All warfare is based on deception, Sun tzu, The Art of War
War is a racket. It always has been. It is possibly the oldest, easily the most profitable, surely the most vicious. It is the only one international in scope. It is the only one in which the profits are reckoned in dollars and the losses in lives, Smedley Butler
Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$
It involves (e.g., $\frac{dy}{dx} = 3x +5y$):
Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:
These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.
If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.
The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:
Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x0, y0), meaning that it satisfies the initial condition y(x0) = y0.
This theorem ensures that under these conditions, the solution exists and is unique near x = x0.
In the study of differential equations, many real-world problems can be modeled using systems of equations that include external inputs or forces. These are known as inhomogeneous systems. Understanding how to solve such systems is crucial in fields like engineering, physics, and applied mathematics.
We will explore how to solve a system of inhomogeneous differential equations using matrix methods.
Consider the following system of first-order linear differential equations: $\begin{cases} x’ = ax + by + r_1(t) \\ y’ = cx + dy + r_2(t) \end{cases}$ where:
This system can be written in matrix form as: $\vec{X’} = A\vec{X}+\vec{r}(t)$ where:
Theorem. General Solution. The general solution to the inhomogeneous system $\vec{X'} = A\vec{X}+\vec{r}(t)$ is given by: $\vec{x_{gen}} = \vec{x_c}+\vec{x_p}$ where:
Therefore, the general solution is the sum of the complementary and particular solutions: $\vec{x_{gen}} = \vec{x_c}+\vec{x_p}$.
Proof:
Let $\vec{x_c},\vec{x_p}$ be complementary and particular solution, respectively, that satisfies: $\vec{X’} = A\vec{X},\text{ and} \vec{X’} = A\vec{X}+\vec{r}(t)$
We propose that the general solution can be expressed as: $\vec{x_{gen}} = \vec{x_c}+\vec{x_p}$
$\vec{x_{gen}} = \vec{x’_c}+\vec{x’_p}$ =
Since $\vec{x_c}$ is a solution to the inhomogeneous equation, and $\vec{x_p}$ is a solution to the homogeneous associated equation:
= $A\vec{x_c}+A\vec{x_p}+\vec{r}(t) = A(\vec{x_c}+\vec{x_p})+\vec{r}(t) = A\vec{x_{gen}}+\vec{r}(t)$ which shows that $\vec{x_{gen}}$ satisfies the inhomogeneous system.
Consider a system involving two interconnected tanks, each with a capacity of 1 liter. The tanks are labeled Tank 1 and Tank 2 (Why Originality is King?) They are connected by pipes that allow liquid to flow between them at specified rates. We are interested in determining how the amount of salt in each tank changes over time.
As it is illustrated, 2 indicates that the liquid flows from the right tank, 2, to the left tank, 1; 3: the liquid flows from the left tank 1 to the right tank 2. Both tanks have a external inflow port to fill in the container and only the one on the right, the 2 container, have a an outflow (drain) to release its content. The flow rates are measured in liters/hour. x and y describes the amount of salt (in grams) in tank 1 and 2 respectively.
Flow Rates:
Rate of change of salt in Tank 1: $\frac{dx}{dt} = Inflow − Outflow = -3x +2y + 5e^{-t}$.
Rate of change of salt in Tank 2: $\frac{dy}{dt} = Inflow − Outflow =3x -4y + 0$.
Thus, we have the system:
$\begin{cases} x’ = -3x +2y + 5e^{-t} \\ y’ = 3x -4y + 0 \end{cases}$ where x(t) represents the amount of salt in tank 1, and y(t) represents the amount of salt in tank 2 (Refer to Figure vi for a visual representation and aid in understanding it).
This is an inhomogeneous system. This system can be written in matrix form as: $\vec{X’} = A\vec{X} + \vec{r}(t) = (\begin{smallmatrix}-3 & 2\\ 3 & -4\end{smallmatrix})\vec{X}+(\begin{smallmatrix}5e^{-t}\\ 0\end{smallmatrix})$
We can use the variation of parameters method to find a particular solution $\vec{x_p}(t)$ to the inhomogeneous system.
To find a particular solution to the inhomogeneous system, we assume that the particular solution has the form: $\vec{x_p} = v_1(t)\vec{x_1}+v_2(t)\vec{x_2}$ where $\vec{x_1}$ and $\vec{x_2}$ are the independent solutions of the homogeneous system (the eigenvectors), and v1(t) and v2(t) are unknown functions to be determined.
