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.
Lecture 21: Appendix B.1 - Introduction to Linear Algebra ↑ Back to top
Definition. A matrix is a rectangular array of numbers arranged in rows and columns. The entries of a matrix are called its elements.
$$ A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} = (a_{ij}) $$the size or order of a matrix is given by the number of rows and columns.
Remark. For this course, we will mostly be interacting with square matrices which are size $n \times n.$
Example.
$$ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} $$This matrix has 2 rows and 3 columns, so its size is $2 \times 3$.
Operations with Matrices
Much like with real numbers, we can perform basic operations with matrices, such as addition, subtraction, and multiplication. Although some operations have a different look than we might expect.
Addition
Matrices of the same size can be added by adding the corresponding entries.
$$
A =
\begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix}, \quad
B =
\begin{pmatrix}
5 & 6 \\
7 & 8
\end{pmatrix}
$$
$$
A + B =
\begin{pmatrix}
1+5 & 2+6 \\
3+7 & 4+8
\end{pmatrix}
=
\begin{pmatrix}
6 & 8 \\
10 & 12
\end{pmatrix}
$$
Subtraction
Matrices of the same size can be subtracted by subtracting the corresponding entries. Continuing with the matrices $A$ and $B$ from the previous example,
$$
A - B =
\begin{pmatrix}
1-5 & 2-6 \\
3-7 & 4-8
\end{pmatrix}
=
\begin{pmatrix}
-4 & -4 \\
-4 & -4
\end{pmatrix}
$$
Scalar Multiplication
We can multiply matrices by a constant by multiplying every entry by that constant. Continuing with the matrix $A$ from the previous example,
$$
2A =
2 \begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix}
=
\begin{pmatrix}
2\cdot1 & 2\cdot2 \\
2\cdot3 & 2\cdot4
\end{pmatrix}
=
\begin{pmatrix}
2 & 4 \\
6 & 8
\end{pmatrix}
$$
Matrix Multiplication
To multiply two matrices, we look at the $i$th row of $A$ and the $j$th column of $B.$ we take the dot product of both of those vectors and that goes into position $(i,j)$ in the resulting matrix. We provide some examples below.
2 × 2 Case (Formula)
$$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad B = \begin{pmatrix} e & f \\ g & h \end{pmatrix} $$ $$ AB = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix} $$
$$
A =
\begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix}, \quad
B =
\begin{pmatrix}
5 & 6 \\
7 & 8
\end{pmatrix}
$$
$$
AB =
\begin{pmatrix}
(1)(5) + (2)(7) & (1)(6) + (2)(8) \\
(3)(5) + (4)(7) & (3)(6) + (4)(8)
\end{pmatrix}
=
\begin{pmatrix}
5 + 14 & 6 + 16 \\
15 + 28 & 18 + 32
\end{pmatrix}
=
\begin{pmatrix}
19 & 22 \\
43 & 50
\end{pmatrix}
$$
Remark. Matrix multiplication is NOT commutative. That is, $AB \ne BA$ most of the time.
Multiplicative Identity
Definition. The identity matrix is a square matrix with 1's on the diagonal and 0's elsewhere.
$$
I_2 =
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$$
$$
I_3 =
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{pmatrix}
$$
The property that the identity matrix has is for any matrix $A,$
$$ AI = IA = A. $$Notice that this is analogous to "multiplication by $1$."
Determinants
We saw determinants when we computed the Wronksian to show linear independence of functions. To review,
$$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ $$ \det(A) = ad - bc $$
$$
A =
\begin{pmatrix}
3 & 4 \\
2 & 5
\end{pmatrix}
$$
$$
\det(A) = (3)(5) - (4)(2) = 15 - 8 = 7
$$
$$
A =
\begin{pmatrix}
1 & 2 & 3 \\
0 & 4 & 5 \\
1 & 0 & 6
\end{pmatrix}
$$
Expand along first row:
$$ \det(A) = 1 \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} - 2 \begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} + 3 \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} $$Compute each minor:
$$ \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} = (4)(6) - (5)(0) = 24 $$ $$ \begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} = (0)(6) - (5)(1) = -5 $$ $$ \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} = (0)(0) - (4)(1) = -4 $$Substitute:
$$ \det(A) = 1(24) - 2(-5) + 3(-4) $$ $$ = 24 + 10 - 12 = 22 $$Multiplicative Inverses
Definition. A matrix $A$ of size $n\times n$ is invertible if there exists $A^{-1}$ such that:
$$ AA^{-1} = A^{-1}A = I_n. $$Note that a square matrix is only invertible when its determinant is nonzero. In the $2 \times 2$ case, we have a formula:
$$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ $$ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} $$Lecture 22: Appendix B.2 - Gaussian Elimination ↑ Back to top
Definition. Gaussian elimination is a systematic method used to solve systems of linear equations. The idea is to use elementary row operations in order to transform the system into an equivalent system that is less complex to solve.
Definition. The allowed elementary row operations that keep the system the same are
- Swap rows
- Multiply a row by a constant
- Add or subtract constant multiples of one row to another row.
The idea is the following:
- Set up the augmented matrix representation for the system of equations.
- Starting with the first column, obtain a pivot and use the pivot position and row operations to eliminate (make \(0\)) the rest of the column.
- Rewrite the system of equations given your reduced augmented matrix.
- Solve this system to obtain solution.
Example. Solve the system:
\[
\begin{cases}
x + y = 5 \\
2x - y = 1
\end{cases}
\]
Write the augmented matrix:
\[
\left[
\begin{array}{cc|c}
1 & 1 & 5 \\
2 & -1 & 1
\end{array}
\right]
\]
Let the top left entry be our first pivot. Now we perform the row operation \(R_2 \to R_2 - 2R_1.\) We get
\[
\left[
\begin{array}{cc|c}
1 & 1 & 5 \\
2-2(1) & -1-2(1) & 1-2(5)
\end{array}
\right]
=
\left[
\begin{array}{cc|c}
1 & 1 & 5 \\
0 & -3 & -9
\end{array}
\right]
\]
Now we need a pivot in the second column. Let the \(-3\) entry be the next pivot position. We perform the row operation \(R_2 \to -\frac{1}{3}R_2.\) We get
\[
\left[
\begin{array}{cc|c}
1 & 1 & 5 \\
0 & 1 & 3
\end{array}
\right]
\]
Lastly, we will do the row operation \(R_1 \to R_1 - R_2\) to get
\[
\left[
\begin{array}{cc|c}
1 & 0 & 2 \\
0 & 1 & 3
\end{array}
\right]
\]
Now we rewrite the system as
\[
x=2, y=3.
