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

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

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., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

- The function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:

- y is the dependent variable (a function of the independent variable t),
- y′ and y′′ are the first and second derivatives of y with respect to t,
- t is the independent variable,
- A and B are constants.

This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.

In 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:

- x(t) and y(t) are the unknown functions of time t.
- a, b, c, and d d are constants.
- r
_{1}(t) and r_{2}(t) are known functions representing external inputs to the system

This system can be written in matrix form as: $\vec{X’} = A\vec{X}+\vec{r}(t)$ where:

- A = $(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})$ is the coefficient matrix.
- $\vec{X} = (\begin{smallmatrix}x\\ y\end{smallmatrix})$ is the vector of unknown functions.
- $\vec{r}(t)= (\begin{smallmatrix}r_1(t)\\ r_2(t)\end{smallmatrix})$ is the vector of nonhomogeneous terms. It represents the external inputs or forces acting on the system.

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:

- $\vec{x_c}$ is the complementary (or homogeneous) solution. This is the the general solution to the associated homogeneous system, $\vec{X'} = A\vec{X}$, where $\vec{r}(t)=0$. This gives us the complementary solution $\vec{x_c}$, which is based on the eigenvalues and eigenvectors of the matrix A. Represents the natural behavior of the system without external inputs.
- $\vec{x_p}$ is a particular solution to the inhomogeneous system. A solution that satisfies the full inhomogeneous system, including the non-homogeneous term $\vec{r}(t)$. It accounts for the response of the system due to external inputs $\vec{r}(t)$.

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: $\vec{x_{gen}} = \vec{x_c}+\vec{x_p}$

$\vec{x’_c}+\vec{x’_p}$ =

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. 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:

- Notice that in the tank 1, 2 is coming in (2 → 1), 3 is coming out (1 → 3), so some liquid need to come in from an external source to keep some balance or equilibrium. Liquid flows from Tank 2 to Tank 1 at a rate of 2 liters/hour.
- Notice that in the tank 2, 2 (2 → 1) + 2 (drain) are coming out, 3 (1 → 3) is coming in, so some liquid also need to come in from an external source. Liquid flows from Tank 1 to Tank 2 at a rate of 3 liters/hour.
- Tank 1 and 2 receive an external inflow of liquid at a rate of 1 liter/hour.
- Tank 2 has an outflow (drain) at a rate of 2 liters/hour.
- The external inflow into Tank 1 contains a salt concentration of 5e
^{-t}. - The external inflow into Tank 2 is pure water (zero salt concentration, that is, pure water).

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 v_{1}(t) and v_{2}(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:

- Right-side: [Since X’(t) = A(t) because each column of X(t) satisfies the homogeneous equation] $\vec{x’_p}(t) = X’(t)\vec{v}(t) + X(t)\vec{v’}(t) = AX(t)\vec{v}(t) + X(t)\vec{v’}(t)$
- Left-side: $A\vec{x_p} + \vec{r}(t) = AX(t)\vec{v} + \vec{r}(t)$

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: λ_{1}=-1, λ_{2} = -6.

**Eigenvectors**: For λ_{1}=-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 λ_{1}=-1 is $\vec{v_1} = (\begin{smallmatrix}1\\ 1\end{smallmatrix})$

For λ_{2}=-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 λ_{2}=-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 C_{1} = C_{2} = 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 c_{1} and c_{2} determined by initial conditions.

- Solve $\begin{cases} x’ = -2x +y\\ y’ = x -2y + 3t \end{cases}$

Step 1: **Write the System in Matrix Form**

This system can be written 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: λ_{1} = -1, λ_{2} = -3.

**Find Eigenvectors**: For λ_{1}=-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 λ_{2}=-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: $v_{2x} +v_{2y} = 0 ⇒ v_{2y} = -v_{2y}, v_{2x}= [\text{We choose for simplicity}] 1 ⇒ v_{2y} = -1$

$v_2 = (\begin{smallmatrix}1\\ -1 \end{smallmatrix})$

**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 $\vec{x_1} = (\begin{smallmatrix}e^{-t}\\ c_1e^{-t}\end{smallmatrix})$ 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 = e^{t}dt. Then du = dt, v = e^{t}.

$\int te^tdt = te^t -\int e^tdt = te^t-e^t + C$

$v_1(t) = \frac{3}{2}\int te^tdt = \frac{3}{2}(te^t-e^t)$. 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 = e^{3t}dt. 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 $\vec{x}(t) = \vec{x_h}(t)+\vec{x_p}(t) = c_1e^{-t}(\begin{smallmatrix}1\\ 1\end{smallmatrix}) + c_2e^{-3t}(\begin{smallmatrix}1\\ -1\end{smallmatrix}) + (\begin{smallmatrix}t-\frac{4}{3}\\ 2t -\frac{5}{3}\end{smallmatrix})$

We aim to find a particular solution $\vec{x_p}(t)$ that mimics the form of the nonhomogeneous term $\vec{r}(t) = (\begin{smallmatrix}0\\ 3t\end{smallmatrix})$

Assume a particular solution $\vec{x_p}(t) = at + b$, appropriate for a linear nonhomogeneous term, where a and b are constant vectors to be determined.

$x’_p(t) =[\text{Since a and b are constants:}] a =[\text{Substitute into the Differential Equation}] A(at + b) + \vec{r}(t) = A(at + b) + t(\begin{smallmatrix}0\\ 3\end{smallmatrix})$

Since a is constant, equate the coefficients of like terms on both sides.

$\begin{cases} Ab = a \\ Aa + (\begin{smallmatrix}0\\ 3\end{smallmatrix}) = 0 \end{cases}$

$Aa + (\begin{smallmatrix}0\\ 3\end{smallmatrix}) = 0 ↭ (\begin{smallmatrix}-2 & 1\\ 1 & -2\end{smallmatrix})a = (\begin{smallmatrix}0\\ -3\end{smallmatrix}) ↭ a = \frac{1}{3}(\begin{smallmatrix}-2 & -1\\ -1 & -2\end{smallmatrix})(\begin{smallmatrix}0\\ -3\end{smallmatrix}) = (\begin{smallmatrix}1\\ 2\end{smallmatrix})$

$b = A^{-1}a = \frac{1}{3}(\begin{smallmatrix} -2 & -1 \\ -1 & -2 \end{smallmatrix})(\begin{smallmatrix}1\\ 2\end{smallmatrix}) = \frac{1}{3}(\begin{smallmatrix}-4\\ -5\end{smallmatrix})$

x_{p}(t) = at + 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.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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