MATH 2410 — Differential Equations

Lecture Notes. This is NOT meant to be a substitute for coming to lecture or consulting the textbook directly, but rather a helpful resource to review for exams/ review what was discussed in lecture.

Appendix B.1

Introduction to Matrices

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Matrices and Basic Operations

Definition A matrix is any rectangular array of numbers (or functions), \[ A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}. \] The size of \(A\) is \(m\times n\) (rows \(\times\) columns). If \(m=n\), \(A\) is called a square matrix of order \(n\).

Definition Two \(m\times n\) matrices \(A=(a_{ij})\) and \(B=(b_{ij})\) are equal if \(a_{ij}=b_{ij}\) for every \(i,j\).

Definition A column matrix (or vector) is any matrix with \(n\) rows and exactly one column.

Definition The zero matrix \(0\) is the matrix (of any size) whose entries are all \(0\). The \(n\times n\) identity matrix \(I_n\) has \(1\)'s down the main diagonal and \(0\)'s elsewhere: \[ I_n = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix}. \] It satisfies \(IA = AI = A\) for every square matrix \(A\) of the same size.

Definition The transpose of an \(m\times n\) matrix \(A=(a_{ij})\), denoted \(A^T\), is the \(n\times m\) matrix obtained by turning the rows of \(A\) into columns: \(A^T = (a_{ji})\). A square matrix with \(A=A^T\) is called symmetric. A square matrix whose only nonzero entries lie on the main diagonal is called a diagonal matrix.

Scalar Multiplication and Addition

For a constant \(k\) and a matrix \(A=(a_{ij})_{m\times n}\), \[ kA = (k\cdot a_{ij})_{m\times n}. \]

Example \[ 5\begin{pmatrix}1&2\\3&4\\0&5\end{pmatrix} = \begin{pmatrix}5&10\\15&20\\0&25\end{pmatrix}, \qquad -2\begin{pmatrix}1&3\\2&7\end{pmatrix} = \begin{pmatrix}-2&-6\\-4&-14\end{pmatrix}, \qquad e^t\begin{pmatrix}1\\3\\5\end{pmatrix} = \begin{pmatrix}e^t\\3e^t\\5e^t\end{pmatrix}. \]

Remark Note that \(kA=Ak\) for every constant \(k\).

For two matrices of the same size, \(A=(a_{ij})_{m\times n}\) and \(B=(b_{ij})_{m\times n}\), \[ A+B = (a_{ij}+b_{ij})_{m\times n}, \qquad A-B = A+(-1)B. \]

Example \[ \begin{pmatrix}0&1\\2&3\end{pmatrix} +\begin{pmatrix}5&4\\-2&1\end{pmatrix} = \begin{pmatrix}5&5\\0&4\end{pmatrix}, \qquad \begin{pmatrix}t^2&t\\0&2\end{pmatrix} +\begin{pmatrix}t^3&2t\\4t&3\end{pmatrix} = \begin{pmatrix}t^3+t^2 & t+2t\\4t&5\end{pmatrix}. \]

Matrix Multiplication

Definition If \(A=(a_{ij})\) is \(m\times n\) and \(B=(b_{ij})\) is \(n\times p\) (the number of columns of \(A\) must equal the number of rows of \(B\)), then \(AB\) is the \(m\times p\) matrix whose \((i,k)\) entry is \[ \sum_{j=1}^n a_{ij}b_{jk}, \] i.e. the \(i\)th row of \(A\) "dotted" with the \(k\)th column of \(B\).

Remark If \(A\) is \(m\times n\) and \(B\) is \(n\times p\), then \(AB\) is \(m\times p\). Unlike multiplication of numbers, matrix multiplication is not commutative in general: \(AB\ne BA\), and often one of the two products isn't even defined.

Example \[ \begin{pmatrix}3&0\\-1&2\end{pmatrix} \begin{pmatrix}1&0\\2&-1\end{pmatrix} = \begin{pmatrix}3(1)+0(2) & 3(0)+0(-1)\\-1(1)+2(2) & -1(0)+2(-1)\end{pmatrix} = \begin{pmatrix}3&0\\3&-2\end{pmatrix}. \]

Example \[ \begin{pmatrix}1&2&3\\3&1&0\\1&1&2\end{pmatrix} \begin{pmatrix}-1&0&3\\0&1&0\\2&1&1\end{pmatrix} = \begin{pmatrix}5&5&3\\-3&1&9\\3&3&5\end{pmatrix}. \]

Example Since a \(3\times2\) matrix times a \(2\times3\) matrix produces a \(3\times3\) matrix, \[ \begin{pmatrix}1&4\\5&10\\8&12\end{pmatrix} \begin{pmatrix}-4&6&-3\\1&3&2\end{pmatrix} = \begin{pmatrix}0&18&5\\-10&60&5\\-20&84&0\end{pmatrix}. \] Multiplying the same two matrices in the opposite order gives a \(2\times2\) result instead — a different size entirely: \[ \begin{pmatrix}-4&6&-3\\1&3&2\end{pmatrix} \begin{pmatrix}1&4\\5&10\\8&12\end{pmatrix} = \begin{pmatrix}2&8\\32&58\end{pmatrix}. \] This illustrates why matrix multiplication is not commutative: here the two orders don't even produce matrices of the same size.

Example Even when both orders are defined and produce the same size, \(AB\) and \(BA\) need not be equal. Let \(A=\begin{pmatrix}1&2\\3&4\end{pmatrix}\) and \(B=\begin{pmatrix}0&1\\1&0\end{pmatrix}\). Then \[ AB = \begin{pmatrix}2&1\\4&3\end{pmatrix}, \qquad BA = \begin{pmatrix}3&4\\1&2\end{pmatrix}, \] so \(AB\ne BA\).