\]
Thus, our final solution is:
\[
(x,y) = (2,3).
\]
Remark. Ideally, every pivot we choose would be a \(1\) so that we can do row operations with integer values and not fractions. The rule of choosing pivots is that we want a pivot in every column of the augmented matrix, but we cannot pick a pivot in the same row as another pivot.
Example. Solve the system:
\[
\begin{cases}
x + y + z = 6 \\
2x - y - z = 3 \\
3x + y + 2z = 10
\end{cases}
\]
Augmented matrix:
\[
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 6 \\
2 & -1 & -1 & 3 \\
3 & 1 & 2 & 10
\end{array}
\right]
\]
We perform row operations. First, \(R_2 \to R_2 - 2R_1, \text{and } R_3 \to R_3 - 3R_1:\)
\[
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 6 \\
0 & -3 & -3 & -9 \\
0 & -2 & -1 & -8
\end{array}
\right]
\]
Next, \(R_2 \to -\frac{1}{3}R_2:\)
\[
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 6 \\
0 & 1 & 1 & 3 \\
0 & -2 & -1 & -8
\end{array}.
\right]
\]
Next, \(R_1 \to R_1-R_2, ~R_3 \to R_3+2R_2:\)
\[
\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 3 \\
0 & 1 & 1 & 3 \\
0 & 0 & 1 & -2
\end{array}.
\right]
\]
Lastly, \(R_2 \to R_2 - R_3:\)
\[
\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 3 \\
0 & 1 & 0 & 5 \\
0 & 0 & 1 & -2
\end{array}.
\right]
\]
Now we rewrite the system and get the solution:
\[
(x,y,z) = (3,5,-2)
\]
Remark. A system of \(n\) linear equations with \(n\) unknowns can have a unique solution OR infinitely many solutions. The following will be an example of the latter.
Example. Solve the system:
\[
\begin{cases}
x + y + z = 2 \\
2x + 2y + 2z = 4 \\
x - y + z = 0
\end{cases}
\]
Augmented matrix:
\[
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 2 \\
2 & 2 & 2 & 4 \\
1 & -1 & 1 & 0
\end{array}
\right]
\]
Now we perform row operations. Notice if we perform \(R_2 \to R_2-2R_1\) and \(R_3 \to R_3 - R_1\) we get
\[
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 2 \\
0 & 0 & 0 & 0 \\
0 & -2 & 0 & -2
\end{array}
\right]
\]
Then we do \(R_3 \to -\frac{1}{2}R_3\) then \(R_1 \to R_1-R_3\) we arrive at a reduced matrix
\[
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1
\end{array}
\right]
\]
\[
\left[
\begin{array}{ccc|c}
1 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1
\end{array}
\right]
\]
Notice there is a row of \(0.\) This means there are more equations than unknowns in the resulting system. We rewrite as
\[
x+z = 1, y=0.
\]
One of \(x\) or \(z\) is a free variable. When we solve for \(x\):
\[
x = 1 - z
\]
Let \(z = t\) (free variable).
\[
x = 1 - t, \quad y = 1, \quad z = t
\]
Solution:
\[
(x,y,z) = (1 - t, 1, t), \quad t \in \mathbb{R}
\]
Lecture 23: Appendix B.3 - Eigenvalues and Eigenvectors ↑ Back to top
Definition. Let \(A\) be an \(n \times n\) matrix. A nonzero vector \(\mathbf{v}\) is called an eigenvector of \(A\) if there exists a scalar \(\lambda\) such that
\[
A\mathbf{v} = \lambda \mathbf{v}.
\]
The scalar \(\lambda\) is called an eigenvalue of \(A\).
Definition. To find eigenvalues, we solve the characteristic equation:
\[
\det(A - \lambda I) = 0.
\]
The process is:
- Compute \(A - \lambda I\)
- Find \(\det(A - \lambda I)\)
- Solve for \(\lambda\)
- For each \(\lambda\), solve \((A - \lambda I)\mathbf{v} = 0\)
Example. Find eigenvalues and eigenvectors of
\[
A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}.
\]
Step 1: Compute \(A - \lambda I\):
\[
A - \lambda I =
\begin{bmatrix}
2 - \lambda & 1 \\
1 & 2 - \lambda
\end{bmatrix}
\]
Step 2: Compute determinant:
\[
\det(A - \lambda I) = (2-\lambda)(2-\lambda) - (1)(1)
\]
\[
= (2-\lambda)^2 - 1
\]
\[
= 4 - 4\lambda + \lambda^2 - 1
\]
\[
= \lambda^2 - 4\lambda + 3
\]
Step 3: Solve:
\[
\lambda^2 - 4\lambda + 3 = 0
\]
\[
(\lambda - 1)(\lambda - 3) = 0
\]
\[
\lambda = 1, \quad \lambda = 3
\]
Step 4: Find eigenvectors
For \(\lambda = 1\):
\[
A - I =
\begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}
\]
Solve \((A - I)\mathbf{v} = 0\):
\[
A - I =
\begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix} = \begin{bmatrix}
x + y\\
x + y
\end{bmatrix} = \begin{bmatrix}
0 \\
0
\end{bmatrix}
\]
Both components give the same equation. That is,
\[
x + y = 0 \Rightarrow y = -x
\]
Eigenvector:
\[
\mathbf{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}
\]
For \(\lambda = 3\):
\[
A - 3I =
\begin{bmatrix}
-1 & 1 \\
1 & -1
\end{bmatrix}
\]
Solve \((A - 3I)\mathbf{v} = 0\):
\[
A - 3I =
\begin{bmatrix}
-1 & 1 \\
1 & -1
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix} = \begin{bmatrix}
-x + y\\
x - y
\end{bmatrix} = \begin{bmatrix}
0 \\
0
\end{bmatrix}
\]
Both components give the same equation. That is,
\[
x - y =0 \Rightarrow y = x
\]
Eigenvector:
\[
\mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}
\]
Remark. Notice that when we try to determine eigenvector, we expect to have a free variable. There are infinitely many eigenvectors that correspond to each eigenvalue, it is up to a choice of free variable. That is, every eigenvector is off by a constant multiple.
