There are two ways to do great mathematics. The first is to be smarter than everybody else. The second way is to be stupider than everybody else - but persistent, Raoul Bott

Arithmetic is where numbers fly like pigeons in and out of your head, Carl Sandburg

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}) .

Definition. A first-order linear differential equation is an ordinary differential equation (ODE) that can be written in the general form: a(x)y' + b(x)y = c(x) where:

- y=y(x) is the unknown function of the independent variable x.
- y′ = $\frac{dy}{dx}$ is the derivative of y with respect to x.
- a(x), b(x), and c(x) are known functions of x.

If c(x) = 0, the equation is called **homogeneous**, i.e., a(x)y’ + b(x)y = 0. These equations are termed **“linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together.**

The general first-order linear ODE can be rearranged into the standard linear form: y’ + p(x)y = q(x) where:

- $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$
- a(x) ≠ 0 for all x in the interval of interest.

Expressing the equation in this standard form makes it easier to apply solution methods, such as using an integrating factor.

First-order linear ODEs are widely used to model various physical phenomena. Two common applications are Newton’s Law of Cooling and diffusion processes.

Newton’s law of cooling describes how the temperature of an object changes over time as it exchanges heat with its surroundings.

This law is expressed by the differential equation: $\frac{dT}{dt}=k(T_e-T)$ where:

- T(t) is temperature of the object at time t.
- T
_{e}is the constant external (ambient) temperature of the environment. - k is a positive constant known as the cooling constant or heat transfer coefficient.
- $\frac{dT}{dt}$ represents the rate of change of the object’s temperature with respect to time.
- T(0) = T
_{0}is the initial temperature of the object.

This equation states that the rate at which the object's temperature changes is proportional to the difference between the object's temperature and the external temperature T_{e}. Specifically,

- If T > T
_{e}, the object is warmer than its surroundings, then $\frac{dT}{dt}< 0$, indicating that the object will cool down (negative rate of change). - If T < T
_{e}, the object is cooler than its surroundings, then $\frac{dT}{dt} > 0$, indicating that the object will warm up (positive rate of change).

To solve the differential equation, we first rewrite it in standard linear form: $\frac{dT}{dt} + kT = kT_e$. This matches the standard form y’ + p(x)y = q(x), where p(x) = k and q(x) = kT_{e}.

- Find the integrating factor: $μ(t)=e^{\int p(t)dt} = e^{\int kdt} = e^{kt}$
- Multiply both sides by the integrating factor: $e^{kt}\frac{dT}{dt} + e^{kt}kT = e^{kt}kT_e ↭ \frac{d}{dt}(e^{kt}T) = e^{kt}kT_e$
- Integrate both sides: $e^{kt}T = \int e^{kt}kT_edt = T_ee^{kt} ↭ e^{kt}T = T_ee^{kt}+C$, where C is the constant of integration.
- Solve for T: $T = T_e + Ce^{-kt}$.
- Apply initial conditions. If T(0) = T
_{0}, then $T_0 = T_e + Ce^{-0·t} = T_e + C ⇒ C = T_0 - T_e$ - Final Solution: $T = T_e + (T_0 - T_e)e^{-kt}$.
- Interpretation of the Solution. The object's temperature T(t) exponentially approaches the external temperature T
_{e}over time. The rate at which T(t) approaches T_{e}depends on the cooling constant k:

- Larger k means faster cooling or heating.
- Smaller k means slower temperature change.

Example: A brewing pot of coffee is initially at a temperature of 180°F. It is placed in a room where the ambient (external) temperature is constant at 76°F. After 5 minutes, the temperature of the coffee drops to 168°F. Determine the cooling constant k using Newton’s Law of Cooling.

Refer to Figure B for a visual representation and aid in understanding it.

