Undetermined Coefficients — Question 5

Question 5

Consider $$(D+1)^2(D-2)y=e^{-x}.$$ An annihilator can simplify the forcing while introducing extra solutions that must later be rejected.

Tasks

1. Apply a minimal constant-coefficient annihilator of the forcing and write the full solution of the resulting fourth-order homogeneous equation.

2. Substitute that enlarged family into the original third-order equation. Find the constraint that removes the extra freedom and write the actual general solution.

3. Find the solution with $y(0)=y'(0)=y''(0)=0$. Determine the coefficient of its growing exponential and its limit after multiplication by $e^{-2x}$.

4. Instead require $y(0)=y'(0)=0$ and decay as $x\to+\infty$. Find this solution and the initial value $y''(0)$ it requires. Explain why decaying forcing does not guarantee that the zero-data response decays.

Question 5 – Solution

Strategy. Retain the original equation as a constraint after annihilating its right-hand side.

Step 1: Solve the enlarged equation. The forcing is annihilated by $D+1$. Applying it gives $(D+1)^3(D-2)y=0$, whose general solution is $$y=e^{-x}(A+Bx+Cx^2)+Ee^{2x}.$$ This family has four constants, although the original equation has order three.

Step 2: Restore the original forcing. Under $y=e^{-x}v$, the original operator becomes $e^{-x}D^2(D-3)v$. For $v=A+Bx+Cx^2$ its value is $-6Ce^{-x}$; the $Ee^{2x}$ term is homogeneous. Thus $C=-1/6$, leaving $$\boxed{y=e^{-x}(A+Bx-x^2/6)+Ee^{2x}.}$$ Annihilation alone only ensures that the original residual is a multiple of $e^{-x}$, not that its coefficient equals one.

Step 3: Fit the zero initial vector. The three equations are $A+E=0$, $B-A+2E=0$, and $A-2B-1/3+4E=0$. They yield $E=1/27$, $A=-1/27$, $B=-1/9$, so $$\boxed{y_0=\frac{e^{2x}}{27}-e^{-x}\left(\frac 1{27}+\frac{x}{9}+\frac{x^2}{6}\right), \qquad \lim_{x\to\infty}e^{-2x}y_0=\frac 1{27}.}$$ The displayed coefficient equations verify the initial data, and Step 2 verifies the forcing.

Step 4: Replace curvature data by a future condition. Decay requires $E=0$; with $y(0)=y'(0)=0$, this gives $A=B=0$. Hence $$\boxed{y_d=-x^2e^{-x}/6,\qquad y_d''(0)=-1/3.}$$ This is the unique decaying response with those first two data. The zero-data response must add a growing homogeneous mode to cancel the particular solution’s initial curvature. Decay of the input alone does not suppress that mode.