Notice that $(\begin{smallmatrix}x_1 & x_2\\ y_1 & y_2\end{smallmatrix})(\begin{smallmatrix}v_1\\ v_2\end{smallmatrix})↭[\text{We can write this as:}] \vec{x_p} = X\vec{v}$ where X is the fundamental matrix formed by the columns $\vec{x_1}$ and $\vec{x_2}$.
Now, differentiate $\vec{x_p}(t)$: $X’(t)\vec{v}(t)+X(t)\vec{v’}(t)$
Substituting into the original system, we get:
Subtract $AX(t)\vec{v}(t)$ from both sides: $X\vec{v’} = \vec{r} ⇒[\text{We solve for } \vec{v’}] \vec{v’} = X^{-1}(t)\vec{r}$ where X-1 is the inverse of the fundamental matrix X(t), which exists because the determinant of X(t) is non zero (the solutions $\vec{x_1}$ and $\vec{x_2}$ are linearly independent).
To find $\vec{v}(t)$, we integrate: $\vec{v}(t) = \int X^{-1}\vec{r}dt$.
Finally, the particular solution is: $\vec{x_p} = X\vec{v} = X \int X^{-1}\vec{r}dt$. We don’t need to add constants of integration because they are already included in the complementary solution.
We are given the system of differential equations representing the amount of salt in two connected tanks: $\begin{cases} x’ = -3x +2y + 5e^{-t} \\ y’ = 3x -4y + 0 \end{cases}$
This system can be written in matrix form as: $\vec{X’} = A\vec{X} + \vec{r}(t)$ where:
$A = (\begin{smallmatrix}-3 & 2\\ 3 & -4\end{smallmatrix}), \vec{r}(t) = (\begin{smallmatrix}5e^{-t}\\ 0\end{smallmatrix})$
Step 1. Solve the Homogeneous System
To solve this, we need to find the eigenvalues and eigenvectors of the matrix A = $(\begin{smallmatrix}-3 & 2\\ 3 & -4\end{smallmatrix})$
The characteristic equation is obtained from: det(A -λI) = 0 where I is the identity matrix. So:
$det(\begin{smallmatrix}-3-λ & 2\\ 3 & -4-λ\end{smallmatrix}) = 0 ↭[\text{Expanding the determinant:}] (−3−λ)(−4−λ)−(2)(3)=0 ↭ λ^2+7λ +6 = 0$
Using the quadratic formula $λ = \frac{-7±\sqrt{49-24}}{2·1} = \frac{-7±\sqrt{25}}{2} = \frac{-7±5}{2}$. Thus, the eigenvalues are: λ1=-1, λ2 = -6.
Eigenvectors: For λ1=-1, we solve $(A-λ_1I)\vec{v_1} = 0 ↭ A + I = 0 ↭ (\begin{smallmatrix}-2 & 2\\ 3 & -3\end{smallmatrix})(\begin{smallmatrix}v_{1x}\\ v_{1y}\end{smallmatrix}) = (\begin{smallmatrix} 0 \\ 0 \end{smallmatrix})$
Set up the homogeneous system: $\begin{cases} -2v_{1x} +2v_{1y} = 0 \\ -3v_{1x} -3v_{1y} = 0 \end{cases}$
From the first row: $-2v_{1x} +2v_{1y} = 0⇒v_{1x} = v_{2x}$. Thus, an eigenvector for λ1=-1 is $\vec{v_1} = (\begin{smallmatrix}1\\ 1\end{smallmatrix})$
For λ2=-6, we solve $(A-λ_1I)\vec{v_2} = 0 ↭ (\begin{smallmatrix}3 & 2\\ 3 & 2\end{smallmatrix})(\begin{smallmatrix}v_{2x}\\ v_{2y}\end{smallmatrix}) = (\begin{smallmatrix} 0 \\ 0 \end{smallmatrix})$
Set up the homogeneous system: $\begin{cases} 3v_{2x} +2v_{2y} = 0 \\ 3v_{2x} +2v_{2y} = 0 \end{cases}$
From the first row: $3v_{2x} +2v_{2y} = 0⇒v_{2x} = \frac{-2}{3}v_{2y}$. Thus, an eigenvector for λ2=-6 is $\vec{v_2} = (\begin{smallmatrix}-2\\ 3\end{smallmatrix})$
The general solution to the homogeneous system is: $\vec{x_c} = c_1e^{-t}(\begin{smallmatrix}1\\ 1\end{smallmatrix}) + c_2e^{-6t}(\begin{smallmatrix}-2\\ 3\end{smallmatrix})$. This represents the complementary solution.