- At time t = 0, the temperature is T(0) = 180°F, T
_{e}= 76°F. $T = T_e + (T_0-T_e)e^{-kt} ↭ T = T = 76 + (180-76)e^{-kt} = 76 +104e^{-kt}$ - Apply the temperature at t = 5 minutes: $ T(t) = 76 +104e^{-5·k} = 168 ↭ 104e^{-5·k} = 168 -76 = 92 ↭ 104e^{-5·k} = 92$
- Solve for $e^{-5·k}$: $e^{-5·k} = \frac{92}{104} = 0.8846.$
- Take the natural logarithm of both sides: $-5k = ln(0.8846)$.
- Solve for k: $k = \frac{-ln(0.8846)}{5}≈ \frac{0.122}{5} ≈ -0.0245$ per minute. The cooling constant k is approximately 0.0244 per minute. This value quantifies the rate at which the coffee cools down in the given environment.

Another application of first-order linear ODEs is modeling diffusion processes, such as the diffusion of salt across a semi-permeable membrane.

Consider a chamber containing a solution with salt concentration C(t) at time t. The chamber is surrounded by a water bath with a constant salt concentration C_{e}. A semi-permeable membrane allows salt to diffuse between the chamber and the water bath. The rate of diffusion is proportional to the concentration difference across the membrane.

The diffusion process is modeled by a first-order ODE: $\frac{dC}{dt} = k(C_e-C)$ where:

- C(t) is the concentration of salt inside the chamber at time t.
- C
_{e}is the constant external concentration of salt in the water bath. - k is a positive constant representing the diffusion rate.
- $\frac{dC}{dt}$ represents the rate of change of the internal concentration.

This equation is similar to the cooling equation. It states that the rate at which the internal salt concentration changes is proportional to the difference between the external and internal concentrations:

- If C < C
_{e}, the internal concentration is lower, so salt diffuses into the chamber ($\frac{dC}{dt}> 0$). - If C > C
_{e}, the internal concentration is higher, so salt diffuses out of the chamber ($\frac{dC}{dt}> 0$). Refer to Figure C for a visual representation and aid in understanding it.

Rewriting the equation in standard linear form: $\frac{dC}{dt}+kC = kC_e$. Again, this matches the previously studied standard form y’ + p(t)y = q(t), where p(t) = k and q(t) = kC_{e}.

- Find the integrating factor: $μ(t)=e^{\int p(t)dt} = e^{\int kdt} = e^{kt}$
- Multiply both sides by the integrating factor: $e^{kt}\frac{dC}{dt} + e^{kt}kC = e^{kt}kC_e ↭ \frac{d}{dt}(e^{kt}C) = e^{kt}kC_e$
- Integrate both sides: $e^{kt}C = \int e^{kt}kC_edt = C_ee^{kt} ↭ e^{kt}C = C_ee^{kt}+C’$, where C is the constant of integration.
- Solve for C: $C = C_e + C’e^{-kt}$.
- Apply initial conditions. If C(0) = C
_{0}, then $C_0 = C_e + C’e^{-0·t} = C_e + C’ ⇒ C’ = C_0 - C_e$ - Final Solution: $C(t) = C_e + (C_0 - C_e)e^{-kt}$.
- Interpretation of the Solution. The internal concentration C(t) exponentially approaches the external concentration C
_{e}over time. The rate of diffusion k determines how quickly C(t) approaches C_{e}:

- Larger k means faster diffusion.
- Smaller k means slower diffusion.

In both applications, the solutions exhibit exponential behavior, reflecting how the quantity of interest approaches equilibrium over time.

- Solving xy’ - y = x
^{3}

- Rewrite the differential equation in standard form (assuming x ≠ 0): $y’ -\frac{1}{x}y = x^2$, where $p(x) = \frac{-1}{x}, q(x) = x^2.$
- Find the integrating factor: $μ = e^{\int p(x)dx} = e^{\int \frac{-1}{x}dx}$ =[Assuming x > 0] $e^{-ln(x)} = e^{ln(x^{-1})} = x^{-1} = \frac{1}{x}$.
By the power rule of logarithms loga

^{b}= b*loga. - Multiplying both sides of the differential equation by the integrating factor μ(x): $\frac{1}{x}y’-\frac{1}{x^2}y = x$
- Recognize the left-hand side as a product derivative μ(x)y, $\frac{d}{dx}(\frac{1}{x}y) = \frac{1}{x}y’-\frac{1}{x^2}y$.
- Integrate both sides of the equation with respect to x: $\frac{1}{x}y = \int xdx = \frac{x^2}{2} + C$ where C is the constant of integration.
- Solve for y, $y = \frac{x^3}{2}+Cx$ where C is an arbitrary constant determined by initial conditions (if any are provided).

