambda x}.$$ Evaluating at zero reads off $A_\lambda$ without solving all four coefficients simultaneously.

Step 2: Extract the unstable coefficients. The relevant polynomials are $$\Pi_1(r)=-\frac{(r+2)(r+1)(r-2)}6 =\frac{-r^3-r^2+4r+4}{6},$$ $$\Pi_2(r)=\frac{(r+2)(r+1)(r-1)}{12} =\frac{r^3+2r^2-r-2}{12}.$$ Hence $$\boxed{A_1=\frac{4a+4b-c-d}{6},\qquad A_2=\frac{d+2c-b-2a}{12}.}$$

Step 3: Describe exactly the decaying initial states. Forward boundedness requires $A_2=A_1=0$; distinct positive exponential rates cannot cancel. The remaining modes decay. Equivalently, $$\boxed{c+3b+2a=0,\qquad d+3c+2b=0.}$$ These follow from $y''+3y'+2y=0$ and its derivative for the two negative-root modes. Conversely, substituting these relations into $A_1,A_2$ makes both zero, so the relations are sufficient as well.

Step 4: Quantify the loss of cancellation. The unperturbed solution is $e^{-x}$. Replacing $d=-1$ by $-1+\varepsilon$ produces $A_1=-\varepsilon/6$ and $A_2=\varepsilon/12$. All modes other than $e^{2x}$ disappear after multiplication by $e^{-2x}$ in the limit, giving $$\boxed{\lim_{x\to\infty}e^{-2x}y_\varepsilon(x)=\varepsilon/12.}$$ Every nonzero error of this form eventually yields exponential growth in magnitude. Exact decay requires precise cancellation of both positive-root modes.