An expert is a person who has made all the mistakes that can be made in a very narrow field, Niels Bohr

Mistakes are a great educator when one is honest enough to admit them and willing to learn from them, Aleksandr Solzhenitsynv

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 the study of linear systems of differential equations, we often encounter the equation: x′(t) = Ax(t) where x(t) is a vector of functions, and A is a constant n × n matrix. When solving such systems, finding the eigenvalues and eigenvectors of the matrix A is a crucial step.

In cases where A has distinct real eigenvalues, λ_{1} and λ_{2} with corresponding eigenvectors v_{1} and v_{2}, the general solution is: $x = C_1e^{λ_1t}v_1 + C_2e^{λ_2t}v_2$.

However, when the eigenvalues of A are complex, the method requires additional steps to obtain real-valued solutions.

Suppose A has a pair of complex conjugate eigenvalues, λ = a ± ib with a corresponding complex eigenvector v. Since the system x′(t) = Ax(t) is real, we seek real-valued solutions. We can achieve this by utilizing Euler’s formula and separating the complex solution into its real and imaginary parts.

Theorem: Given a real n x n matrix A with a complex eigenvalue λ = a + ib and corresponding complex eigenvector $\vec{v}$, the general real solution to the system of differential equations x′ = Ax is: $x(t) = e^{at}[C_1· Re(e^{ibt}\vec{v}) + C_2Im(e^{ibt}\vec{v})]$.

- Finding Eigenvalues and Eigenvectors. Solve the characteristic equation det(A− λI) = 0 to find eigenvalues λ. If λ = a +ib is a complex eigenvalue, its complex conjugate λ = a −ib is also an eigenvalue of A because A has real entries. Find a corresponding complex eigenvector v by solving (A −λI)v=0.
- Constructing the Complex Solution. The complex solution to the system is x
_{complex}(t) = e^{λt}v = e^{at}(e^{ibt}v). - Separating into Real and Imaginary Parts. Use Euler’s Formula : e
^{ibt}= cos(bt) + isin(bt). Express v in terms of its real and imaginary parts: v = v_{real}+ iv_{imaginary}. Multiply e^{ibt}and v to obtain the real and imaginary components.x

_{complex}(t) = e^{λt}v = e^{at}(e^{ibt}v) = e^{at}(cos(bt) + isin(bt))(v_{real}+ iv_{imaginary}). - Forming Real-Valued Solutions. The real part x
_{real}(t) = $e^{at}[cos(bt)v_{real}-sin(bt)v_{imaginary}]$ is a real solution. The imaginary part x_{imaginary}(t) = $e^{at}[sin(bt)v_{real}+cos(bt)v_{imaginary}]$ is also a real solution. Both x_{real}(t) and x_{imaginary}(t) satisfy the differential equation. - General Solution. The general real solution is a linear combination of the real and imaginary parts: x(t) = C
_{1}x_{real}(t) + C_{2}x_{imaginary}(t).

$\begin{cases} x’ = 2x + -5y \\ y’ = x -2y \end{cases}$

Our goal is to solve this system and find expressions for x(t) and y(t).

We can express this system as a matrix equation: x’ = Ax where x = $\left( \begin{smallmatrix}x\\ y\end{smallmatrix} \right)$ and the matrix A is $\left( \begin{smallmatrix}2 & -5\\ 1 & -2\end{smallmatrix} \right)$

The first step in solving this system is to find the eigenvalues of matrix A. The characteristic equation is given by: |A - λI| = $\left| \begin{smallmatrix}2-λ & -5\\ 1 & -2-λ\end{smallmatrix} \right| = (2-λ)(-2-λ)+5 = -4-2λ +2λ +λ^2 + 5 = λ^2 + 1 = 0 ⇒[\text{Solving for λ}] λ = ± i$. The eigenvalues are complex conjugates.

Now we solve for the eigenvector to λ = i:

$(A−λI)v=0 ↭ \left( \begin{smallmatrix}2-i & -5\\ 1 & -2-i\end{smallmatrix} \right)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

This leads to the following system of equations:

$\begin{cases} (2-i)a_1 + -5a_2 = 0 \\ a_1 + (-2-i)a_2 = 0 \end{cases}$

From the second equation (ii), $a_1 = (2+i)a_2 ⇒[\text{Replace this in (i)}] (2-i)(2+i)a_2 -5a_2 = 0 ↭ (4 -i^2)a_2 -5a_2 = 0 ↭ 5a_2 -5a_2 = 0,$ so any a_2 is a solution. Therefore, for simplicity, let’s choose $a_2 = 1, a_1 = (2 +i)$. Eigenvector Corresponding to λ=i: $v = (\begin{smallmatrix}2+i\\1\end{smallmatrix})$

The complex solution to the system is given by: $x(t)= e^{λt}v = e^{it}(\begin{smallmatrix}2+i\\1\end{smallmatrix}) =[\text{Using Euler’s formula}] (cos(t)+isin(t))(\begin{smallmatrix}2+i\\1\end{smallmatrix}) = (\begin{smallmatrix}(cos(t)+isin(t))(2+i)\\cos(t)+isin(t)\end{smallmatrix}) = \text{Simplifying, we get} (\begin{smallmatrix}(2cos(t)-sin(t))+i(cos(t)+2sin(t))\\cos(t)+isin(t)\end{smallmatrix}) $