Determinants

Definition The determinant of a \(2\times2\) matrix is \[ \det\begin{pmatrix}a&b\\c&d\end{pmatrix} = \begin{vmatrix}a&b\\c&d\end{vmatrix} = ad-bc. \] The determinant of a larger square matrix can be computed by cofactor expansion along any row or column; expanding along the first row of a \(3\times3\) matrix, \[ \begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{vmatrix} = a_{11}\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix} - a_{12}\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix} + a_{13}\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}. \]

Example \[ \begin{vmatrix}1&2&3\\0&1&2\\1&3&3\end{vmatrix} = 1\begin{vmatrix}1&2\\3&3\end{vmatrix} - 2\begin{vmatrix}0&2\\1&3\end{vmatrix} + 3\begin{vmatrix}0&1\\1&3\end{vmatrix}. \] Each \(2\times2\) minor is \[ \begin{vmatrix}1&2\\3&3\end{vmatrix} = 3-6=-3, \qquad \begin{vmatrix}0&2\\1&3\end{vmatrix} = 0-2=-2, \qquad \begin{vmatrix}0&1\\1&3\end{vmatrix} = 0-1=-1. \] So \[ \boxed{\begin{vmatrix}1&2&3\\0&1&2\\1&3&3\end{vmatrix} = 1(-3)-2(-2)+3(-1) = -3+4-3 = -2.} \]

Appendix B.2

Gaussian Elimination

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Solving Systems of Equations by Gaussian Elimination

To solve a system of linear equations, we represent it as an augmented matrix (the coefficients of the system, together with a final column of constants), and simplify it using three types of row operations:

Row Operations

Multiply a row by a nonzero constant.
Interchange any two rows.
Add a constant multiple of one row to another row.

The goal of Gaussian elimination is to use these operations to reduce the augmented matrix to a form where each row has a leading \(1\), and every other entry in that leading \(1\)'s column is \(0\) — at which point the solution can be read directly from the matrix.

Example Solve the system \[ \begin{aligned} 2x+6y+z &= 7\\ x+2y-z &= -1\\ 5x+7y-4z &= 9. \end{aligned} \]

Step 1 (Form the augmented matrix).

\[ \left(\begin{array}{ccc|c} 2&6&1&7\\1&2&-1&-1\\5&7&-4&9 \end{array}\right). \]

Step 2 (Swap \(R_1\) and \(R_2\) so the top-left entry is \(1\)).

\[ \left(\begin{array}{ccc|c} 1&2&-1&-1\\2&6&1&7\\5&7&-4&9 \end{array}\right). \]

Step 3 (Eliminate the rest of column 1 using \(R_2-2R_1\), \(R_3-5R_1\), then scale \(R_2\) by \(\tfrac12\)).

\[ \left(\begin{array}{ccc|c} 1&2&-1&-1\\0&1&3/2&9/2\\0&-3&1&14 \end{array}\right). \]

Step 4 (Eliminate the rest of column 2 using \(R_1-2R_2\), \(R_3+3R_2\), then scale \(R_3\) by \(\tfrac{2}{11}\)).

\[ \left(\begin{array}{ccc|c} 1&0&-4&-10\\0&1&3/2&9/2\\0&0&1&5 \end{array}\right). \]

Step 5 (Read off the system).

\(x-4z=-10\), \(y+\tfrac32z=\tfrac92\), \(z=5\). Back-substituting, \[ y = \frac{9-3(5)}{2} = -3, \qquad x = -10+4(5) = 10. \] This system has exactly one solution: \[ \boxed{x=10,\quad y=-3,\quad z=5.} \]

Remark Not every system has a unique solution. If row reduction produces a row of all zeros (including in the constant column), the system has infinitely many solutions, with one or more variables free to take any value. If instead a row reduces to all zeros except the constant column (e.g. \(0=1\)), the system has no solution.

Example Solve the system \[ \begin{aligned} y+z &= -5\\ 5x+4y-16z &= -10\\ x-y-5z &= 7. \end{aligned} \] Forming the augmented matrix and swapping \(R_1\) and \(R_3\) so the top-left entry is \(1\): \[ \left(\begin{array}{ccc|c} 1&-1&-5&7\\5&4&-16&-10\\0&1&1&-5 \end{array}\right). \] Eliminate column 1 using \(R_2-5R_1\), then scale by \(\tfrac19\): \[ \left(\begin{array}{ccc|c} 1&-1&-5&7\\0&1&1&-5\\0&1&1&-5 \end{array}\right). \] Now \(R_3-R_2\) gives a row of all zeros: \[ \left(\begin{array}{ccc|c} 1&-1&-5&7\\0&1&1&-5\\0&0&0&0 \end{array}\right). \] The system reduces to just two equations in three unknowns: \[ x-y-5z=7, \qquad y+z=-5, \] so \(z\) is a free variable. Let \(z=t\). Then \(y=-5-t\), and \[ x = 7+y+5z = 7+(-5-t)+5t = 2+4t. \] This system has infinitely many solutions: \[ \boxed{x=2+4t,\quad y=-5-t,\quad z=t, \qquad t\in\mathbb{R}.} \]

