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Section 3.4 Eigenvalue method

Note: 2 lectures, §5.2 in [EP], part of §7.3, §7.5, and §7.6 in [BD]

In this section we will learn how to solve linear homogeneous constant coefficient systems of ODEs by the eigenvalue method. Suppose we have such a system

\begin{equation*} {\vec{x}}' = P\vec{x} , \end{equation*}

where \(P\) is a constant square matrix. We wish to adapt the method for the single constant coefficient equation by trying the function \(e^{\lambda t}\text{.}\) However, \(\vec{x}\) is a vector. So we try \(\vec{x} = \vec{v} e^{\lambda t}\text{,}\) where \(\vec{v}\) is an arbitrary constant vector. We plug this \(\vec{x}\) into the equation to get

\begin{equation*} \underbrace{\lambda \vec{v} e^{\lambda t}}_{{\vec{x}}'} = \underbrace{P\vec{v} e^{\lambda t}}_{P\vec{x}} . \end{equation*}

We divide by \(e^{\lambda t}\) and notice that we are looking for a scalar \(\lambda\) and a vector \(\vec{v}\) that satisfy the equation

\begin{equation*} \lambda \vec{v} = P\vec{v} . \end{equation*}

To solve this equation we need a little bit more linear algebra, which we now review.

Subsection 3.4.1 Eigenvalues and eigenvectors of a matrix

Let \(A\) be a constant square matrix. Suppose there is a scalar \(\lambda\) and a nonzero vector \(\vec{v}\) such that

\begin{equation*} A \vec{v} = \lambda \vec{v}. \end{equation*}

We call \(\lambda\) an eigenvalue of \(A\) and we call \(\vec{v}\) a corresponding eigenvector.

Example 3.4.1.

The matrix \(\left[ \begin{smallmatrix} 2 & 1 \\ 0 & 1 \end{smallmatrix} \right]\) has an eigenvalue \(\lambda = 2\) with a corresponding eigenvector \(\left[ \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right]\) as

\begin{equation*} \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 0 \end{bmatrix} . \end{equation*}

Let us see how to compute eigenvalues for any matrix. Rewrite the equation for an eigenvalue as

\begin{equation*} (A - \lambda I)\vec{v} = \vec{0} . \end{equation*}

This equation has a nonzero solution \(\vec{v}\) only if \(A - \lambda I\) is not invertible. Were it invertible, we could write \({(A - \lambda I)}^{-1}(A - \lambda I)\vec{v} = {(A-\lambda I)}^{-1}\vec{0}\text{,}\) which implies \(\vec{v} = \vec{0}\text{.}\) Therefore, \(A\) has the eigenvalue \(\lambda\) if and only if \(\lambda\) solves the equation

\begin{equation*} \det (A-\lambda I) = 0 . \end{equation*}

Consequently, we will be able to find an eigenvalue of \(A\) without finding a corresponding eigenvector. An eigenvector will have to be found later, once \(\lambda\) is known.

Example 3.4.2.

Find all eigenvalues of \(\left[ \begin{smallmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{smallmatrix} \right]\text{.}\)

We write

\begin{multline*} \det \left( \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) = \det \left( \begin{bmatrix} 2-\lambda & 1 & 1 \\ 1 & 2-\lambda & 0 \\ 0 & 0 & 2-\lambda \end{bmatrix} \right) = \\ = (2-\lambda) \bigl({(2-\lambda)}^2 - 1\bigr) = -(\lambda -1)(\lambda -2)(\lambda-3) . \end{multline*}

So the eigenvalues are \(\lambda = 1\text{,}\) \(\lambda = 2\text{,}\) and \(\lambda = 3\text{.}\)

For an \(n \times n\) matrix, the polynomial we get by computing \(\det(A - \lambda I)\) is of degree \(n\text{,}\) and hence in general, we have \(n\) eigenvalues. Some may be repeated, some may be complex.