Combining the real and imaginary parts, we get two real-valued solutions:

x_{real}(t) = $(\begin{smallmatrix}2cos(t)-sin(t)\\cos(t)\end{smallmatrix})$

x_{imaginary}(t) = $(\begin{smallmatrix}cos(t)+2sin(t)\\sin(t)\end{smallmatrix})$

Thus, the general solution to the system is a linear combination of the real and imaginary parts:

x(t) = c_{1}x_{real}(t) + c_{2}x_{imaginary}(t) = $c_1(\begin{smallmatrix}2cos(t)-sin(t)\\cos(t)\end{smallmatrix})+c_2(\begin{smallmatrix}cos(t)+2sin(t)\\sin(t)\end{smallmatrix})$

Intuition:

- The eigenvalues are purely imaginary (λ = ±i), indicating that the system exhibits oscillatory behavior without exponential growth or decay.
- The solutions involve sine and cosine functions, reflecting periodic oscillations.
- c
_{1}and c_{2}are constants determined by the initial conditions of the system.

Consider a model of a love affair between Susan and George. Let x(t) represents Susan’s love for George, and y(t) represents George’s love for Susan. Their love dynamics are modeled by the following system of differential equations:

$\begin{cases} x’ = x + 2y \\ y’ = -x -y \end{cases}$

Our goal is to solve this system and find expressions for x(t) and y(t).

We can express this system as a matrix equation: x’ = Ax where x = $\left( \begin{smallmatrix}x\\ y\end{smallmatrix} \right)$ and the matrix A is $\left( \begin{smallmatrix}1 & +2\\ -1 & -1\end{smallmatrix} \right)$

Thus, the system becomes: $\left( \begin{smallmatrix}x’\\ y’\end{smallmatrix} \right) = \left( \begin{smallmatrix}1 & +2\\ -1 & -1\end{smallmatrix} \right)\left( \begin{smallmatrix}x\\ y\end{smallmatrix} \right)$

The first step in solving this system is to find the eigenvalues of matrix A. The characteristic equation is given by: |A - λI| = $\left| \begin{smallmatrix}1-λ & +2\\ -1 & -1-λ\end{smallmatrix} \right| = (1-λ)(-1-λ)+2 = -1+λ -λ +λ^2 + 2 = λ^2 + 1 = 0 ⇒[\text{Solving for λ}] λ = ± i$. The eigenvalues are complex conjugates.

**Finding the eigenvectors for**λ = i

We solve the system: $(A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

which is equivalent to solving:

$\left( \begin{smallmatrix}1-i & +2\\ -1 & -1-i\end{smallmatrix} \right)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$.

This leads to the following system of equations:

$\begin{cases} (1-i)a_1 + 2a_2 = 0 \\ -a_1 + (-1-i)a_2 = 0 \end{cases}$

From the first equation: $a_1 = \frac{-2a_2}{1-i}$. Substitute in the second equation $-\frac{-2a_2}{1-i} + (-1-i)a_2 = 0 ↭ \frac{2a_2(1+i)}{(1-i)(1+i)}+ (-1-i)a_2 = 0 ↭ a_2(1+i)+ (-1-i)a_2 = 0 ↭ a_2(1+i-1-i) = 0↭ a_2·0 = 0$. Since this equation is always true, a_{2} can be any non-zero value. Let’s choose a_{1} = 1 for simplicity. Then $a_1 = 1 = \frac{-2a_2}{1-i} ⇒a_2 = \frac{1-i}{-2} = \frac{-1 +i}{2}$

The solution is: $(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = (\begin{smallmatrix}1\\ \frac{-1+i}{2} \end{smallmatrix})$

The complex solution to the system is given by: $(\begin{smallmatrix}1\\ \frac{-1+i}{2} \end{smallmatrix})e^{it} = [(\begin{smallmatrix}1\\ \frac{-1}{2} \end{smallmatrix})+i(\begin{smallmatrix}0\\ \frac{1}{2} \end{smallmatrix})](cos(t) + isin(t))$

Combining the real and imaginary parts, we get two real-valued solutions:

x_{real}(t) = $(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (\begin{smallmatrix}1\\ \frac{-1}{2} \end{smallmatrix})cos(t)-(\begin{smallmatrix}0\\ \frac{1}{2} \end{smallmatrix})sin(t)$

x_{imaginary}(t) = $(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (\begin{smallmatrix}1\\ \frac{-1}{2} \end{smallmatrix})sin(t)+(\begin{smallmatrix}0\\\ \frac{1}{2} \end{smallmatrix})cos(t)$

Thus, the general solution to the system is a linear combination of these:

x(t) = c_{1}x_{real}(t) + c_{2}x_{imaginary}(t) = $c_1((\begin{smallmatrix}1\\ \frac{-1}{2} \end{smallmatrix})cos(t)-(\begin{smallmatrix}0\\ \frac{1}{2} \end{smallmatrix})sin(t))+c_2((\begin{smallmatrix}1\\ \frac{-1}{2} \end{smallmatrix})sin(t)+(\begin{smallmatrix}0\\ \frac{1}{2} \end{smallmatrix})cos(t))$

$x(t) = (\begin{smallmatrix}c_1cos(t)+c_2sin(t)\\ \frac{1}{2}·(c_2-c_1)·cos(t)+\frac{1}{2}·(c_1-c_2)·sin(t)\end{smallmatrix})$

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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