Example Solve the system \[ \begin{aligned} 3x+y+z &= 4\\ 4x+2y-z &= 7\\ x+y-3z &= 6. \end{aligned} \] Forming the augmented matrix and swapping \(R_1\) and \(R_3\): \[ \left(\begin{array}{ccc|c} 1&1&-3&6\\4&2&-1&7\\3&1&1&4 \end{array}\right). \] Eliminate column 1 with \(R_2-4R_1\), \(R_3-3R_1\): \[ \left(\begin{array}{ccc|c} 1&1&-3&6\\0&-2&11&-17\\0&-2&10&-14 \end{array}\right). \] Scale \(R_2\) by \(-\tfrac12\), then eliminate column 2 with \(R_1-R_2\), \(R_3+2R_2\): \[ \left(\begin{array}{ccc|c} 1&0&5/2&-5/2\\0&1&-11/2&17/2\\0&0&-1&3 \end{array}\right). \] Scale \(R_3\) by \(-1\), then eliminate column 3 with \(R_1-\tfrac52R_3\), \(R_2+\tfrac{11}{2}R_3\): \[ \left(\begin{array}{ccc|c} 1&0&0&5\\0&1&0&-8\\0&0&1&-3 \end{array}\right). \] This system has exactly one solution: \[ \boxed{x=5,\quad y=-8,\quad z=-3.} \]

Appendix B.3

Eigenvalues and Eigenvectors

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Eigenvalues and Eigenvectors

Definition Let \(A\) be an \(n\times n\) square matrix. A number \(\lambda\) is called an eigenvalue of \(A\) if there exists a nonzero vector \(K\) such that \[ AK = \lambda K. \] The vector \(K\) is called an eigenvector of \(A\) corresponding to \(\lambda\).

Example Verify that \(K=\begin{pmatrix}1\\0\\2\end{pmatrix}\) is an eigenvector of \(A=\begin{pmatrix}3&0&1\\0&1&0\\6&7&2\end{pmatrix}\).

Computing \(AK\),

\[ AK = \begin{pmatrix}3&0&1\\0&1&0\\6&7&2\end{pmatrix} \begin{pmatrix}1\\0\\2\end{pmatrix} = \begin{pmatrix}3(1)+0(0)+1(2)\\0(1)+1(0)+0(2)\\6(1)+7(0)+2(2)\end{pmatrix} = \begin{pmatrix}5\\0\\10\end{pmatrix} = 5\begin{pmatrix}1\\0\\2\end{pmatrix} = 5K. \] So \(K\) is indeed an eigenvector of \(A\), with corresponding eigenvalue \(\boxed{\lambda=5}\).

Finding Eigenvalues: The Characteristic Equation

How do we find eigenvalues without being handed a candidate eigenvector? Starting from the definition, \[ AK = \lambda K \;\Longleftrightarrow\; AK - \lambda K = 0 \;\Longleftrightarrow\; (A-\lambda I)K = 0. \] This is a linear system in \(K\) with coefficient matrix \(A-\lambda I\). Since we want a nonzero solution \(K\), this system must have more than the trivial solution — which happens exactly when \(\det(A-\lambda I)=0\).

Definition The equation \(\det(A-\lambda I)=0\) is called the characteristic equation of \(A\). Its roots are exactly the eigenvalues of \(A\).

Finding Eigenvalues and Eigenvectors

Form \(A-\lambda I\).
Solve the characteristic equation \(\det(A-\lambda I)=0\) for \(\lambda\); these are the eigenvalues.
For each eigenvalue \(\lambda\), solve \((A-\lambda I)K=0\) (via Gaussian elimination) to find the corresponding eigenvector(s) \(K\).

Example Find all eigenvalues and eigenvectors of \(A=\begin{pmatrix}-1&2\\-7&8\end{pmatrix}\).

Eigenvalues.

\[ A-\lambda I = \begin{pmatrix}-1-\lambda&2\\-7&8-\lambda\end{pmatrix}. \] \[ \det(A-\lambda I) = (-1-\lambda)(8-\lambda)-(2)(-7) = (\lambda^2-7\lambda-8)+14 = \lambda^2-7\lambda+6 = (\lambda-1)(\lambda-6). \] So the eigenvalues are \(\lambda_1=1\) and \(\lambda_2=6\).

Eigenvector for \(\lambda_1=1\). Solve \((A-I)K=0\):

\[ \begin{pmatrix}-2&2\\-7&7\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix} \;\Longrightarrow\; -2x+2y=0 \;\Longrightarrow\; x=y. \] Choosing \(x=1\) gives \(y=1\), so \(K_1=\begin{pmatrix}1\\1\end{pmatrix}\). (Check: \(AK_1 = \begin{pmatrix}-1+2\\-7+8\end{pmatrix} =\begin{pmatrix}1\\1\end{pmatrix}=1\cdot K_1. \checkmark\))

Eigenvector for \(\lambda_2=6\). Solve \((A-6I)K=0\):

\[ \begin{pmatrix}-7&2\\-7&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix} \;\Longrightarrow\; -7x+2y=0 \;\Longrightarrow\; x=\frac27y. \] Choosing \(y=7\) gives \(x=2\), so \(K_2=\begin{pmatrix}2\\7\end{pmatrix}\). (Check: \(AK_2 = \begin{pmatrix}-2+14\\-14+56\end{pmatrix} =\begin{pmatrix}12\\42\end{pmatrix}=6\begin{pmatrix}2\\7\end{pmatrix} =6K_2. \checkmark\)) \[ \boxed{\lambda_1=1,\ K_1=\begin{pmatrix}1\\1\end{pmatrix}; \qquad \lambda_2=6,\ K_2=\begin{pmatrix}2\\7\end{pmatrix}.} \]

Example Find all eigenvalues and eigenvectors of \(A=\begin{pmatrix}3&0&0\\0&2&0\\4&0&1\end{pmatrix}\).