Step 2. Find the particular solution using variation of parameters.
We assume the particular solution is of the form: $\vec{x_p} = v_1(t)\vec{x_1}+v_2(t)\vec{x_2}$ where $\vec{x_1} = (\begin{smallmatrix}e^{-t}\\ e^{-t}\end{smallmatrix})$ and $\vec{x_2} = (\begin{smallmatrix}-2e^{-6t}\\ 3e^{-6t}\end{smallmatrix})$ are the solutions to the homogeneous system, $\vec{v} = (\begin{smallmatrix}v_1(t)\\ v_2(t)\end{smallmatrix})$
Thus, $\vec{x_p}(t) = (\begin{smallmatrix}e^{-t} & -2e^{-6t}\\ e^{-t} & 3e^{-6t}\end{smallmatrix})(\begin{smallmatrix}v_1(t)\\ v_2(t)\end{smallmatrix}) = X(\begin{smallmatrix}v_1(t)\\ v_2(t)\end{smallmatrix}) = X(t)\vec{v}(t)$
Differentiate $\vec{x_p}(t): \vec{x_p’}(t) = X’(t)\vec{v}(t) + X(t)\vec{v’}(t)$
Since each column of X(t) is a solution to the homogeneous system, we have X’(t) = AX(t). Therefore, $\vec{x_p’}(t) = X’(t)\vec{v}(t) + X(t)\vec{v’}(t) = AX(t)\vec{v}(t) + X(t)\vec{v’}(t)$
Substitute $\vec{x_p}(t), \vec{x_p’}(t)$ into the nonhomogeneous equation: $AX(t)\vec{v}(t) + X(t)\vec{v’}(t) = AX(t)\vec{v}(t) + \vec{r}(t) ↭ X(t)\vec{v’}(t) = \vec{r}(t) ↭ \vec{v’}(t) = X^{-1}(t)\vec{r}(t)$
Compute X-1(t)
det(X(t)) = $e^{-t}·3e^{-6t}+2e^{-6t}·e^{-t} = 3e^{-7t}+2e^{-7t} = 5e^{-7t}$
$X^{-1}(t) = \frac{1}{det(X(t))}(\begin{smallmatrix}x_{22}(t) & -x_{12}(t)\\ -x_{21}(t) & x_{11}(t)\end{smallmatrix}) = \frac{1}{5e^{-7t}} (\begin{smallmatrix}3e^{-6t} & 2e^{-6t}\\ -e^{-t} & e^{-t}\end{smallmatrix})$
Compute $\vec{v’}(t) = X^{-1}(t)\vec{r}(t) = \frac{1}{5e^{-7t}} (\begin{smallmatrix}3e^{-6t} & 2e^{-6t}\\ -e^{-t} & e^{-t}\end{smallmatrix})(\begin{smallmatrix}5e^{-t}\\ 0\end{smallmatrix})$
First component: $v_1(t) = \int v_1’(t)dt = \int \frac{1}{5e^{-7t}}·3e^{-6t}·5e^{-t}dt = 3 \int dt = 3t + C_1$
Second component: $v_2(t) = \int v_2’(t)dt = \int \frac{1}{5e^{-7t}}·(-e^{-t})·5e^{-t}dt = \int -e^{5t}dt = \frac{-1}{5}e^{5t}+ C_2$. We can set C1 = C2 = 0 since any constants will be absorbed into the homogeneous solution.