Remark. Since the characteristic equation is a polynomial, there are a couple of cases to consider. On the one hand, we could have distinct eigenvalues. Alternatively, we could have repeated eigenvalues or complex eigenvalues.
Example. Find eigenvalues and eigenvectors of
\[
A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}.
\]
Step 1:
\[
A - \lambda I =
\begin{bmatrix}
2 - \lambda & 1 \\
0 & 2 - \lambda
\end{bmatrix}
\]
Step 2:
\[
\det(A - \lambda I) = (2-\lambda)^2
\]
Step 3:
\[
(2-\lambda)^2 = 0 \Rightarrow \lambda = 2
\]
Step 4:
\[
A - 2I =
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
\]
Solve \((A - 2I)\mathbf{v} = 0\):
\[
A - 2I =
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix} = \begin{bmatrix}
y\\
0
\end{bmatrix} = \begin{bmatrix}
0 \\
0
\end{bmatrix}
\]
Only the top equation gives information about the variables. That is,
\[
y=0.
\]
Since an eigenvector must be nonzero, x is a free variable and we get an eigenvector of
\[
\mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
\]
Remark. When an eigenvalue is repeated, there may be fewer independent eigenvectors than its multiplicity. This is important in determining whether a matrix is diagonalizable.
Example. Find eigenvalues and eigenvectors of
\[
A =
\begin{bmatrix}
3 & 0 & 0 \\
0 & 2 & 0 \\
4 & 0 & 1
\end{bmatrix}.
\]
Step 1:
\[
A - \lambda I =
\begin{bmatrix}
3-\lambda & 0 & 0 \\
0 & 2-\lambda & 0 \\
4 & 0 & 1-\lambda
\end{bmatrix}
\]
Step 2: Compute the determinant
\[
\det(A - \lambda I) =
\begin{vmatrix}
3-\lambda & 0 & 0 \\
0 & 2-\lambda & 0 \\
4 & 0 & 1-\lambda
\end{vmatrix}
\]
Expand along the first row:
\[
= (3-\lambda)
\begin{vmatrix}
2-\lambda & 0 \\
0 & 1-\lambda
\end{vmatrix}
\]
\[
= (3-\lambda)\left[(2-\lambda)(1-\lambda)\right]
\]
Step 3: Solve for eigenvalues
\[
(3-\lambda)(2-\lambda)(1-\lambda) = 0
\]
\[
\lambda = 3, \quad 2, \quad 1
\]
Step 4: Find eigenvectors
First, for \(\lambda = 3\):
\[
A - 3I =
\begin{bmatrix}
0 & 0 & 0 \\
0 & -1 & 0 \\
4 & 0 & -2
\end{bmatrix}
\]
Solve \((A - 3I)\mathbf{v} = 0\):
\[
\begin{bmatrix}
0 & 0 & 0 \\
0 & -1 & 0 \\
4 & 0 & -2
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
0 \\
-y \\
4x - 2z
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}
\]
This gives:
\[
-y = 0 \Rightarrow y = 0
\]
\[
4x - 2z = 0 \Rightarrow 2x - z = 0 \Rightarrow z = 2x
\]
Let \(x = t\). Then:
\[
\mathbf{v}_1 =
\begin{bmatrix}
1 \\
0 \\
2
\end{bmatrix}
\]
---
Next, for \(\lambda = 2\):
\[
A - 2I =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
4 & 0 & -1
\end{bmatrix}
\]
Solve \((A - 2I)\mathbf{v} = 0\):
\[
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
4 & 0 & -1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
x \\
0 \\
4x - z
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}
\]
This gives:
\[
x = 0
\]
\[
4x - z = 0 \Rightarrow z = 0
\]
So \(y\) is free. Let \(y = t\). Then:
\[
\mathbf{v}_2 =
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}
\]
---
Finally, for \(\lambda = 1\):
\[
A - I =
\begin{bmatrix}
2 & 0 & 0 \\
0 & 1 & 0 \\
4 & 0 & 0
\end{bmatrix}
\]
Solve \((A - I)\mathbf{v} = 0\):
\[
\begin{bmatrix}
2 & 0 & 0 \\
0 & 1 & 0 \\
4 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
2x \\
y \\
4x
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}
\]
This gives:
\[
2x = 0 \Rightarrow x = 0
\]
\[
y = 0
\]
No restriction on \(z\), so let \(z = t\). Then:
\[
\mathbf{v}_3 =
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}
\]
Remark. This is only one way to solve for eigenvalues and eigenvectors. We can also use Gaussian elimination to do this process of finding an eigenvector once we have all the eigenvalues.
These values and vectors will become very important in the coming sections.
Lecture 24: Section 8.1 - System of Linear Equations (revisited) ↑ Back to top
The goal of this section is to gain a different perspective on all system problems we saw in Chapter 3. That is, we will be able to restate all linear differential equation problems in terms of matrices and use linear algebra techniques to solve them.
Suppose we have a system of first-order linear differential equations:
\[ \begin{cases} x_1' = a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n + f_1(t) \\ x_2' = a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n + f_2(t) \\ \vdots \\ x_n' = a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n + f_n(t) \end{cases} \]We can rewrite this system in matrix form by using the following remark.