Eigenvalues.

\[ A-\lambda I = \begin{pmatrix}3-\lambda&0&0\\0&2-\lambda&0\\4&0&1-\lambda\end{pmatrix}. \] Expanding along the first row, \[ \det(A-\lambda I) = (3-\lambda)\begin{vmatrix}2-\lambda&0\\0&1-\lambda\end{vmatrix} = (3-\lambda)(2-\lambda)(1-\lambda). \] So the eigenvalues are \(\lambda_1=1\), \(\lambda_2=2\), \(\lambda_3=3\).

Eigenvector for \(\lambda_1=1\). Solve \((A-I)K=0\):

\[ \begin{pmatrix}2&0&0\\0&1&0\\4&0&0\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = 0 \;\Longrightarrow\; x=0,\ y=0,\ z \text{ free}. \] Choosing \(z=1\): \(K_1=\begin{pmatrix}0\\0\\1\end{pmatrix}\).

Eigenvector for \(\lambda_2=2\). Solve \((A-2I)K=0\):

\[ \begin{pmatrix}1&0&0\\0&0&0\\4&0&-1\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = 0 \;\Longrightarrow\; x=0,\ z=4x=0,\ y \text{ free}. \] Choosing \(y=1\): \(K_2=\begin{pmatrix}0\\1\\0\end{pmatrix}\).

Eigenvector for \(\lambda_3=3\). Solve \((A-3I)K=0\):

\[ \begin{pmatrix}0&0&0\\0&-1&0\\4&0&-2\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = 0 \;\Longrightarrow\; y=0,\ z=2x,\ x \text{ free}. \] Choosing \(x=1\): \(K_3=\begin{pmatrix}1\\0\\2\end{pmatrix}\). \[ \boxed{\lambda_1=1,\ K_1=\begin{pmatrix}0\\0\\1\end{pmatrix}; \quad \lambda_2=2,\ K_2=\begin{pmatrix}0\\1\\0\end{pmatrix}; \quad \lambda_3=3,\ K_3=\begin{pmatrix}1\\0\\2\end{pmatrix}.} \]

Section 8.1

Systems of Linear Differential Equations

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Systems of Linear Differential Equations in Matrix Form

Our goal is to restate systems of linear differential equations in matrix form. A general (linear) first-order system of \(n\) equations has the form \[ \begin{aligned} \frac{dx_1}{dt} &= a_{11}(t)x_1+a_{12}(t)x_2+\cdots+a_{1n}(t)x_n+f_1(t)\\ \frac{dx_2}{dt} &= a_{21}(t)x_1+a_{22}(t)x_2+\cdots+a_{2n}(t)x_n+f_2(t)\\ &\ \,\vdots\\ \frac{dx_n}{dt} &= a_{n1}(t)x_1+a_{n2}(t)x_2+\cdots+a_{nn}(t)x_n+f_n(t). \end{aligned} \] Setting \(X=(x_1,\dots,x_n)^T\), so that \(X'=(x_1',\dots,x_n')^T\), this system can be written compactly as \[ \boxed{X' = AX+F,} \] where \(A=(a_{ij}(t))\) is the \(n\times n\) coefficient matrix and \(F=(f_1(t),\dots,f_n(t))^T\).

Remark If \(F=0\) (the zero vector), the system \(X'=AX\) is called homogeneous; otherwise it is nonhomogeneous.

Example The system \(\dfrac{dx}{dt}=3x-5y\), \(\dfrac{dy}{dt}=4x+8y\) is already linear with constant coefficients, so with \(X=(x,y)^T\), \[ X' = \begin{pmatrix}3&-5\\4&8\end{pmatrix}X. \]

Example For the system \(2\dfrac{dx}{dt}=5x-14y+9t\), \(4\dfrac{dy}{dt}=-x+6y-2\), first divide each equation by the coefficient of its derivative: \[ \frac{dx}{dt} = \frac52x-7y+\frac92t, \qquad \frac{dy}{dt} = -\frac14x+\frac32y-\frac12. \] So \[ X' = \begin{pmatrix}5/2&-7\\-1/4&3/2\end{pmatrix}X + \begin{pmatrix}9t/2\\-1/2\end{pmatrix}. \]

Example Conversely, the matrix equation \(X' = \begin{pmatrix}4&2\\-1&3\end{pmatrix}X+\begin{pmatrix}1\\-1\end{pmatrix}e^t\) corresponds to the scalar system \[ \frac{dx}{dt}=4x+2y+e^t, \qquad \frac{dy}{dt}=-x+3y-e^t. \]

Converting a Higher-Order DE to a System

A single \(n\)th-order linear DE \(y^{(n)} = f(t,y,y',\dots,y^{(n-1)})\) can always be rewritten as a system of \(n\) first-order equations. Let \(x_1=y\), \(x_2=y'\), \(\dots\), \(x_n=y^{(n-1)}\). Then \(x_1'=x_2\), \(x_2'=x_3\), \(\dots\), \(x_{n-1}'=x_n\), and \(x_n'=y^{(n)}=f(t,x_1,\dots,x_n)\).

Example Convert \(y'''-3y''+5y'-y=2e^{-2t}\) to a system in matrix form.