To find an eigenvector corresponding to an eigenvalue \(\lambda\text{,}\) we write

\begin{equation*} (A-\lambda I) \vec{v} = \vec{0} , \end{equation*}

and solve for a nontrivial (nonzero) vector \(\vec{v}\text{.}\) If \(\lambda\) is an eigenvalue, there will be at least one free variable, and so for each distinct eigenvalue \(\lambda\text{,}\) we can always find an eigenvector.

Example 3.4.3.

Find an eigenvector of \(\left[ \begin{smallmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{smallmatrix} \right]\) corresponding to the eigenvalue \(\lambda = 3\text{.}\)

We write

\begin{equation*} (A-\lambda I) \vec{v} = \left( \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \vec{0} . \end{equation*}

It is easy to solve this system of linear equations. We write down the augmented matrix

\begin{equation*} \left[ \begin{array}{ccc|c} -1 & 1 & 1 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \end{array} \right] , \end{equation*}

and perform row operations (exercise: which ones?) until we get:

\begin{equation*} \left[ \begin{array}{ccc|c} 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] . \end{equation*}

The entries of \(\vec{v}\) have to satisfy the equations \(v_1 - v_2 = 0\text{,}\) \(v_3 = 0\text{,}\) and \(v_2\) is a free variable. We can pick \(v_2\) to be arbitrary (but nonzero), let \(v_1 = v_2\text{,}\) and of course \(v_3 = 0\text{.}\) For example, if we pick \(v_2 = 1\text{,}\) then \(\vec{v} = \left[ \begin{smallmatrix} 1 \\ 1 \\ 0 \end{smallmatrix} \right]\text{.}\) Let us verify that \(\vec{v}\) really is an eigenvector corresponding to \(\lambda = 3\text{:}\)

\begin{equation*} \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 0 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} . \end{equation*}

Yay! It worked.

Exercise 3.4.1.

(easy)   Are eigenvectors unique? Can you find a different eigenvector for \(\lambda = 3\) in the example above? How are the two eigenvectors related?

Exercise 3.4.2.

When the matrix is \(2 \times 2\) you do not need to do row operations when computing an eigenvector, you can read it off from \(A-\lambda I\) (if you have computed the eigenvalues correctly). Can you see why? Explain. Try it for the matrix \(\left[ \begin{smallmatrix} 2 & 1 \\ 1 & 2 \end{smallmatrix} \right]\text{.}\)

Subsection 3.4.2 The eigenvalue method with distinct real eigenvalues

OK. We have the system of equations

\begin{equation*} {\vec{x}}' = P\vec{x} . \end{equation*}

We find the eigenvalues \(\lambda_1\text{,}\) \(\lambda_2\text{,}\) ..., \(\lambda_n\) of the matrix \(P\text{,}\) and corresponding eigenvectors \(\vec{v}_1\text{,}\) \(\vec{v}_2\text{,}\) ..., \(\vec{v}_n\text{.}\) Now we notice that the functions \(\vec{v}_1 e^{\lambda_1 t}\text{,}\) \(\vec{v}_2 e^{\lambda_2 t}\text{,}\) ..., \(\vec{v}_n e^{\lambda_n t}\) are solutions of the system of equations and hence \(\vec{x} = c_1 \vec{v}_1 e^{\lambda_1 t} + c_2 \vec{v}_2 e^{\lambda_2 t} + \cdots + c_n \vec{v}_n e^{\lambda_n t}\) is a solution.

The corresponding fundamental matrix solution is

\begin{equation*} X(t) = \bigl[\, \vec{v}_1 e^{\lambda_1 t} \quad \vec{v}_2 e^{\lambda_2 t} \quad \cdots \quad \vec{v}_n e^{\lambda_n t} \,\bigr]. \end{equation*}

That is, \(X(t)\) is the matrix whose \(j^{\text{th}}\) column is \(\vec{v}_j e^{\lambda_j t}\text{.}\)

Example 3.4.4.

