Real & Distinct Roots — Question 9

Question 9 – Solution

Strategy. Factorization creates a sequence of two first-order equations, not a rule requiring one of two first-order residuals to vanish.

Step 1: Expand the operator composition. Applying $(D-3)$ to $(D-1)y=y'-y$ gives $$(y'-y)'-3(y'-y)=y''-4y'+3y.$$ Thus the factorization is valid and its characteristic roots are 1 and 3.

Step 2: Solve for the intermediate quantity. With $z=y'-y$, the equation is $z'-3z=0$. The data imply $z(0)=4-2=2$. Separating or using an integrating factor yields $$\boxed{z=2e^{3t}}.$$ The intermediate residual is nonzero, yet it is annihilated by $D-3$.

Step 3: Recover the original solution. We must solve $y'-y=2e^{3t}$. Multiplication by $e^{-t}$ gives $$(e^{-t}y)'=2e^{2t}.$$ Integrating from 0 to $t$ and using $y(0)=2$ gives $e^{-t}y-2=e^{2t}-1$. Hence $$\boxed{y=e^t+e^{3t}}.$$ Indeed $y(0)=2$, $y'(0)=1+3=4$, and $$y''-4y'+3y=(1-4+3)e^t+(9-12+3)e^{3t}=0.$$ The construction recovers both modes by first-order methods and gives the unique solution with the prescribed data.

Step 4: Refute the incorrect either/or inference. For this solution, $$(D-1)y=2e^{3t}\ne 0,\qquad (D-3)y=-2e^t\ne 0.$$ Nevertheless, $(D-3)((D-1)y)=(D-3)(2e^{3t})=0$. An operator composition first produces a new function and then applies another operator to it. A nonzero function can be sent to zero by differentiation combined with multiplication. The scalar zero-product rule therefore does not imply the proposed either/or claim. Sums of the two modes must be retained, not just the two separate pure-mode families.