Let \(x_1=y\), \(x_2=y'\), \(x_3=y''\). Then \(x_1'=x_2\), \(x_2'=x_3\), and solving the DE for \(y'''\),

\[ x_3' = y''' = 2e^{-2t}+y-5y'+3y'' = x_1-5x_2+3x_3+2e^{-2t}. \] So \(X=(x_1,x_2,x_3)^T\) satisfies \[ X' = \begin{pmatrix}0&1&0\\0&0&1\\1&-5&3\end{pmatrix}X + \begin{pmatrix}0\\0\\2e^{-2t}\end{pmatrix}. \]

Example Convert \(y''+4y=t\cos t\) to a system in matrix form.

Let \(x_1=y\), \(x_2=y'\). Then \(x_1'=x_2\), and \(x_2'=y''=t\cos t-4y=t\cos t-4x_1\). So

\[ X' = \begin{pmatrix}0&1\\-4&0\end{pmatrix}X + \begin{pmatrix}0\\t\cos t\end{pmatrix}. \]

Solution Vectors and Fundamental Sets

Definition A solution vector \(X\) on an interval \(I\) is a column matrix \(X=(x_1,\dots,x_n)^T\) whose entries satisfy the system \(X'=AX+F\) for all \(t\) in \(I\).

Example Verify that \(X_1=\begin{pmatrix}e^{-2t}\\-e^{-2t}\end{pmatrix}\) and \(X_2=\begin{pmatrix}3e^{6t}\\5e^{6t}\end{pmatrix}\) are both solution vectors of \(X'=\begin{pmatrix}1&3\\5&3\end{pmatrix}X\).

For \(X_1\):

\[ \begin{pmatrix}1&3\\5&3\end{pmatrix} \begin{pmatrix}e^{-2t}\\-e^{-2t}\end{pmatrix} = \begin{pmatrix}e^{-2t}-3e^{-2t}\\5e^{-2t}-3e^{-2t}\end{pmatrix} = \begin{pmatrix}-2e^{-2t}\\2e^{-2t}\end{pmatrix} = \frac{d}{dt}\begin{pmatrix}e^{-2t}\\-e^{-2t}\end{pmatrix}. \checkmark \]

For \(X_2\):

\[ \begin{pmatrix}1&3\\5&3\end{pmatrix} \begin{pmatrix}3e^{6t}\\5e^{6t}\end{pmatrix} = \begin{pmatrix}3e^{6t}+15e^{6t}\\15e^{6t}+15e^{6t}\end{pmatrix} = \begin{pmatrix}18e^{6t}\\30e^{6t}\end{pmatrix} = \frac{d}{dt}\begin{pmatrix}3e^{6t}\\5e^{6t}\end{pmatrix}. \checkmark \]

Definition A set of \(n\) linearly independent solution vectors \(X_1,\dots,X_n\) of an \(n\)th-order system on an interval \(I\) is called a fundamental set of solutions on \(I\).

Remark To test linear independence of \(X_1,\dots,X_n\), either:

show that \(c_1X_1+\cdots+c_nX_n=0\) forces \(c_1=\cdots=c_n=0\), or
compute the Wronskian \(W(X_1,\dots,X_n)=\det(X_1\ \cdots\ X_n)\) (the matrix whose columns are \(X_1,\dots,X_n\)) and show it is nonzero.

Example Show that \(\begin{pmatrix}1\\1\end{pmatrix}\) and \(\begin{pmatrix}1\\0\end{pmatrix}\) are linearly independent.

Method 1. If \(c_1\begin{pmatrix}1\\1\end{pmatrix} +c_2\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\), then \(c_1+c_2=0\) and \(c_1=0\), forcing \(c_1=c_2=0\).

Method 2.

\[ W = \det\begin{pmatrix}1&1\\1&0\end{pmatrix} = -1 \ne 0. \] Either way, the two vectors are linearly independent.

Remark For \(X'=\begin{pmatrix}1&3\\5&3\end{pmatrix}X\), one can check that \(\{(1,-1)^Te^{-2t},\,(3,5)^Te^{6t}\}\) is a fundamental set of solutions, giving general solution \(X=c_1(1,-1)^Te^{-2t}+c_2(3,5)^Te^{6t}\). Notice that \(X_1=(1,-1)^Te^{-2t}\) satisfies \[ X_1' = (1,-1)^T(-2e^{-2t}) = \begin{pmatrix}1&3\\5&3\end{pmatrix} (1,-1)^Te^{-2t} = AX_1, \] so \(-2(1,-1)^T = \begin{pmatrix}1&3\\5&3\end{pmatrix}(1,-1)^T\): that is, \(\lambda=-2\) is an eigenvalue of \(A\) with corresponding eigenvector \((1,-1)^T\). This is exactly the pattern we will exploit in the next lecture: solutions of \(X'=AX\) can be found by computing the eigenvalues and eigenvectors of \(A\).

Section 8.2

Homogeneous Linear Systems

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Solving \(X'=AX\) via Eigenvalues

As we saw at the end of the last lecture, a solution vector of the form \(X=Ke^{\lambda t}\) (with \(K\) a constant, nonzero column vector) satisfies \(X'=AX\) exactly when \[ \lambda Ke^{\lambda t} = AKe^{\lambda t} \;\Longleftrightarrow\; AK=\lambda K, \] i.e. exactly when \(\lambda\) is an eigenvalue of \(A\) with corresponding eigenvector \(K\).