Thus, the particular solution is: $\vec{x_p} = X(t)\vec{v}(t) = (\begin{smallmatrix}e^{-t} & -2e^{-6t}\\ e^{-t} & 3e^{-6t}\end{smallmatrix})(\begin{smallmatrix}3t\\ \frac{-1}{5}e^{5t}\end{smallmatrix}) = (\begin{smallmatrix}3te^{-t}+(-2e^{-6t})(\frac{-1}{5}e^{5t})\\ 3te^{-t}+3e^{-6t}(\frac{-1}{5}e^{5t})\end{smallmatrix})= (\begin{smallmatrix}3te^{-t}+\frac{2}{5}e^{-t}\\ 3te^{-t}+\frac{-3}{5}e^{-t}\end{smallmatrix}) = e^{-t} (\begin{smallmatrix}3t+\frac{2}{5}\\ 3t+\frac{-3}{5}\end{smallmatrix})$
Step 3: General Solution
The general solution to the inhomogeneous system is the sum of the complementary and particular solutions:
$\vec{x_{gen}} = \vec{x_c} + \vec{x_p} = c_1e^{-t}(\begin{smallmatrix}1\\ 1\end{smallmatrix}) + c_2e^{-6t}(\begin{smallmatrix}-2\\ 3\end{smallmatrix}) + e^{-t} (\begin{smallmatrix}3t+\frac{2}{5}\\ 3t+\frac{-3}{5}\end{smallmatrix})= e^{-t}(\begin{smallmatrix}c_1+3t+\frac{2}{5}\\ c_1+3t+\frac{-3}{5}\end{smallmatrix})+ c_2e^{-6t}(\begin{smallmatrix}-2\\ 3\end{smallmatrix})$
The general solution provides the amounts of salt x(t) and y(t) in tanks 1 and 2 at time t, with constant c1 and c2 determined by initial conditions.
Step 1: Write the System in Matrix Form
We can express the system in matrix form as: $\vec{X’} = A\vec{X} + \vec{r}(t)$ where:
$\vec{X} = (\begin{smallmatrix}x\\ y\end{smallmatrix}), A = (\begin{smallmatrix}-2 & 1\\ 1 & -2\end{smallmatrix}), \vec{r}(t) = (\begin{smallmatrix}0\\ 3t\end{smallmatrix})$
Step 2. Solve the Homogeneous System
Compute the eigenvalues λ by solving the characteristic equation: det(A −λI) = 0 ↭ $det(\begin{smallmatrix}-2-λ & 1\\ 1 & -2-λ\end{smallmatrix}) = 0 ↭ (λ+2)^2-1 = 0 ↭ (λ+2−1)(λ+2+1)=(λ+1)(λ+3)=0.$.
Therefore, the eigenvalues are: λ1 = -1, λ2 = -3.
Find the Eigenvectors Corresponding to Each Eigenvalue: For λ1=-1, we solve $(A-λ_1I)\vec{v_1} = \vec{0} ↭ A + I\vec{v_1} = \vec{0} ↭ (\begin{smallmatrix}-1 & 1\\ 1 & -1\end{smallmatrix})(\begin{smallmatrix}v_{1x}\\ v_{1y}\end{smallmatrix}) = (\begin{smallmatrix} 0 \\ 0 \end{smallmatrix})$
Set up the homogeneous system: $\begin{cases} -v_{1x} +v_{1y} = 0 \\ v_{1x} -v_{1y} = 0 \end{cases}$
Form the first equation: $-v_{1x} +v_{1y} = 0 ⇒ v_{1x} = v_{1y} =[\text{Simplicity}] 1$
$v_1 = (\begin{smallmatrix}1\\1\end{smallmatrix})$
For λ2=-3, we solve $(A-λ_2I)\vec{v_2} = \vec{0} ↭ A + 3I\vec{v_2} = \vec{0} ↭ (\begin{smallmatrix}1 & 1\\ 1 & 1\end{smallmatrix})(\begin{smallmatrix}v_{2x}\\ v_{2y}\end{smallmatrix}) = (\begin{smallmatrix} 0 \\ 0 \end{smallmatrix})$
Set up the homogeneous system: $\begin{cases} v_{2x} +v_{2y} = 0 \\ v_{2y} + v_{1y} = 0 \end{cases}$
Form the first equation: $v_{2x} +v_{2y} = 0 ⇒ v_{2y} = -v_{2y}, v_{2x}= [\text{We choose for simplicity}] 1 ⇒ v_{2y} = -1$
$v_2 = (\begin{smallmatrix}1\\ -1 \end{smallmatrix})$
Write the General Solution to the Homogeneous System
The homogeneous system has a general solution, $\vec{X_h}(t) = c_1e^{λ_1t}\vec{v_1} + c_2e^{λ_2t}\vec{v_2} = c_1e^{-t}(\begin{smallmatrix}1\\ 1\end{smallmatrix}) + c_2e^{-3t}(\begin{smallmatrix}1\\ -1\end{smallmatrix})$
Thus, the fundamental matrix X(t) is: $X(t) = (\begin{smallmatrix}x_1^{1}(t) & x_2^{1}(t)\\ x_1^{2}(t) & x_2{2}(t)\end{smallmatrix}) = (\begin{smallmatrix}e^{-t} & e^{-3t}\\ e^{-t} & -e^{-3t}\end{smallmatrix})$
Step 3. Find the particular solution using variation of parameters.