- Solve $x^2y’ + xy = 3x$

This equation is not immediately in the standard linear form, so we will manipulate it to fit that form and then apply the method of integrating factors.

To bring the equation into the standard linear form y′ +P(x)y=Q(x), we divide both sides by x^{2}: $\frac{1}{x^2}(x^2y’ + xy) = \frac{1}{x^2}(3x) ↭ y’ + \frac{1}{x}y = \frac{3}{x}$

Compute the integrating factor, μ(x). It is defined as $μ(x) = e^{\int P(x)dx} = e^{\int \frac{1}{x}dx} = e^{ln|x|} = |x|$ Since x is in the denominator and we need x ≠ 0, we’ll consider x > 0 for simplicity, so μ(x) = x.

We write the differential equation in the form $\frac{d}{dx}(μ·y)=μ·Q(x)$ by multiplying both sides of the differential equation by the integrating factor, $xy’ + y = 3 ↭ \frac{d}{dx}(x·y) = 3$.

Next, we can solve the differential equation integrating with respect to x, $xy = \int 3dx = 3x + C ⇒[\text{Solve for y}] y = 3 + \frac{C}{x}$ where C is the constant of integration determined by initial conditions.

- Solve (1+cos(x))y’ -sin(x)y = 2x with y(0) = 1

- Rewrite the differential equation in standard form: $y’ -\frac{sin(x)}{1+cos(x)}y = \frac{2x}{1+cos(x)}$ where $p(x) = -\frac{sin(x)}{1+cos(x)}, q(x) = \frac{2x}{1+cos(x)}$
- Compute the integrating factor: $μ = e^{\int p(x)dx} = e^{\int -\frac{sin(x)}{1+cos(x)}dx} = e^{ln(1+cos(x))} = 1 + cos(x)$.
$\int -\frac{sin(x)}{1+cos(x)}dx$ =[Let u = 1+cos(x), then du = −sin(x)dx] $\int \frac{du}{u} = ln|u| = ln|1+cos(x)|=$[Since 1 + cos(x) ≥ 0 for all real x, we can drop the absolute value:] ln(1 + cos(x))

- Multiplying both sides of the differential equation by the integrating factor μ(x): $(1+cos(x))y’ -sin(x)y = 2x.$
- By construction, the left-hand side becomes the derivative of the product μ(x)y, i.e., $(1+cos(x)y)’ = 2x$
- Integrate both sides of the equation with respect to x: $1+cos(x)y = \int 2xdx = x^2 + C$ where C is the constant of integration.
- Solve for y, $y = \frac{x^2+C}{1+cos(x)}$
- Apply the initial condition $y(0) = 1, 1 = \frac{C}{1+cos(0)} ↭ 1 = \frac{C}{2} ↭[\text{Solving for C}] C = 2$. The particular solution to the differential equation, satisfying y(0) = 1, is: $y = \frac{x^2+2}{1+cos(x)}$

- Solve $\frac{dy}{dx}+\frac{3y}{x} = \frac{e^x}{x^3}$

- Write the differential equation in standard form. It is already in standard form where $p(x) = \frac{3}{x}, q(x) = \frac{e^x}{x^3}$
- Compute the integrating factor: $μ = e^{\int p(x)dx} = e^{\int \frac{3}{x}} = e^{3ln(x)} = e^{ln(x^3)} = x^3$ where x > 0 (to avoid division by zero and ensure μ(x) is defined).
- Multiplying both sides of the differential equation by the integrating factor μ(x): $x^3\frac{dy}{dx} +3x^2y = e^x.$
- By construction, the left-hand side becomes the derivative of the product μ(x)y, i.e., $(x^3y)’ = e^x$
- Integrate both sides of the equation with respect to x: $x^3y = \int e^xdx = e^x + C$ where C is the constant of integration.
- Solve for y, $y = \frac{e^x+C}{x^3}$ where C is an arbitrary constant determined by initial conditions.

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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