Remark. To take the derivative or integral of a matrix \(A = (a_{ij})\),
\[\frac{d}{dt}A = A' = (\frac{d}{dt}(a)_{ij})\]
or
\[\int A dt = \left(\int a_{ij}dt\right).\]
Using this remark, we can rewrite the system above as:
\[
X =
\begin{bmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{bmatrix}, \quad
A =
\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{bmatrix}, \quad
F(t) =
\begin{bmatrix}
f_1(t) \\
f_2(t) \\
\vdots \\
f_n(t)
\end{bmatrix}
\]
Then the system becomes:
\[ X' = AX + F(t) \] This is what we call the matrix representation of the system or the matrix equation.Homogeneous vs Nonhomogeneous Systems
Definition. Let a system of linear differential equations have matrix representation \(X' = AX + F\).
If \(F = 0\) then the system is called homogeneous. Otherwise, the system is considered nonhomogeneous.
Example (Homogeneous). Convert the system to matrix form:
\[
\begin{cases}
x' = 2x + y \\
y' = 5x + 3y
\end{cases}
\]
Step 1: Define the vector
\[
X =
\begin{bmatrix}
x \\
y
\end{bmatrix}
\]
Step 2: Write coefficient matrix
\[
A =
\begin{bmatrix}
2 & 1 \\
5 & 3
\end{bmatrix}
\]
Step 3: Write matrix equation
\[
X' = AX =
\begin{bmatrix}
2 & 1 \\
5 & 3
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
\]
Example (Nonhomogeneous). Convert the system to matrix form:
\[
\begin{cases}
x' = 2x + y + t \\
y' = 5x + 3y + e^t
\end{cases}
\]
Step 1: Define vectors
\[
X =
\begin{bmatrix}
x \\
y
\end{bmatrix}, \quad
F(t) =
\begin{bmatrix}
t \\
e^t
\end{bmatrix}
\]
Step 2: Matrix form
\[
X' = AX + F(t) =
\begin{bmatrix}
2 & 1 \\
5 & 3
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
+
\begin{bmatrix}
t \\
e^t
\end{bmatrix}
\]
Higher-Order Differential Equations in Matrix Form
We can convert higher-order ODEs into systems of first-order equations.
Example. Convert
\[
y''' - 3y'' + 5y' - y = 2e^{-2t}
\]
Step 1: Define new variables
\[
x_1 = y, \quad x_2 = y', \quad x_3 = y''
\]
Then:
\[
x_1' = x_2, \quad x_2' = x_3
\]
Step 2: Express \(x_3'\)
\[
x_3' = y''' = 3y'' - 5y' + y + 2e^{-2t}
\]
Substitute:
\[
x_3' = 3x_3 - 5x_2 + x_1 + 2e^{-2t}
\]
Step 3: Write system
\[
\begin{cases}
x_1' = x_2 \\
x_2' = x_3 \\
x_3' = x_1 - 5x_2 + 3x_3 + 2e^{-2t}
\end{cases}
\]
Step 4: Matrix form
\[
X' =
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & -5 & 3
\end{bmatrix}
X +
\begin{bmatrix}
0 \\
0 \\
2e^{-2t}
\end{bmatrix}
\]
We now want to extend the definitions learned in Chapter 4 into terms for systems of linear differential equations. We start with the definition of a solution vector.
Definition. A vector function \(X(t)\) is called a solution vector on an interval \(I\) if it satisfies the matrix equation \(X' = AX+F\) for all \(t \in I\).
Example (Verification). Verify that
\[
X_1 =
\begin{bmatrix}
e^{-2t} \\
-e^{2t}
\end{bmatrix}
\]
and
\[
X_2 =
\begin{bmatrix}
3e^{6t} \\
5e^{6t}\end{bmatrix}
\]
is a solution to
\[
X' =
\begin{bmatrix}
1 & 3 \\
5 & 3
\end{bmatrix}
X.
\]
Step 1: Compute derivative
\[
X_1' =
\begin{bmatrix}
-2e^{-2t} \\
2e^{-2t}
\end{bmatrix}
\]
\[
X_2' = \begin{bmatrix}
18e^{6t} \\
30e^{6t}
\end{bmatrix}
\]
Step 2: Compute \(AX\)
\[
AX_1 =
\begin{bmatrix}
1 & 3 \\
5 & 3
\end{bmatrix}
\begin{bmatrix}
e^{-2t} \\
-e^{-2t}
\end{bmatrix}
=
\begin{bmatrix}
e^{-2t} -3e^{-2t} \\
5e^{-2t} -3e^{-2t}
\end{bmatrix}
=
\begin{bmatrix}
-2e^{-2t} \\
2e^{-2t}
\end{bmatrix} = X_1'
\]
and
\[
AX_2 =
\begin{bmatrix}
1 & 3 \\
5 & 3
\end{bmatrix}
\begin{bmatrix}
3e^{6t} \\
5e^{6t}
\end{bmatrix}
=
\begin{bmatrix}
3e^{6t}+15e^{6t} \\
15e^{6t} + 15e^{6t}
\end{bmatrix}
=
\begin{bmatrix}
18e^{-2t} \\
30e^{-2t}
\end{bmatrix} = X_2'
\]
Since \(X_1' = AX_1\) and \(X_2' = AX_2\), then both \(X_1\) and \(X_2\) are solutions.
Definition. A set of solutions \(\{X_1(t), X_2(t), \dots, X_n(t)\}\) is called a fundamental set of solutions if the vectors are linearly independent.
Remark. We test linear independence the same way as before (using determinant / Wronskian ideas).
Example. Consider the system from the previous example.
\[
X' =
\begin{bmatrix}
1 & 3 \\
5 & 3
\end{bmatrix}
X.
\]
We will show the set \(\left\{X_1 =
\begin{bmatrix}
e^{-2t} \\
-e^{-2t}
\end{bmatrix}, \quad
X_2 =
\begin{bmatrix}
3e^{6t} \\
5e^{6t}
\end{bmatrix}\right\}\) form a fundamental set of solutions to the system.