Consider the system

\begin{equation*} {\vec{x}}' = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \vec{x} . \end{equation*}

Find the general solution.

Earlier, we found the eigenvalues are \(1,2,3\text{.}\) We found the eigenvector \(\left[ \begin{smallmatrix} 1 \\ 1 \\ 0 \end{smallmatrix} \right]\) for the eigenvalue 3. Similarly we find the eigenvector \(\left[ \begin{smallmatrix} 1 \\ -1 \\ 0 \end{smallmatrix} \right]\) for the eigenvalue 1, and \(\left[ \begin{smallmatrix} 0 \\ 1 \\ -1 \end{smallmatrix} \right]\) for the eigenvalue 2 (exercise: check). Hence our general solution is

\begin{equation*} \vec{x} = c_1 \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} e^t + c_2 \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} e^{2t} + c_3 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} e^{3t} = \begin{bmatrix} c_1 e^t+c_3 e^{3t} \\ -c_1 e^t + c_2 e^{2t} + c_3 e^{3t} \\ - c_2 e^{2t} \end{bmatrix} . \end{equation*}

In terms of a fundamental matrix solution,

\begin{equation*} \vec{x} = X(t)\, \vec{c} = \begin{bmatrix} e^t & 0 & e^{3t} \\ -e^t & e^{2t} & e^{3t} \\ 0 & -e^{2t} & 0 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} . \end{equation*}
Exercise 3.4.3.

Check that this \(\vec{x}\) really solves the system.

Note: If we write a single homogeneous linear constant coefficient \(n^{\text{th}}\) order equation as a first order system (as we did in Section 3.1), then the eigenvalue equation

\begin{equation*} \det(P - \lambda I) = 0 \end{equation*}

is essentially the same as the characteristic equation we got in Section 2.2 and Section 2.3.

Subsection 3.4.3 Complex eigenvalues

A matrix may very well have complex eigenvalues even if all the entries are real. Take, for example,

\begin{equation*} {\vec{x}}' = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \vec{x} . \end{equation*}

Let us compute the eigenvalues of the matrix \(P = \left[ \begin{smallmatrix} 1 & 1 \\ -1 & 1 \end{smallmatrix} \right]\text{.}\)

\begin{equation*} \det(P - \lambda I) = \det\left( \begin{bmatrix} 1-\lambda & 1 \\ -1 & 1-\lambda \end{bmatrix} \right) = {(1-\lambda)}^2 + 1 = \lambda^2 - 2 \lambda + 2 = 0 . \end{equation*}

Thus \(\lambda = 1 \pm i\text{.}\) Corresponding eigenvectors are also complex. Start with \(\lambda = 1-i\text{.}\)

\begin{equation*} \begin{aligned} \bigl(P-(1-i) I\bigr) \vec{v} & = \vec{0} , \\ \begin{bmatrix} i & 1 \\ -1 & i \end{bmatrix} \vec{v} = \vec{0}. \end{aligned} \end{equation*}

The equations \(i v_1 + v_2 = 0\) and \(-v_1 + iv_2 = 0\) are multiples of each other. So we only need to consider one of them. After picking \(v_2 = 1\text{,}\) for example, we have an eigenvector \(\vec{v} = \left[ \begin{smallmatrix} i \\ 1 \end{smallmatrix} \right]\text{.}\) In similar fashion we find that \(\left[ \begin{smallmatrix} -i \\ 1 \end{smallmatrix} \right]\) is an eigenvector corresponding to the eigenvalue \(1+i\text{.}\)

We could write the solution as

\begin{equation*} \vec{x} = c_1 \begin{bmatrix} i \\ 1 \end{bmatrix} e^{(1-i)t} + c_2 \begin{bmatrix} -i \\ 1 \end{bmatrix} e^{(1+i)t} = \begin{bmatrix} c_1 i e^{(1-i)t} - c_2 i e^{(1+i)t} \\ c_1 e^{(1-i)t} + c_2 e^{(1+i)t} \end{bmatrix} . \end{equation*}

We would then need to look for complex values \(c_1\) and \(c_2\) to solve any initial conditions. It is perhaps not completely clear that we get a real solution. After solving for \(c_1\) and \(c_2\text{,}\) we could use Euler's formula and do the whole song and dance we did before, but we will not. We will apply the formula in a smarter way first to find independent real solutions.

