Eigenvalues/Eigenvectors and a Linear System Initial Value Problem: 04-BS-1, December 2017, Q3

Question 3: Eigenvalues/Eigenvectors and a Linear System Initial Value Problem (a) 8, (b) 12 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. (a) Matrix $A=\begin{pmatrix}4&2\\3&-1\end{pmatrix}$. (b) The linear system $\mathbf x'=A\mathbf x$ with the same $A$, ICs $x(0)=0,\ y(0)=7$.

Find. (a) The eigenvalues and eigenvectors of $A$. (b) The unique solution $x(t),y(t)$, built directly from part (a)'s eigenpairs.

Approach. (a) Solve $\det(A-\lambda I)=0$ for the eigenvalues, then find each eigenvector from $(A-\lambda I)\mathbf v=0$. (b) Since $A$ has two real, distinct eigenvalues, the general solution of $\mathbf x'=A\mathbf x$ is $\mathbf x(t)=a\mathbf v_1e^{\lambda_1t}+b\mathbf v_2e^{\lambda_2t}$ using exactly the eigenpairs from (a); fix $a,b$ from the initial conditions.

(a) Eigenvalues. $$\det(A-\lambda I)=(4-\lambda)(-1-\lambda)-6=\lambda^2-3\lambda-10=(\lambda-5)(\lambda+2)=0\;\Rightarrow\;\lambda_1=5,\ \lambda_2=-2.$$
(a) Eigenvectors. For $\lambda_1=5$: $(A-5I)\mathbf v=0\Rightarrow\begin{pmatrix}-1&2\\3&-6\end{pmatrix}\mathbf v=0\Rightarrow v_1=2v_2$, so $\mathbf v_1=(2,1)$. For $\lambda_2=-2$: $(A+2I)\mathbf v=0\Rightarrow\begin{pmatrix}6&2\\3&1\end{pmatrix}\mathbf v=0\Rightarrow v_2=-3v_1$, so $\mathbf v_2=(1,-3)$.
(b) General solution from the eigenpairs. Reusing $\lambda_1=5,\mathbf v_1=(2,1)$ and $\lambda_2=-2,\mathbf v_2=(1,-3)$ from part (a) directly (no new eigenanalysis needed), $$\begin{pmatrix}x(t)\\y(t)\end{pmatrix}=a\begin{pmatrix}2\\1\end{pmatrix}e^{5t}+b\begin{pmatrix}1\\-3\end{pmatrix}e^{-2t}.$$
(b) Apply the initial conditions. At $t=0$: $2a+b=0$ and $a-3b=7$. From the first, $b=-2a$; substituting, $a-3(-2a)=7\Rightarrow7a=7\Rightarrow a=1,\ b=-2$.

$$x(t)=\boxed{2e^{5t}-2e^{-2t}},\qquad y(t)=\boxed{e^{5t}+6e^{-2t}}$$

Quantity	Result
Eigenvalues	$\lambda_1=5,\ \lambda_2=-2$
Eigenvectors	$\mathbf v_1=(2,1),\ \mathbf v_2=(1,-3)$
$x(t)$	$2e^{5t}-2e^{-2t}$
$y(t)$	$e^{5t}+6e^{-2t}$

Concept & Theory

Topic. Eigenvalue/eigenvector method for a linear homogeneous system of ODEs.

Key relations. For $\mathbf x'=A\mathbf x$ with $A$ having distinct real eigenvalues $\lambda_i$ and eigenvectors $\mathbf v_i$, the general solution is $\mathbf x(t)=\sum_ic_i\mathbf v_ie^{\lambda_it}$; the constants $c_i$ are fixed by the initial vector $\mathbf x(0)$.

Why this works. Each $\mathbf v_ie^{\lambda_it}$ satisfies the system individually because $A\mathbf v_i=\lambda_i\mathbf v_i$ makes both sides of $\mathbf x'=A\mathbf x$ reduce to $\lambda_i\mathbf v_ie^{\lambda_it}$; distinct eigenvalues guarantee the eigenvectors are linearly independent, so their combination spans all solutions, and part (b) is simply this formula reusing part (a)'s already-computed eigenpairs.

Common pitfall. Re-deriving the eigenvalues from scratch in part (b) instead of reusing part (a); mixing up which eigenvector pairs with which eigenvalue when assembling the general solution.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 4 (§4.3–4.4 Systems of ODEs, Eigenvalue Method).

Question 3 of 8: Eigenvalues/Eigenvectors and a Linear System Initial Value Problem

Question 3: Eigenvalues/Eigenvectors and a Linear System Initial Value Problem (a) 8, (b) 12 marks