Eigenvalues/Eigenvectors and a Nonhomogeneous Linear System: 04-BS-1, May 2013, Q3

Question 3: Eigenvalues/Eigenvectors and a Nonhomogeneous Linear System (a) 6, (b) 14 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. Matrix $A=\begin{pmatrix}4&3\\-1&0\end{pmatrix}$; the linear system in (b) has the same coefficient matrix $A$, plus a forcing term $t$ on the $y$-equation, with $x(0)=2,\ y(0)=-1$.

Find. (a) eigenvalues and eigenvectors of $A$. (b) $x(t),y(t)$ satisfying the system and initial conditions.

Approach. (a) standard characteristic-polynomial eigen-decomposition. (b) since the system's matrix is exactly $A$ from part (a), reuse its eigenpairs to build the homogeneous solution, then find a polynomial particular solution matching the linear forcing $(0,t)^T$, and fix constants from the ICs.

(a) Eigenvalues. $$\det(A-\lambda I)=(4-\lambda)(-\lambda)-(3)(-1)=\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)=0 \;\Rightarrow\; \boxed{\lambda_1=1,\ \lambda_2=3}.$$ Eigenvector for $\lambda_1=1$: $(A-I)v=0 \Rightarrow \begin{pmatrix}3&3\\-1&-1\end{pmatrix}v=0 \Rightarrow v_1+v_2=0$, so $\boxed{\mathbf v_1=(1,-1)}$. Eigenvector for $\lambda_2=3$: $(A-3I)v=0 \Rightarrow \begin{pmatrix}1&3\\-1&-3\end{pmatrix}v=0 \Rightarrow v_1+3v_2=0$, so $\boxed{\mathbf v_2=(3,-1)}$.
(b) Homogeneous solution. Reusing $\lambda_1=1,\mathbf v_1=(1,-1)$ and $\lambda_2=3,\mathbf v_2=(3,-1)$ from (a), $$\begin{pmatrix}x\\y\end{pmatrix}_h=C_1e^{t}\begin{pmatrix}1\\-1\end{pmatrix}+C_2e^{3t}\begin{pmatrix}3\\-1\end{pmatrix}.$$
Particular solution. The forcing is linear in $t$, so try $x_p=at+b,\ y_p=ct+d$. Substituting into $x_p'=4x_p+3y_p$ and $y_p'=-x_p+t$ and matching powers of $t$ gives four equations: $4a+3c=0$ (t-coeff, eq.1), $a=4b+3d$ (const, eq.1), $1-a=0$ (t-coeff, eq.2), $c=-b$ (const, eq.2). Solving in order: $a=1$; $c=-\tfrac{4}{3}$ (from eq.1's t-coeff); $b=-c=\tfrac{4}{3}$; $d=\tfrac{a-4b}{3}=\tfrac{1-16/3}{3}=-\tfrac{13}{9}$. So $$x_p(t)=t+\tfrac{4}{3},\qquad y_p(t)=-\tfrac{4}{3}t-\tfrac{13}{9}.$$
General solution and initial conditions. $$x(t)=C_1e^{t}+3C_2e^{3t}+t+\tfrac43,\qquad y(t)=-C_1e^{t}-C_2e^{3t}-\tfrac43 t-\tfrac{13}{9}.$$ At $t=0$: $x(0)=C_1+3C_2+\tfrac43=2\Rightarrow C_1+3C_2=\tfrac23$; $y(0)=-C_1-C_2-\tfrac{13}{9}=-1\Rightarrow C_1+C_2=-\tfrac49$. Subtracting, $2C_2=\tfrac23+\tfrac49=\tfrac{10}{9}\Rightarrow C_2=\tfrac{5}{9}$, and $C_1=-\tfrac49-\tfrac59=-1$.

$$x(t)=\boxed{-e^{t}+\tfrac{5}{3}e^{3t}+t+\tfrac43},\qquad y(t)=\boxed{e^{t}-\tfrac{5}{9}e^{3t}-\tfrac43 t-\tfrac{13}{9}}$$

Quantity	Result
Eigenvalues of $A$	$\lambda_1=1,\ \lambda_2=3$
Eigenvectors of $A$	$\mathbf v_1=(1,-1)$ for $\lambda_1$; $\mathbf v_2=(3,-1)$ for $\lambda_2$
$x(t)$	$-e^{t}+\tfrac{5}{3}e^{3t}+t+\tfrac43$
$y(t)$	$e^{t}-\tfrac{5}{9}e^{3t}-\tfrac43 t-\tfrac{13}{9}$

Concept & Theory

Topic. Eigen-decomposition and its use in solving linear systems of ODEs ($\dot{\mathbf x}=A\mathbf x+\mathbf f(t)$).

Key relations. $\det(A-\lambda I)=0$ gives eigenvalues; $(A-\lambda I)v=0$ gives eigenvectors; the homogeneous solution of $\dot{\mathbf x}=A\mathbf x$ is $\sum_i C_i e^{\lambda_i t}\mathbf v_i$; for polynomial forcing, try a matching-degree polynomial particular solution (undetermined coefficients extended to systems).

Why this works. Because $A$ is diagonalizable with two real distinct eigenvalues, its eigenvectors form a basis in which the homogeneous system decouples into two independent scalar exponentials — this is exactly the same eigen-machinery used in (a). The particular solution follows the same undetermined-coefficients logic as scalar ODEs, just carried out on a vector of unknowns, because the forcing $(0,t)^T$ is polynomial and not resonant with either $e^{t}$ or $e^{3t}$.

Common pitfall. Sign errors when back-substituting between the two matched-coefficient equations (there are four unknowns $a,b,c,d$ solved from four scalar equations); also swapping which eigenvector belongs to which eigenvalue when assembling the homogeneous solution.

Source. Strang, Introduction to Linear Algebra, Ch. 6 (Eigenvalues and Eigenvectors); Kreyszig, Advanced Engineering Mathematics, Ch. 4 (Systems of ODEs, §4.4 nonhomogeneous systems).

Question 3 of 8: Eigenvalues/Eigenvectors and a Nonhomogeneous Linear System

Question 3: Eigenvalues/Eigenvectors and a Nonhomogeneous Linear System (a) 6, (b) 14 marks