We saw in the previous example both \(X_1\) and \(X_2\) are solutions to the system. Now we show linear independence.
Step 2: Compute Wronksian
\[
W(X_1,X_2) = \det\begin{bmatrix}
e^{-2t} & 3e^{6t} \\
-e^{-2t} & 5e^{6t}
\end{bmatrix} = 5e^{4t}+3e^{4t} = 3e^{4t} \ne 0.
\]
Since the Wronksian is never zero, the solutions are linearly independent.
Therefore, \(\{X_1, X_2\}\) forms a fundamental set of solutions.
Now we arrive at the definition of general solution
Definition. Given a fundamental set of solutions \(\{X_1, \cdots , X_n\}\) to the system \(X' = AX+F\). Then, the general solution to the system is
\[X = c_1X_1 + \cdots + c_nX_n.\]
Connection
Here is the thing to notice about the connection between the linear algebra techniques learned in the past lectures and solutions vectors.
Example. Continuing with this example
\[
X' =
\begin{bmatrix}
1 & 3 \\
5 & 3
\end{bmatrix}
X,
\]
notice \(X_1 =
\begin{bmatrix}
e^{-2t} \\
-e^{-2t}
\end{bmatrix} =
e^{-2t}\begin{bmatrix}
1 \\
-1
\end{bmatrix}\) is a solution. Then
\[
X_1' = -2e^{-2t}\begin{bmatrix}
1 \\
-1
\end{bmatrix} = AX_1 = \begin{bmatrix}
1 & 3 \\
5 & 3
\end{bmatrix}e^{-2t}\begin{bmatrix}
1 \\
-1
\end{bmatrix}.
\]
We conclude that
\[
-2\begin{bmatrix}
1 \\
-1
\end{bmatrix}= \begin{bmatrix}
1 & 3 \\
5 & 3
\end{bmatrix}\begin{bmatrix}
1 \\
-1
\end{bmatrix}.
\]
In other words, the value \(-2\) is an eigenvalue with corresponding eigenvector \(\begin{bmatrix}
1 \\
-1
\end{bmatrix}\).
So there is clearly a relationship between finding eigenvalues and eigenvectors to the matrix representation for a system of linear differential equations, and finding solution vectors.
Lecture 25: Section 8.2 - General Solution to Homogeneous Systems of DEs ↑ Back to top
As we saw in the connection above, there is a way take the matrix representation for a system \(X' = AX+F\) and relate the solution vectors of that equation to the eigenvalues and eigenvectors for the system. For this section, we will consider a homogeneous system first, that is the matrix representations we are looking at come in the form \(X' = AX\).
Definition.
Consider the homogeneous system \[ X' = AX \] where \(A\) is an \(n \times n\) constant matrix.
If \(A\) has n distinct eigenvalues \(\lambda_1, \dots, \lambda_n\) with corresponding eigenvectors \(\mathbf{v}_1, \dots, \mathbf{v}_n\), then the general solution is
\[ X(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2 + \cdots + c_n e^{\lambda_n t}\mathbf{v}_n. \]Each term \(e^{\lambda t}\mathbf{v}\) is a solution vector, and linear independence follows from distinct eigenvalues.
Example.
For the system
\[
X' =
\begin{bmatrix}
1 & 3 \\
5 & 3
\end{bmatrix}
X.
\]
We proved the set \(\left\{X_1 =
\begin{bmatrix}
e^{-2t} \\
-e^{-2t}
\end{bmatrix}, \quad
X_2 =
\begin{bmatrix}
3e^{6t} \\
5e^{6t}
\end{bmatrix}\right\}\) form a fundamental set of solutions to the system. Therefore, the general solution for this system by the thorem above is
\[
X = c_1
\begin{bmatrix}
1 \\
-1
\end{bmatrix}e^{-2t}+c_2
\begin{bmatrix}
3 \\
5
\end{bmatrix}e^{6t}.
\]
The idea is the following:
- Create the matrix representation for the system.
- Find all the eigenvalues, and check how the characteristic equation factors
- Find all corresponding eigenvectors (when possible)
- Create the general solution
Example.
Solve
\[
\begin{cases}
x' = 2x + 3y \\
y' = 2x + y
\end{cases}
\]
The question is the following: in Chapter 4, we saw that the auxiliary equation corresponding to a higher order differential equation did not have to give us all distinct roots. What happens
if we obtain a repeated root in the characteristic equation? That is, what if we have an \(n \times n\) matrix with less than \(n\) distinct eigenvalues.
Step 1: Matrix form.
\[ X' = \begin{bmatrix} 2 & 3 \\ 2 & 1 \end{bmatrix} X \]Step 2: Eigenvalues.
\[ \det(A-\lambda I) = \begin{vmatrix} 2-\lambda & 3 \\ 2 & 1-\lambda \end{vmatrix} \] \[ = (2-\lambda)(1-\lambda)-6 \] \[ = \lambda^2 -3\lambda -4 = (\lambda-4)(\lambda+1) \] \[ \lambda=4,-1 \]Step 3: Eigenvectors.
For \(\lambda=4\): \[ A-4I= \begin{bmatrix} -2 & 3 \\ 2 & -3 \end{bmatrix} \] \[ -2x+3y=0 \Rightarrow y=\frac{2}{3}x \] Choose \(x=3\), \(y=2\). \[ \mathbf{v}_1= \begin{bmatrix} 3\\ 2 \end{bmatrix} \] For \(\lambda=-1\): \[ A+I= \begin{bmatrix} 3 & 3 \\ 2 & 2 \end{bmatrix} \] \[ 3x+3y=0 \Rightarrow y=-x \] \[ \mathbf{v}_2= \begin{bmatrix} 1\\ -1 \end{bmatrix} \]Step 4: General solution.
\[ X(t)= c_1 e^{4t} \begin{bmatrix} 3\\ 2 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1\\ -1 \end{bmatrix} \]
Remark.