Theorem Let \(\lambda_1,\dots,\lambda_n\) be \(n\) distinct eigenvalues of the \(n\times n\) matrix \(A\), with corresponding eigenvectors \(K_1,\dots,K_n\). Then the general solution of \(X'=AX\) on \((-\infty,\infty)\) is \[ X = c_1K_1e^{\lambda_1t}+c_2K_2e^{\lambda_2t}+\cdots+c_nK_ne^{\lambda_nt}. \]

Example Solve the system \(\dfrac{dx}{dt}=2x+3y\), \(\dfrac{dy}{dt}=2x+y\).

In matrix form, \(X'=\begin{pmatrix}2&3\\2&1\end{pmatrix}X=AX\).

\[ \det(A-\lambda I) = \det\begin{pmatrix}2-\lambda&3\\2&1-\lambda\end{pmatrix} = \lambda^2-3\lambda-4 = (\lambda-4)(\lambda+1). \] So \(\lambda_1=4\), \(\lambda_2=-1\).

For \(\lambda_1=4\): \((A-4I\mid0) = \begin{pmatrix}-2&3\\2&-3\end{pmatrix}\) row-reduces to \(2x-3y=0\), so \(y=\tfrac23x\); choosing \(x=3\) gives eigenvector \(\begin{pmatrix}3\\2\end{pmatrix}\).

For \(\lambda_2=-1\): \((A+I\mid0) = \begin{pmatrix}3&3\\2&2\end{pmatrix}\) row-reduces to \(x+y=0\), so \(y=-x\); choosing \(x=1\) gives eigenvector \(\begin{pmatrix}1\\-1\end{pmatrix}\).

Thus

\[ \boxed{X = c_1\begin{pmatrix}1\\-1\end{pmatrix}e^{-t} + c_2\begin{pmatrix}3\\2\end{pmatrix}e^{4t}.} \]

Example Solve the system \(\dfrac{dx}{dt}=x+y-z\), \(\dfrac{dy}{dt}=2y\), \(\dfrac{dz}{dt}=y-z\).

In matrix form, \(X'=\begin{pmatrix}1&1&-1\\0&2&0\\0&1&-1\end{pmatrix}X=AX\). Expanding \(\det(A-\lambda I)\) along the first column,

\[ \det(A-\lambda I) = (2-\lambda)(1-\lambda)(-1-\lambda). \] So \(\lambda_1=2\), \(\lambda_2=1\), \(\lambda_3=-1\).

For \(\lambda_1=2\): \((A-2I\mid0)\) reduces to \(-x+y-z=0\), \(y=3z\); choosing \(z=1\) gives \(y=3\), \(x=y-z=2\), so \(K_1=\begin{pmatrix}2\\3\\1\end{pmatrix}\).

For \(\lambda_2=1\): \((A-I\mid0)\) reduces to \(y-z=0\), \(y=0\), so \(y=z=0\); choosing \(x=1\) gives \(K_2=\begin{pmatrix}1\\0\\0\end{pmatrix}\).

For \(\lambda_3=-1\): \((A+I\mid0)\) reduces to \(2x+y-z=0\), \(y=0\); choosing \(x=1\) gives \(z=2x=2\), so \(K_3=\begin{pmatrix}1\\0\\2\end{pmatrix}\).

Thus

\[ \boxed{X = c_1\begin{pmatrix}2\\3\\1\end{pmatrix}e^{2t} + c_2\begin{pmatrix}1\\0\\0\end{pmatrix}e^{t} + c_3\begin{pmatrix}1\\0\\2\end{pmatrix}e^{-t}.} \]

Repeated Eigenvalues

What if the characteristic equation has a repeated root? If a repeated eigenvalue \(\lambda\) (of multiplicity \(2\)) has only one corresponding eigenvector \(K\), we still need a second, linearly independent solution. It always has the form \[ X_2 = Kte^{\lambda t} + Pe^{\lambda t}, \] where \(P\) is found by solving \((A-\lambda I)P = K\).

Example Solve the system \(X' = \begin{pmatrix}3&-18\\2&-9\end{pmatrix}X\). \[ \det(A-\lambda I) = \det\begin{pmatrix}3-\lambda&-18\\2&-9-\lambda\end{pmatrix} = \lambda^2+6\lambda+9 = (\lambda+3)^2. \] So \(\lambda=-3\) is the only eigenvalue, but we need two linearly independent solution vectors.

Eigenvector. \((A+3I\mid0) = \begin{pmatrix}6&-18\\2&-6\end{pmatrix}\) row-reduces to \(x-3y=0\), so \(x=3y\); choosing \(y=1\) gives \(K=\begin{pmatrix}3\\1\end{pmatrix}\). This gives one solution, \(X_1 = \begin{pmatrix}3\\1\end{pmatrix}e^{-3t}\).

Second solution vector. Solve \((A+3I)P=K\):

\[ \begin{pmatrix}6&-18\\2&-6\end{pmatrix}\begin{pmatrix}p_1\\p_2\end{pmatrix} = \begin{pmatrix}3\\1\end{pmatrix}. \] Both rows reduce to the single equation \(2p_1-6p_2=1\); choosing \(p_2=0\) gives \(p_1=\tfrac12\), so \(P=\begin{pmatrix}1/2\\0\end{pmatrix}\). Thus \[ X_2 = \begin{pmatrix}3\\1\end{pmatrix}te^{-3t} + \begin{pmatrix}1/2\\0\end{pmatrix}e^{-3t}. \]

The general solution is

\[ \boxed{X = c_1\begin{pmatrix}3\\1\end{pmatrix}e^{-3t} + c_2\left[\begin{pmatrix}3\\1\end{pmatrix}te^{-3t} + \begin{pmatrix}1/2\\0\end{pmatrix}e^{-3t}\right].} \]

Remark Just as with scalar constant-coefficient equations (Section 4.3), a matrix \(A\) can also have complex conjugate eigenvalues. In that case, the two corresponding (complex) solution vectors can be combined — via Euler's formula, exactly as in the scalar case — to produce two real-valued, linearly independent solutions involving \(\cos\) and \(\sin\).

