General Solution of a Resonant Linear System: 04-BS-1, December 2019, Q4

Question 4: General Solution of a Resonant Linear System (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\mathbf x'=A\mathbf x+\mathbf F(t)$ with $A=\begin{pmatrix}4&2\\3&-1\end{pmatrix}$, $\mathbf F(t)=(0,\,e^{-2t})^T$.

Find. The general solution $(x(t),y(t))$.

Approach. Find the eigenpairs of $A$ for the homogeneous solution. Since the forcing rate $-2$ turns out to equal one of the eigenvalues, this is a resonant case: try $\mathbf x_p=(\mathbf a t+\mathbf b)e^{-2t}$ rather than a plain $\mathbf a e^{-2t}$.

Eigenvalues. $\det(A-\lambda I)=(4-\lambda)(-1-\lambda)-6=\lambda^2-3\lambda-10=0\Rightarrow\lambda=5,\,-2$.
Eigenvectors. For $\lambda=5$: $-x+2y=0\Rightarrow\mathbf v_1=(2,1)$. For $\lambda=-2$: $6x+2y=0\Rightarrow\mathbf v_2=(1,-3)$. $$\mathbf x_h(t)=C_1e^{5t}(2,1)+C_2e^{-2t}(1,-3).$$
Resonance check. The forcing $e^{-2t}$ shares its rate with the eigenvalue $\lambda=-2$, so the plain trial $\mathbf a e^{-2t}$ would duplicate $\mathbf x_h$. Use $\mathbf x_p=(\mathbf a t+\mathbf b)e^{-2t}$. Matching the $t$-coefficient forces $(A+2I)\mathbf a=0$, so $\mathbf a=k(1,-3)$ is a multiple of the $\lambda=-2$ eigenvector — the hallmark of resonance.
Solve for $k$ and $\mathbf b$. Matching the constant term gives $\mathbf a=(A+2I)\mathbf b+(0,1)^T$ with $A+2I=\begin{pmatrix}6&2\\3&1\end{pmatrix}$ (rank 1, consistent with $\lambda=-2$ being an eigenvalue). Writing both rows of this equation and eliminating $\mathbf b$ pins down $k=-\tfrac27$; choosing the free component $b_1=0$ then gives $b_2=-\tfrac17$: $$\mathbf a=\left(-\tfrac27,\,\tfrac67\right),\qquad \mathbf b=\left(0,\,-\tfrac17\right).$$
Assemble the particular and general solutions. $$x_p(t)=-\tfrac{2t}7e^{-2t},\qquad y_p(t)=\tfrac{6t-1}7e^{-2t}.$$ $$x(t)=2C_1e^{5t}+C_2e^{-2t}-\tfrac{2t}7e^{-2t},\qquad y(t)=C_1e^{5t}-3C_2e^{-2t}+\tfrac{6t-1}7e^{-2t}.$$

$$\boxed{x(t)=2C_1e^{5t}+C_2e^{-2t}-\tfrac{2t}7e^{-2t},\quad y(t)=C_1e^{5t}-3C_2e^{-2t}+\tfrac{6t-1}7e^{-2t}}$$

Quantity	Result
Eigenvalues	$5,\,-2$
Eigenvectors	$(2,1)$ for $5$; $(1,-3)$ for $-2$
Resonant particular solution	$x_p=-\tfrac{2t}7e^{-2t}$, $y_p=\tfrac{6t-1}7e^{-2t}$
General solution	$x=2C_1e^{5t}+C_2e^{-2t}-\tfrac{2t}7e^{-2t}$; $y=C_1e^{5t}-3C_2e^{-2t}+\tfrac{6t-1}7e^{-2t}$

Concept & Theory

Topic. Resonance in a nonhomogeneous linear system of ODEs.

Key relations. When the forcing $\mathbf F(t)=\mathbf f_0e^{\mu t}$ has a rate $\mu$ equal to an eigenvalue of $A$, the plain trial $\mathbf a e^{\mu t}$ fails (it solves the homogeneous system) and must be replaced by $(\mathbf at+\mathbf b)e^{\mu t}$, exactly analogous to the scalar $t e^{\mu t}$ fix for a repeated/matching root.

Why this works. Matching the $t$-coefficient of the trial always forces $(A-\mu I)\mathbf a=0$; a nonzero solution exists only because $\mu$ is an eigenvalue, which is precisely the resonance condition — the matrix $A-\mu I$ is then singular by construction, so solving for $\mathbf b$ requires the forcing vector to satisfy a solvability (consistency) condition tied to the same null space that supplied $\mathbf a$.

Common pitfall. Trying the non-resonant form $\mathbf a e^{-2t}$ first and getting a singular system with no solution, then not recognizing this failure as the signal to switch to the $(\mathbf at+\mathbf b)e^{-2t}$ trial.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 4 (§4.6 Nonhomogeneous Linear Systems).

Question 4 of 8: General Solution of a Resonant Linear System

Question 4: General Solution of a Resonant Linear System (20 marks)