If \(A\) is (for example) a \(2 \times 2\) matrix with a repeated eigenvalue \(\lambda\) and only one eigenvector \(K\), then the second solution has the form
\[ X_2 = K t e^{\lambda t} + P e^{\lambda t} \]where \(P\) satisfies \[ (A-\lambda I)P = K. \]
Together, the set \(\{K, X_2\}\) form a fundamental set of solutions. This process can be mimicked in the higher order cases as well.
Example.
\[
X'=
\begin{bmatrix}
3 & -18 \\
2 & -9
\end{bmatrix}
X
\]
If eigenvalues are complex, solutions involve sine and cosine terms.
See textbook for full treatment.
Step 1: Eigenvalues.
\[ \det(A-\lambda I) = \lambda^2+6\lambda+9 = (\lambda+3)^2 \] \[ \lambda=-3 \]Step 2: Eigenvector.
\[ A+3I= \begin{bmatrix} 6 & -18 \\ 2 & -6 \end{bmatrix} \] \[ 6x-18y=0 \Rightarrow x=3y \] \[ K= \begin{bmatrix} 3\\ 1 \end{bmatrix} \]Step 3: Find \(P\).
Solve \[ (A+3I)P=K. \] Let \(P=\begin{bmatrix}p_1\\p_2\end{bmatrix}\). \[ 6p_1-18p_2=3 \] Choose \(p_2=0\): \[ 6p_1=3 \Rightarrow p_1=\frac12 \] \[ P= \begin{bmatrix} \frac12\\ 0 \end{bmatrix} \]General solution:
\[ X(t)= c_1 e^{-3t} \begin{bmatrix} 3\\ 1 \end{bmatrix} + c_2 e^{-3t} \left( t \begin{bmatrix} 3\\ 1 \end{bmatrix} + \begin{bmatrix} \frac12\\ 0 \end{bmatrix} \right) \]
Example. Find the general solution to the system
\[
\begin{cases}
x' = x+y-z \\
y' = 2y \\
z' = y-z
\end{cases}.
\]
Step 1: Matrix form.
\[ A= \begin{bmatrix} 1 & 1 & -1\\ 0 & 2 & 0\\ 0 & 1 & -1 \end{bmatrix} \]Step 2: Eigenvalues.
\[ \det(A-\lambda I) = (2-\lambda) \begin{vmatrix} 1-\lambda & -1\\ 0 & -1-\lambda \end{vmatrix} \] \[ = (2-\lambda)(1-\lambda)(-1-\lambda) \] \[ \lambda=2,1,-1 \]Step 3: Eigenvectors.
For \(\lambda=2\): \[ A-2I= \begin{bmatrix} -1 & 1 & -1\\ 0 & 0 & 0\\ 0 & 1 & -3 \end{bmatrix} , \quad \mathbf{v}_1= \begin{bmatrix} 2\\ 3\\ 1 \end{bmatrix} \] For \(\lambda=1\): \[ A-I= \begin{bmatrix} 0 & 1 & -1\\ 0 & 1 & 0\\ 0 & 1 & 0 \end{bmatrix} , \quad \mathbf{v}_2= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \] For \(\lambda=-1\): \[ A-(-1)I= \begin{bmatrix} 2 & 1 & -1\\ 0 & 3 & 0\\ 0 & 1 & 0 \end{bmatrix} , \quad \mathbf{v}_3= \begin{bmatrix} 1\\ 0\\ 2 \end{bmatrix} \]General solution:
\[ X(t)= c_1 e^{2t}\mathbf{v}_1 + c_2 e^{t}\mathbf{v}_2 + c_3 e^{-t}\mathbf{v}_3 \]Math Modeling Systems problem.
For our last example, lets connect this to the models we were looking at in chapter 3. We will look at the interaction between two tanks of fluid and solve the corresponding homoegenous system.
Example. Two tanks each contain 100 gallons of brine. Let \(x_A(t)\) and \(x_B(t)\) be the number of pounds of salt in tanks A and B, respectively.
Flow rates:
- Pure water into tank A: \(2\) gal/min
- Out of tank A (to drain): \(1\) gal/min
- From A to B: \(2\) gal/min
- From B to A: \(1\) gal/min
- Out of tank B (to drain): \(1\) gal/min
- From B: \(1\) gal/min with concentration \(\frac{x_B}{100}\)
- Pure water contributes \(0\) salt
- To drain: \(1\) gal/min
- To B: \(2\) gal/min
- Total outflow: \(3\) gal/min with concentration \(\frac{x_A}{100}\)
- From A: \(2\) gal/min with concentration \(\frac{x_A}{100}\)
- To A: \(1\) gal/min
- To drain: \(1\) gal/min
- Total: \(2\) gal/min with concentration \(\frac{x_B}{100}\)
Lecture 26: Section 8.3 - Solving Nonhomogeneous Systems ↑ Back to top
Definition. Consider the nonhomogeneous system
\[
X' = AX + F(t),
\]
where \(A\) is an \(n \times n\) constant matrix and \(F(t)\) is a vector-valued function.
The general solution is given by
\[
X(t) = X_c(t) + X_p(t),
\]
where:
To find the general solution to a nonhomogeneous system \(X'=AX+F\) is the following:
- \(X_c(t)\) is the general solution to the associated homogeneous system \[ X' = AX, \]
- \(X_p(t)\) is any particular solution to the nonhomogeneous system \[ X' = AX + F(t). \]
- Find general solution to \(X' = AX\) via method used in previous section (eigenvalues/ eigenvectors). This is \(X_c\)
- Use a method to find a particular solution \(X_p\)
- Form the general solution
Methods for Finding \(X_p(t)\): In this section, we will isolate the method to finding the particular solution \(X_p\). We have two choices,
- Method of Undetermined Coefficients
- Method of Variation of Parameters
Method of Undetermined Coefficients
This method works when \(F(t)\) consists of functions such as polynomials, exponentials, sines, or cosines (or combinations of these).
The way to solve this is by inspecting \(F\) and choosing the most general form (using the guesses from 4.4) of each component.
