Nonhomogeneous Systems — Question 3

Question 3 – Solution

Strategy. Resonance is a compatibility problem for the forcing direction, not merely a comparison of two scalar exponents.

Step 1: Test the simple exponential trial. Substitution gives $(I-A)v=(a,b)^T$, or $0=a$, $3v_2=b$. Thus the trial works exactly when $a=0$, with $\boxed{v=(c,b/3)^T,\quad c\in\mathbb R}$. The free first component is a homogeneous contribution, not a new forced effect.

Step 2: Construct the general forced response. A particular solution for all parameters is $X_p=(at e^t,(b/3)e^t)^T$. Adding $(c_1e^t,c_2e^{-2t})^T$ and imposing zero initial data gives $$\boxed{x=at e^t,\qquad y=\tfrac b3(e^t-e^{-2t}).}$$ Differentiation gives $x'-x=ae^t$ and $y'+2y=be^t$, verifying both forcing components and the initial values.

Step 3: Identify the resonant component. The resonant eigendirection is $(1,0)^T$. Its forced scalar equation is $(e^{-t}x)'=a$, so a nonzero $a$ necessarily creates $at e^t$. When $a=0$, there is no forcing in that direction and no polynomial factor is needed. No choice of homogeneous constants can cancel a nonzero $at e^t$ for all time. The condition is directional, not just spectral.

Step 4: Separate forced growth from resonance. For $a=0,b\ne 0$, the zero-state solution has $x=0$ and $y/e^t\to b/3$. It grows because its forcing grows, although the forced second component has homogeneous rate $-2$ and is not resonant. The first eigenvalue $1$ is present in the matrix but receives no input. Thus growth alone does not prove resonance, and the student’s claim fails for every nonzero forcing vector of the form $(0,b)^T$.