Section 8.3

Nonhomogeneous Linear Systems

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Nonhomogeneous Systems

When we have a nonhomogeneous system \(X'=AX+F(t)\), the general solution has the form \[ X = X_c+X_p, \] where \(X_c=c_1X_1+\cdots+c_nX_n\) is the complementary function (the general solution of the associated homogeneous system \(X'=AX\)) and \(X_p\) is any single particular solution of \(X'=AX+F(t)\). As with scalar equations, we have two methods for finding \(X_p\): the method of undetermined coefficients, and variation of parameters.

Undetermined Coefficients

This method can be used when (1) \(A\) has constant entries, and (2) the entries of \(F(t)\) are constants, polynomials, exponentials, or sine/cosine functions.

Example Solve the system \(X'=\begin{pmatrix}2&2\\1&3\end{pmatrix}X +\begin{pmatrix}-8\\3\end{pmatrix}\) on \((-\infty,\infty)\).

Homogeneous part.

\[ \det(A-\lambda I) = \det\begin{pmatrix}2-\lambda&2\\1&3-\lambda\end{pmatrix} = \lambda^2-5\lambda+4 = (\lambda-1)(\lambda-4). \] For \(\lambda=1\): \((A-I\mid0)\) gives \(x+2y=0\), eigenvector \(\begin{pmatrix}-2\\1\end{pmatrix}\). For \(\lambda=4\): \((A-4I\mid0)\) gives \(x-y=0\), eigenvector \(\begin{pmatrix}1\\1\end{pmatrix}\). So \[ X_c = c_1\begin{pmatrix}-2\\1\end{pmatrix}e^{t} + c_2\begin{pmatrix}1\\1\end{pmatrix}e^{4t}. \]

Particular solution. Since \(F=\begin{pmatrix}-8\\3\end{pmatrix}\) has constant entries, guess \(X_p=\begin{pmatrix}a\\b\end{pmatrix}\), a constant vector, so \(X_p'=0\). Substituting into the DE,

\[ 0 = AX_p+F \;\Longrightarrow\; \begin{pmatrix}2&2\\1&3\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}8\\-3\end{pmatrix}. \] This gives \(2a+2b=8\) and \(a+3b=-3\). From the first equation, \(a+b=4\), so \(a=4-b\). Substituting, \[ (4-b)+3b = -3 \;\Longrightarrow\; 2b=-7 \;\Longrightarrow\; b=-\frac72, \qquad a = 4-\left(-\frac72\right)=\frac{15}{2}. \] So \(X_p = \begin{pmatrix}15/2\\-7/2\end{pmatrix}\), and \[ \boxed{X = c_1\begin{pmatrix}-2\\1\end{pmatrix}e^{t} + c_2\begin{pmatrix}1\\1\end{pmatrix}e^{4t} + \begin{pmatrix}15/2\\-7/2\end{pmatrix}.} \]

Example Solve the system \(X'=\begin{pmatrix}6&1\\4&3\end{pmatrix}X +\begin{pmatrix}6t\\-10t+4\end{pmatrix}\).

Homogeneous part.

\[ \det(A-\lambda I) = \det\begin{pmatrix}6-\lambda&1\\4&3-\lambda\end{pmatrix} = \lambda^2-9\lambda+14 = (\lambda-2)(\lambda-7). \] For \(\lambda=2\): eigenvector \(\begin{pmatrix}1\\-4\end{pmatrix}\). For \(\lambda=7\): eigenvector \(\begin{pmatrix}1\\1\end{pmatrix}\). So \[ X_c = c_1\begin{pmatrix}1\\-4\end{pmatrix}e^{2t} + c_2\begin{pmatrix}1\\1\end{pmatrix}e^{7t}. \]

Particular solution. Since \(F\) is linear in \(t\), guess \(X_p = \begin{pmatrix}a_1t+a_2\\b_1t+b_2\end{pmatrix}\), so \(X_p' = \begin{pmatrix}a_1\\b_1\end{pmatrix}\). Substituting into \(X_p'=AX_p+F\) and matching coefficients of \(t\) and the constant term in each row gives the system

\[ \begin{aligned} 6a_1+b_1+6&=0, &4a_1+3b_1-10&=0,\\ 6a_2+b_2-a_1&=0, &4a_2+3b_2+4-b_1&=0. \end{aligned} \] Solving the first pair for \(a_1,b_1\) and then the second pair for \(a_2,b_2\), \[ a_1=-2, \qquad b_1=6, \qquad a_2=-\frac47, \qquad b_2=\frac{10}{7}. \] So \(X_p = \begin{pmatrix}-2t-4/7\\6t+10/7\end{pmatrix}\), and \[ \boxed{X = c_1\begin{pmatrix}1\\-4\end{pmatrix}e^{2t} + c_2\begin{pmatrix}1\\1\end{pmatrix}e^{7t} + \begin{pmatrix}-2t-4/7\\6t+10/7\end{pmatrix}.} \]

Variation of Parameters

Let \(X_1,\dots,X_n\) be a fundamental set of solutions of \(X'=AX\), so the general solution is \(X=c_1X_1+\cdots+c_nX_n\). Writing this as \[ X = \Phi(t)\,C, \qquad \Phi(t) = \big(X_1\ X_2\ \cdots\ X_n\big), \qquad C = \begin{pmatrix}c_1\\\vdots\\c_n\end{pmatrix}, \] the matrix \(\Phi(t)\) (whose columns are the \(X_i\)) is called a fundamental matrix. Since \(\{X_i\}\) are linearly independent, \(\det(\Phi(t))\ne0\), so \(\Phi(t)\) is invertible, and one can check \(\Phi'(t)=A\Phi(t)\).