Table of Guesses
| \(g(x)\) | Guess for \(y_p\) |
|---|---|
| Polynomial degree n | General polynomial degree n |
| \(e^{ax}\) | \(Ae^{ax}\) |
| \(\sin kx\), \(\cos kx\) | \(A\cos kx + B\sin kx\) |
| Sum/Difference/Product or functions above | Product of guesses |
Example. Solve the system
\[
X' =
\begin{bmatrix}
2 & 2 \\
1 & 3
\end{bmatrix}
X
+
\begin{bmatrix}
-8 \\
3
\end{bmatrix}.
\]
Notice the general solution to the homogeneous system \(X' = AX\) is
\[
X_c =
c_1 e^t
\begin{bmatrix}
-2 \\
1
\end{bmatrix}
+
c_2 e^{4t}
\begin{bmatrix}
1 \\
1
\end{bmatrix}.
\]
Now, we want to find a particular solution \(X_p\).
Since the forcing term is constant, we try a constant solution:
\[
X_p =
\begin{bmatrix}
a \\
b
\end{bmatrix}.
\]
Then
\[
X_p' =
\begin{bmatrix}
0 \\
0
\end{bmatrix}.
\]
Using the guess and the derivative of the guess, we plug into the differential equation to get
\[
X_p' = AX_p + F
\]
\[
\begin{bmatrix}
0 \\
0
\end{bmatrix}
=
\begin{bmatrix}
2 & 2 \\
1 & 3
\end{bmatrix}
\begin{bmatrix}
a \\
b
\end{bmatrix}
+
\begin{bmatrix}
-8 \\
3
\end{bmatrix}.
\]
Now we simplify:
\[
\begin{bmatrix}
2 & 2 \\
1 & 3
\end{bmatrix}
\begin{bmatrix}
a \\
b
\end{bmatrix}
=
\begin{bmatrix}
2a + 2b \\
a + 3b
\end{bmatrix}.
\]
So the equation becomes
\[
\begin{bmatrix}
0 \\
0
\end{bmatrix}
=
\begin{bmatrix}
2a + 2b \\
a + 3b
\end{bmatrix}
+
\begin{bmatrix}
-8 \\
3
\end{bmatrix}
=
\begin{bmatrix}
2a + 2b - 8 \\
a + 3b + 3
\end{bmatrix}.
\]
To find \(a\) and \(b\), we solve the system
\[
2a + 2b - 8 = 0 \quad \Rightarrow \quad 2a + 2b = 8
\]
\[
a + 3b + 3 = 0 \quad \Rightarrow \quad a + 3b = -3
\]
Simplify first equation:
\[
a + b = 4
\]
Now subtract:
\[
(a + 3b) - (a + b) = -3 - 4
\]
\[
2b = -7 \Rightarrow b = -\frac{7}{2}
\]
Substitute into \(a + b = 4\):
\[
a - \frac{7}{2} = 4 \Rightarrow a = \frac{15}{2}.
\]
Particular solution
\[
X_p =
\begin{bmatrix}
\frac{15}{2} \\
-\frac{7}{2}
\end{bmatrix}.
\]
General solution
\[
X =
c_1 e^t
\begin{bmatrix}
-2 \\
1
\end{bmatrix}
+
c_2 e^{4t}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
+
\begin{bmatrix}
\frac{15}{2} \\
-\frac{7}{2}
\end{bmatrix}.
\]
If we have a sum or product of functions, then we use the sum or product of the guesses.
Example. Solve the system
\[
X' =
\begin{bmatrix}
6 & 1 \\
4 & 3
\end{bmatrix}
X
+
\begin{bmatrix}
6t \\
-10t + 4
\end{bmatrix}.
\]
Step 1: Homogeneous solution
\[
X_c =
c_1 e^{7t}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
+
c_2 e^{2t}
\begin{bmatrix}
1 \\
-4
\end{bmatrix}.
\]
Step 2: Form of particular solution
Since \(F\) is polynomial, try
\[
X_p =
\begin{bmatrix}
at + b \\
ct + d
\end{bmatrix}.
\]
Compute \(X_p'\)
\[
X_p' =
\begin{bmatrix}
a \\
c
\end{bmatrix}.
\]
Compute \(AX_p\)
\[
AX_p =
\begin{bmatrix}
6 & 1 \\
4 & 3
\end{bmatrix}
\begin{bmatrix}
at + b \\
ct + d
\end{bmatrix}
=
\begin{bmatrix}
6(at+b) + (ct+d) \\
4(at+b) + 3(ct+d)
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
(6a+c)t + (6b+d) \\
(4a+3c)t + (4b+3d)
\end{bmatrix}.
\]
Plug into DE
\[
X_p' = AX_p + F
\]
\[
\begin{bmatrix}
a \\
c
\end{bmatrix}
=
\begin{bmatrix}
(6a+c)t + (6b+d) \\
(4a+3c)t + (4b+3d)
\end{bmatrix}
+
\begin{bmatrix}
6t \\
-10t + 4
\end{bmatrix}.
\]
Match coefficients
\[
0 = 6a + c + 6
\]
\[
a = 6b + d
\]
From bottom:
\[
0 = 4a + 3c - 10
\]
\[
c = 4b + 3d + 4
\]
Solve system
\[
6a + c + 6 = 0 \Rightarrow c = -6a - 6
\]
Substitute into
\[
4a + 3c - 10 = 0
\]
\[
4a + 3(-6a - 6) - 10 = 0
\]
\[
4a - 18a - 18 - 10 = 0
\]
\[
-14a - 28 = 0 \Rightarrow a = -2
\]
Then
\[
c = -6(-2) - 6 = 12 - 6 = 6
\]
Now solve for \(b,d\):
\[
a = 6b + d \Rightarrow -2 = 6b + d
\]
\[
c = 4b + 3d + 4 \Rightarrow 6 = 4b + 3d + 4
\]
\[
2 = 4b + 3d
\]
Solve system:
\[
-2 = 6b + d
\Rightarrow d = -2 - 6b
\]
Substitute:
\[
2 = 4b + 3(-2 - 6b)
\]
\[
2 = 4b - 6 - 18b
\]
\[
2 = -14b - 6
\Rightarrow 8 = -14b
\Rightarrow b = -\frac{4}{7}
\]
Then
\[
d = -2 - 6\left(-\frac{4}{7}\right)
= -2 + \frac{24}{7}
= \frac{10}{7}
\]
Step 3: Particular solution
\[
X_p =
\begin{bmatrix}
-2t - \frac{4}{7} \\
6t + \frac{10}{7}
\end{bmatrix}.