We claim that we did not have to look for a second eigenvector (nor for the second eigenvalue). All complex eigenvalues come in pairs (because the matrix \(P\) is real).

First a small detour. The real part of a complex number \(z\) can be computed as \(\frac{z + \bar{z}}{2}\text{,}\) where the bar above \(z\) means \(\overline{a+ib} = a -ib\text{.}\) This operation is called the complex conjugate. If \(a\) is a real number, then \(\bar{a} = a\text{.}\) Similarly we bar whole vectors or matrices by taking the complex conjugate of every entry. Suppose a matrix \(P\) is real. Then \(\overline{P} = P\text{,}\) and so \(\overline{P\vec{x}} = \overline{P} \, \overline{\vec{x}} = P \overline{\vec{x}}\text{.}\) Also the complex conjugate of 0 is still 0, therefore,

\begin{equation*} \vec{0} = \overline{\vec{0}} = \overline{(P-\lambda I)\vec{v}} = (P-\bar{\lambda} I)\overline{\vec{v}} . \end{equation*}

In other words, if \(\lambda = a+ib\) is an eigenvalue, then so is \(\bar{\lambda} = a-ib\text{.}\) And if \(\vec{v}\) is an eigenvector corresponding to the eigenvalue \(\lambda\text{,}\) then \(\overline{\vec{v}}\) is an eigenvector corresponding to the eigenvalue \(\bar{\lambda}\text{.}\)

Suppose \(a + ib\) is a complex eigenvalue of \(P\text{,}\) and \(\vec{v}\) is a corresponding eigenvector. Then

\begin{equation*} \vec{x}_1 = \vec{v} e^{(a+ib)t} \end{equation*}

is a solution (complex-valued) of \({\vec{x}}' = P \vec{x}\text{.}\) Euler's formula shows that \(\overline{e^{a+ib}} = e^{a-ib}\text{,}\) and so

\begin{equation*} \vec{x}_2 = \overline{\vec{x}_1} = \overline{\vec{v}} e^{(a-ib)t} \end{equation*}

is also a solution. As \(\vec{x}_1\) and \(\vec{x}_2\) are solutions, the function

\begin{equation*} \vec{x}_3 = \operatorname{Re} \vec{x}_1 = \operatorname{Re} \vec{v} e^{(a+ib)t} = \frac{\vec{x}_1 + \overline{\vec{x}_1}}{2} = \frac{\vec{x}_1 + \vec{x}_2}{2} = \frac{1}{2} \vec{x}_1 + \frac{1}{2}\vec{x}_2 \end{equation*}

is also a solution. And \(\vec{x}_3\) is real-valued! Similarly as \(\operatorname{Im} z = \frac{z-\bar{z}}{2i}\) is the imaginary part, we find that

\begin{equation*} \vec{x}_4 = \operatorname{Im} \vec{x}_1 = \frac{\vec{x}_1 - \overline{\vec{x}_1}}{2i} = \frac{\vec{x}_1 - \vec{x}_2}{2i} . \end{equation*}

is also a real-valued solution. It turns out that \(\vec{x}_3\) and \(\vec{x}_4\) are linearly independent. We will use Euler's formula to separate out the real and imaginary part.