We assume the particular solution is of the form: $\vec{x_p} = v_1(t)\vec{x_1}(t)+v_2(t)\vec{x_2}(t)$ where $\vec{x_1} = (\begin{smallmatrix}e^{-t}\\ e^{-t}\end{smallmatrix})$ and $\vec{x_2} = (\begin{smallmatrix}e^{-3t}\\ -e^{-3t}\end{smallmatrix})$ are the solutions to the homogeneous system.
$\vec{x_p} = (\begin{smallmatrix}e^{-t} & e^{-3t}\\ e^{-t} & -e^{-3t}\end{smallmatrix})(\begin{smallmatrix}v_1(t)\\ v_2(t)\end{smallmatrix}) = X\vec{v}$ where X is the fundamental matrix formed by the columns $\vec{x_1}$ and $\vec{x_2}$
We know (Method to find a particular solution to $\vec{X’} = A\vec{X} + \vec{r}$): $\vec{v’} = X^{-1}·\vec{r}$
Compute X-1(t)
det(X(t)) = $e^{-t}·(-e^{-3t}) -e^{-t}·e^{-3t} = -e^{-4t}-e^{-4t} = -2e^{-4t}$
Since det(X(t)) ≠ 0 for all t, X(t) is invertible.
For a 2 x 2 matrix X = $(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})$, the inverse is $\frac{1}{det(X)}(\begin{smallmatrix}d & -b\\ -c & a\end{smallmatrix})$
Thus, $X^{-1}(t) = \frac{1}{-2e^{-4t}}(\begin{smallmatrix}-e^{-3t} & -e^{-3t}\\ -e^{-t} & e^{-t}\end{smallmatrix})$
Compute $\vec{v’} = X^{-1}·\vec{r} = \frac{1}{-2e^{-4t}}(\begin{smallmatrix}-e^{-3t} & -e^{-3t}\\ -e^{-t} & e^{-t}\end{smallmatrix})·(\begin{smallmatrix}0\\ 3t\end{smallmatrix}) = \frac{1}{-2e^{-4t}}(\begin{smallmatrix}-3te^{-3t}\\ 3te^{-t}\end{smallmatrix})$
$v_1(t) = \int v_1’(t) = \int \frac{-3te^{-3t}}{-2e^{-4t}}dt = \int \frac{3te^{t}}{2}dt = \frac{3}{2} \int te^tdt$
Use integration by parts u = t, dv = etdt. Then du = dt, v = et.
$\int te^tdt = te^t -\int e^tdt = te^t-e^t + C$
$v_1(t) = \frac{3}{2}\int te^tdt = \frac{3}{2}(te^t-e^t)$. We can set the constant C = 0 because constants are accounted for in the homogeneous solution.
$v_2(t) = \int v_2’(t) = \int \frac{3te^{-t}}{-2e^{-4t}}dt = \int \frac{-3te^{3t}}{2}dt = \frac{-3}{2} \int te^{3t}dt$
Use integration by parts u = t, dv = e3tdt. Then du = dt, v = $\frac{e^{3t}}{3}$
$\int te^{3t}dt = t\frac{e^{3t}}{3} -\int \frac{e^{3t}}{3}dt = \frac{te^{3t}}{3} -\frac{e^{3t}}{9} + C$
$v_2(t) = \frac{-3}{2} \int te^{3t}dt = \frac{-3}{2}(\frac{te^{3t}}{3} -\frac{e^{3t}}{9}) = -(\frac{te^{3t}}{2}-\frac{e^{3t}}{6})$. We can set the constant C = 0 because constants are accounted for in the homogeneous solution.