\]
Step 4: General solution
\[
X =
c_1 e^{7t}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
+
c_2 e^{2t}
\begin{bmatrix}
1 \\
-4
\end{bmatrix}
+
\begin{bmatrix}
-2t - \frac{4}{7} \\
6t + \frac{10}{7}
\end{bmatrix}.
\]
Variation of Parameters
Here is the idea behind variation of parameters. Once again, this is to find the particular solution \(X_p\). Let \(\{X_1, \cdots, X_n\}\) be a fundamental set of solutions to the system \(X' = AX\). Then the general solution can be written as \[ X = c_1X_1 + \cdots + c_nX_n = \begin{pmatrix} x_{11} & x_{12} & \cdots & x_{1n} \\ x_{21} & x_{22} & \cdots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{n1} & x_{n2} & \cdots & x_{nn} \end{pmatrix} \begin{pmatrix} c_1\\ c_2 \\ \vdots\\ c_n \end{pmatrix} \] Let \(\Phi(t) = \begin{pmatrix} x_{11} & x_{12} & \cdots & x_{1n} \\ x_{21} & x_{22} & \cdots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{n1} & x_{n2} & \cdots & x_{nn} \end{pmatrix}\) and suppose \(\det(\Phi(t)) \ne 0\). Then \[ X_p = \Phi(t)\int \Phi^{-1}(t)F(t) ~dt. \]
Remark.
Since we are assuming \(\det(\Phi(t)) \ne 0\) then linear algebra tells us the inverse matrix \(\Phi^{-1}(t)\) exists. In the \(2 \times 2\) case, there is a formula for the inverse we saw from the previous
lectures.
$$
A =
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
$$
$$
A^{-1} = \frac{1}{ad - bc}
\begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix}
$$
Example. Use variation of parameters to find a particular solution \(X_p\) to
\[
X' =
\begin{bmatrix}
2 & 2 \\
1 & 3
\end{bmatrix}
X
+
\begin{bmatrix}
-8 \\
3
\end{bmatrix}.
\]
Step 1: Fundamental matrix
From the homogeneous system (given previously), we have two solutions:
\[
X_1 = e^{4t}
\begin{bmatrix}
1 \\
1
\end{bmatrix}, \quad
X_2 = e^{t}
\begin{bmatrix}
2 \\
-1
\end{bmatrix}.
\]
Thus the fundamental matrix is
\[
\Phi(t) =
\begin{bmatrix}
e^{4t} & 2e^{t} \\
e^{4t} & -e^{t}
\end{bmatrix}.
\]
Step 2: Compute \(\Phi^{-1}(t)\)
\[
\det(\Phi) =
e^{4t}(-e^{t}) - 2e^{t}(e^{4t})
= -e^{5t} - 2e^{5t}
= -3e^{5t}.
\]
So,
\[
\Phi^{-1}(t)
= \frac{1}{-3e^{5t}}
\begin{bmatrix}
- e^{t} & -2e^{t} \\
- e^{4t} & e^{4t}
\end{bmatrix}.
\]
Step 3: Compute \(\Phi^{-1}(t)F(t)\)
\[
F(t) =
\begin{bmatrix}
-8 \\
3
\end{bmatrix}.
\]
\[
\Phi^{-1}(t)F(t)
=
\frac{1}{-3e^{5t}}
\begin{bmatrix}
- e^{t} & -2e^{t} \\
- e^{4t} & e^{4t}
\end{bmatrix}
\begin{bmatrix}
-8 \\
3
\end{bmatrix}.
\]
\[
\Phi^{-1}(t)F(t)
=
\frac{1}{-3e^{5t}}
\begin{bmatrix}
2e^{t} \\
11e^{4t}
\end{bmatrix}
=
\begin{bmatrix}
-\frac{2}{3}e^{-4t} \\
-\frac{11}{3}e^{-t}
\end{bmatrix}.
\]
Step 4: Integrate
\[
\int \Phi^{-1}(t)F(t)\,dt
=
\begin{bmatrix}
\int -\frac{2}{3}e^{-4t} dt \\
\int -\frac{11}{3}e^{-t} dt
\end{bmatrix}.
\]
\[
=
\begin{bmatrix}
\frac{1}{6}e^{-4t} \\
\frac{11}{3}e^{-t}
\end{bmatrix}.
\]
Step 5: Multiply by \(\Phi(t)\)
\[
X_p = \Phi(t)
\begin{bmatrix}
\frac{1}{6}e^{-4t} \\
\frac{11}{3}e^{-t}
\end{bmatrix}.
\]
\[
=
\begin{bmatrix}
e^{4t} & 2e^{t} \\
e^{4t} & -e^{t}
\end{bmatrix}
\begin{bmatrix}
\frac{1}{6}e^{-4t} \\
\frac{11}{3}e^{-t}
\end{bmatrix}.
\]
Compute:
Top:
\[
e^{4t}\cdot \frac{1}{6}e^{-4t} + 2e^{t}\cdot \frac{11}{3}e^{-t}
= \frac{1}{6} + \frac{22}{3}
= \frac{45}{6} = \frac{15}{2}.
\]
Bottom:
\[
e^{4t}\cdot \frac{1}{6}e^{-4t} - e^{t}\cdot \frac{11}{3}e^{-t}
= \frac{1}{6} - \frac{11}{3}
= -\frac{21}{6} = -\frac{7}{2}.
\]
Step 6: Particular solution
\[
X_p =
\begin{bmatrix}
\frac{15}{2} \\
-\frac{7}{2}
\end{bmatrix}.
\]