Returning to our problem,

\begin{equation*} \vec{x}_1 = \begin{bmatrix} i \\ 1 \end{bmatrix} e^{(1-i)t} = \begin{bmatrix} i \\ 1 \end{bmatrix} \left( e^t \cos t - i e^t \sin t \right) = \begin{bmatrix} i e^t \cos t + e^t \sin t \\ e^t \cos t - i e^t \sin t \end{bmatrix} = \begin{bmatrix} e^t \sin t \\ e^t \cos t \end{bmatrix} + i \begin{bmatrix} e^t \cos t \\ - e^t \sin t \end{bmatrix} . \end{equation*}

Then

\begin{equation*} \operatorname{Re} \vec{x}_1 = \begin{bmatrix} e^t \sin t \\ e^t \cos t \end{bmatrix} , \qquad \text{and} \qquad \operatorname{Im} \vec{x}_1 = \begin{bmatrix} e^t \cos t \\ - e^t \sin t \end{bmatrix} , \end{equation*}

are the two real-valued linearly independent solutions we seek.

Exercise 3.4.4.

Check that these really are solutions.

The general solution is

\begin{equation*} \vec{x} = c_1 \begin{bmatrix} e^t \sin t \\ e^t \cos t \end{bmatrix} + c_2 \begin{bmatrix} e^t \cos t \\ -e^t \sin t \end{bmatrix} = \begin{bmatrix} c_1 e^t \sin t + c_2 e^t \cos t \\ c_1 e^t \cos t - c_2 e^t \sin t \end{bmatrix} . \end{equation*}

This solution is real-valued for real \(c_1\) and \(c_2\text{.}\) We now solve for any initial conditions we may have.

Let us summarize as a theorem.

For each pair of complex eigenvalues \(a+ib\) and \(a-ib\text{,}\) we get two real-valued linearly independent solutions. We then go on to the next eigenvalue, which is either a real eigenvalue or another complex eigenvalue pair. If we have \(n\) distinct eigenvalues (real or complex), then we end up with \(n\) linearly independent solutions. If we had only two equations (\(n=2\)) as in the example above, then once we found two solutions we are finished, and our general solution is

\begin{equation*} \vec{x} = c_1 \vec{x}_1 + c_2 \vec{x}_2 = c_1 \bigl( \operatorname{Re} \vec{v} e^{(a+ib)t} \bigr) + c_2 \bigl( \operatorname{Im} \vec{v} e^{(a+ib)t} \bigr) . \end{equation*}

We can now find a real-valued general solution to any homogeneous system where the matrix has distinct eigenvalues. When we have repeated eigenvalues, matters get a bit more complicated and we will look at that situation in Section 3.7.

Subsection 3.4.4 Exercises

Exercise 3.4.5.

(easy)   Let \(A\) be a \(3 \times 3\) matrix with an eigenvalue of 3 and a corresponding eigenvector \(\vec{v} = \left[ \begin{smallmatrix} 1 \\ -1 \\ 3 \end{smallmatrix} \right]\text{.}\) Find \(A \vec{v}\text{.}\)

Exercise 3.4.6.

  1. Find the general solution of \(x_1' = 2 x_1\text{,}\) \(x_2' = 3 x_2\) using the eigenvalue method (first write the system in the form \({\vec{x}}' = A \vec{x}\)).

  2. Solve the system by solving each equation separately and verify you get the same general solution.

Exercise 3.4.7.

Find the general solution of \(x_1' = 3 x_1 + x_2\text{,}\) \(x_2' = 2 x_1 + 4 x_2\) using the eigenvalue method.

Exercise 3.4.8.

Find the general solution of \(x_1' = x_1 -2 x_2\text{,}\) \(x_2' = 2 x_1 + x_2\) using the eigenvalue method. Do not use complex exponentials in your solution.

Exercise 3.4.9.

  1. Compute eigenvalues and eigenvectors of \(A = \left[ \begin{smallmatrix} 9 & -2 & -6 \\ -8 & 3 & 6 \\ 10 & -2 & -6 \end{smallmatrix} \right]\text{.}\)

  2. Find the general solution of \({\vec{x}}' = A \vec{x}\text{.}\)

Exercise 3.4.10.