Theorem For the nonhomogeneous system \(X'=AX+F(t)\) with fundamental matrix \(\Phi(t)\), a particular solution is given by \[ X_p = \Phi(t)\int \Phi(t)^{-1}F(t)\,dt. \]

Remark Since \(\det(\Phi(t))\ne0\), \(\Phi(t)^{-1}\) always exists; for a \(2\times2\) matrix \(\Phi(t)=\begin{pmatrix}a&b\\c&d\end{pmatrix}\), \[ \Phi(t)^{-1} = \frac{1}{\det(\Phi(t))}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}. \]

Example Solve the system \(X' = \begin{pmatrix}-3&1\\2&-4\end{pmatrix}X + \begin{pmatrix}3t\\e^{-t}\end{pmatrix}\).

Homogeneous part. \(\det(A-\lambda I) = \lambda^2+7\lambda+10=(\lambda+2)(\lambda+5)\), so \(\lambda=-2\) (eigenvector \((1,1)^T\)) and \(\lambda=-5\) (eigenvector \((1,-2)^T\)). So

\[ \Phi(t) = \begin{pmatrix}e^{-2t}&e^{-5t}\\e^{-2t}&-2e^{-5t}\end{pmatrix}. \]

Fundamental matrix inverse. \(\det(\Phi(t)) = -2e^{-7t}-e^{-7t}=-3e^{-7t}\), so

\[ \Phi(t)^{-1} = \begin{pmatrix}\frac23e^{2t}&\frac13e^{2t}\\ \frac13e^{5t}&-\frac13e^{5t}\end{pmatrix}. \]

Particular solution.

\[ \Phi(t)^{-1}F(t) = \begin{pmatrix}2te^{2t}+\frac13e^{t}\\ te^{5t}-\frac13e^{4t}\end{pmatrix}. \] Integrating each entry (using \(\int te^{at}dt = \tfrac{t}{a}e^{at}-\tfrac{1}{a^2}e^{at}\)), \[ \int\Phi(t)^{-1}F(t)\,dt = \begin{pmatrix}te^{2t}-\frac12e^{2t}+\frac13e^{t}\\ \frac15te^{5t}-\frac1{25}e^{5t}-\frac1{12}e^{4t}\end{pmatrix}. \] Multiplying by \(\Phi(t)\) and simplifying each component, \[ X_p = \Phi(t)\int\Phi(t)^{-1}F(t)\,dt = \begin{pmatrix}\frac65t-\frac{27}{50}+\frac14e^{-t}\\ \frac35t-\frac{21}{50}+\frac12e^{-t}\end{pmatrix}. \] Thus \[ \boxed{X = c_1\begin{pmatrix}1\\1\end{pmatrix}e^{-2t} + c_2\begin{pmatrix}1\\-2\end{pmatrix}e^{-5t} + \begin{pmatrix}\frac65t-\frac{27}{50}+\frac14e^{-t}\\ \frac35t-\frac{21}{50}+\frac12e^{-t}\end{pmatrix}.} \]

Example Solve the system \(\dfrac{dx}{dt}=3x-3y+4\), \(\dfrac{dy}{dt}=2x-2y-1\).

In matrix form, \(X'=\begin{pmatrix}3&-3\\2&-2\end{pmatrix}X +\begin{pmatrix}4\\-1\end{pmatrix}\).

Homogeneous part. \(\det(A-\lambda I) = \lambda^2-\lambda = \lambda(\lambda-1)\), so \(\lambda=0\) (eigenvector \((1,1)^T\)) and \(\lambda=1\) (eigenvector \((3,2)^T\)). So

\[ \Phi(t) = \begin{pmatrix}1&3e^{t}\\1&2e^{t}\end{pmatrix}. \]

Fundamental matrix inverse. \(\det(\Phi(t)) = 2e^t-3e^t=-e^t\), so

\[ \Phi(t)^{-1} = \begin{pmatrix}-2&3\\e^{-t}&-e^{-t}\end{pmatrix}. \]

Particular solution.

\[ \Phi(t)^{-1}F(t) = \begin{pmatrix}-2&3\\e^{-t}&-e^{-t}\end{pmatrix} \begin{pmatrix}4\\-1\end{pmatrix} = \begin{pmatrix}-11\\5e^{-t}\end{pmatrix}. \] Integrating, \[ \int\Phi(t)^{-1}F(t)\,dt = \begin{pmatrix}-11t\\-5e^{-t}\end{pmatrix}. \] Multiplying by \(\Phi(t)\), \[ X_p = \begin{pmatrix}1&3e^t\\1&2e^t\end{pmatrix} \begin{pmatrix}-11t\\-5e^{-t}\end{pmatrix} = \begin{pmatrix}-11t-15\\-11t-10\end{pmatrix}. \] Thus \[ \boxed{X = c_1\begin{pmatrix}1\\1\end{pmatrix} + c_2\begin{pmatrix}3\\2\end{pmatrix}e^{t} + \begin{pmatrix}-11t-15\\-11t-10\end{pmatrix}.} \]