Construct the Particular Solution
$\vec{x_p} = X(t)\vec{v}(t) = (\begin{smallmatrix}e^{-t} & e^{-3t}\\ e^{-t} & -e^{-3t}\end{smallmatrix})(\begin{smallmatrix}\frac{3}{2}(te^t-e^t)\\ -(\frac{te^{3t}}{2}-\frac{e^{3t}}{6})\end{smallmatrix})$
First component: $e^{-t}·\frac{3}{2}(te^t-e^t) + e^{-3t}·(-(\frac{te^{3t}}{2}-\frac{e^{3t}}{6})) = \frac{3}{2}(t-1)+ (\frac{-t}{2}+\frac{1}{6}) = \frac{9t-9-3t+1}{6} = \frac{6t-8}{6} = t-\frac{4}{3}$
Second component: $e^{-t}·\frac{3}{2}(te^t-e^t)+e^{-3t}·(\frac{te^{3t}}{2}-\frac{e^{3t}}{6}) = \frac{3}{2}(t-1)+(\frac{1}{2}t-\frac{1}{6}) = \frac{9t-9+3t-1}{6} = \frac{12t-10}{6} = 2t -\frac{5}{3}$
Step 4. Write the General Solution
The general solution $\vec{x}(t) = \vec{x_h}(t)+\vec{x_p}(t) = c_1e^{-t}(\begin{smallmatrix}1\\ 1\end{smallmatrix}) + c_2e^{-3t}(\begin{smallmatrix}1\\ -1\end{smallmatrix}) + (\begin{smallmatrix}t-\frac{4}{3}\\ 2t -\frac{5}{3}\end{smallmatrix})$. This is the general solution to the original system.
We aim to find a particular solution $\vec{x_p}(t)$ that mimics the form of the nonhomogeneous term $\vec{r}(t) = (\begin{smallmatrix}0\\ 3t\end{smallmatrix})$ which is linear in t.
Assume a particular solution $\vec{x_p}(t) = \vec{a}t + \vec{b}$, appropriate for a linear nonhomogeneous term, where $\vec{a}$ and $\vec{b}$ are constant vectors to be determined.
$x’_p(t) =[\text{Since a and b are constants:}] \vec{a} =[\text{Substitute into the Differential Equation}] A(\vec{a}t + \vec{b}) + \vec{r}(t) = A(\vec{a}t + \vec{b}) + t(\begin{smallmatrix}0\\ 3\end{smallmatrix}) ↭ \vec{a} = A(\vec{a}t + \vec{b}) + t(\begin{smallmatrix}0\\ 3\end{smallmatrix})$
Since $\vec{a}$ is constant, group like terms on both sides.
$\begin{cases} A\vec{b} = \vec{a} \\ A\vec{a} + (\begin{smallmatrix}0\\ 3\end{smallmatrix}) = 0 \end{cases}$
$A\vec{a} + (\begin{smallmatrix}0\\ 3\end{smallmatrix}) = 0 ↭ (\begin{smallmatrix}-2 & 1\\ 1 & -2\end{smallmatrix})\vec{a} = (\begin{smallmatrix}0\\ -3\end{smallmatrix}) ↭ \vec{a} = A^{-1}(\begin{smallmatrix}0\\ -3\end{smallmatrix}) = \frac{1}{3}(\begin{smallmatrix}-2 & -1\\ -1 & -2\end{smallmatrix})(\begin{smallmatrix}0\\ -3\end{smallmatrix}) = (\begin{smallmatrix}1\\ 2\end{smallmatrix})$
$A\vec{b} = \vec{a} ~(ii) ⇒\vec{b} = A^{-1}\vec{a} = \frac{1}{3}(\begin{smallmatrix} -2 & -1 \\ -1 & -2 \end{smallmatrix})(\begin{smallmatrix}1\\ 2\end{smallmatrix}) = \frac{1}{3}(\begin{smallmatrix}-4\\ -5\end{smallmatrix})$
xp(t) $= \vec{a}t + \vec{b} = (\begin{smallmatrix}1\\ 2\end{smallmatrix})t + (\begin{smallmatrix} \frac{-4}{3} \\ \frac{-5}{3} \end{smallmatrix})$. This particular solution matches the one found earlier.