Compute eigenvalues and eigenvectors of \(\left[ \begin{smallmatrix} -2 & -1 & -1 \\ 3 & 2 & 1 \\ -3 & -1 & 0 \\ \end{smallmatrix} \right]\text{.}\)

Exercise 3.4.11.

Let \(a,b,c,d,e,f\) be numbers. Find the eigenvalues of \(\left[ \begin{smallmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \\ \end{smallmatrix} \right]\text{.}\)

Exercise 3.4.101.

  1. Compute eigenvalues and eigenvectors of \(A= \left[ \begin{smallmatrix} 1 & 0 & 3 \\ -1 & 0 & 1 \\ 2 & 0 & 2 \end{smallmatrix}\right]\text{.}\)

  2. Solve the system \(\vec{x}\,' = A \vec{x}\text{.}\)

Answer

a) Eigenvalues: \(4,0,-1\)     Eigenvectors: \(\left[ \begin{smallmatrix} 1 \\ 0 \\ 1 \end{smallmatrix}\right]\text{,}\) \(\left[ \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix}\right]\text{,}\) \(\left[ \begin{smallmatrix} 3 \\ 5 \\ -2 \end{smallmatrix}\right]\)
b) \(\vec{x} = C_1 \left[ \begin{smallmatrix} 1 \\ 0 \\ 1 \end{smallmatrix}\right] e^{4t} + C_2 \left[ \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix}\right] + C_3 \left[ \begin{smallmatrix} 3 \\ 5 \\ -2 \end{smallmatrix}\right] e^{-t}\)

Exercise 3.4.102.

  1. Compute eigenvalues and eigenvectors of \(A=\left[ \begin{smallmatrix} 1 & 1 \\ -1 & 0 \end{smallmatrix}\right]\text{.}\)

  2. Solve the system \(\vec{x}\,' = A\vec{x}\text{.}\)

Answer

a) Eigenvalues: \(\frac{1+\sqrt{3}i}{2}, \frac{1-\sqrt{3}i}{2}\text{,}\)     Eigenvectors: \(\left[ \begin{smallmatrix} -2 \\ 1-\sqrt{3}i \end{smallmatrix}\right]\text{,}\) \(\left[ \begin{smallmatrix} -2 \\ 1+\sqrt{3}i \end{smallmatrix}\right]\)
b) \(\vec{x} = C_1 e^{t/2} \left[ \begin{smallmatrix} -2\cos\bigl(\frac{\sqrt{3}t}{2}\bigr) \\ \cos\bigl(\frac{\sqrt{3}t}{2}\bigr) + \sqrt{3}\sin\bigl(\frac{\sqrt{3}t}{2}\bigr) \end{smallmatrix}\right] + C_2 e^{t/2} \left[ \begin{smallmatrix} - 2\sin\bigl(\frac{\sqrt{3}t}{2}\bigr) \\ \sin\bigl(\frac{\sqrt{3}t}{2}\bigr) -\sqrt{3}\cos\bigl(\frac{\sqrt{3}t}{2}\bigr) \end{smallmatrix}\right]\)

Exercise 3.4.103.

Solve \(x_1' = x_2\text{,}\) \(x_2' = x_1\) using the eigenvalue method.

Answer

\(\vec{x} = C_1 \left[ \begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\right] e^{t} + C_2 \left[ \begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right] e^{-t}\)

Exercise 3.4.104.

Solve \(x_1' = x_2\text{,}\) \(x_2' = -x_1\) using the eigenvalue method.

Answer

\(\vec{x} = C_1 \left[ \begin{smallmatrix} \cos(t) \\ -\sin(t) \end{smallmatrix}\right] + C_2 \left[ \begin{smallmatrix} \sin(t) \\ \cos(t) \end{smallmatrix}\right]\)

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