Since Newton, mankind has come to realize that the laws of physics are always expressed in the language of differential equations, Steven Strogatz
The real problem of humanity is the following: We have Paleolithic emotions, medieval institutions and godlike technology. And it is terrifically dangerous, and it is now approaching a point of crisis overall, Edward O. Wilson.
The problem is not the problem. The problem is your attitude about the problem, Captain Jack Sparrow.
Recall
Antiderivatives are fundamental concepts in calculus. They are the inverse operation of derivatives.
Given a function f(x), an antiderivative, also known as indefinite integral, F is the function that can be differentiated to obtain the original function, that is, F’ = f, e.g., 3x2 -1 is the antiderivative of x3 -x +7. Symbolically, we write F(x) = $\int f(x)dx$.
The process of finding antiderivatives is called integration.
Differential equations
An understanding of differential equations is essential in many fields, including physics, engineering, economics, biology, and more. They are powerful tools for modeling and analyzing systems where change is a fundamental aspect.
Algebraic Equations
An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:
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Dependent and independent variables. Variables are symbols representing unknown quantities. Variables can be:
Independent Variables: Variables that can be chosen freely.
Dependent Variables: Variables that depend on the independent variables.
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Constants. Fixed numerical values that do not change.
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Algebraic operations. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.
An algebraic equation typically has the following form: Expression1 = Expression2, where the two expressions are set equal to each other.
Examples: y = 2x + 4 (simple linear equation, represents a straight line with a slope of 2 and a y-intercept of 4), ax2 + bx + c = 0 (quadratic equation, it represents a parabola), $ln(xy + t) = z·sin(\sqrt{x})$.
Analytical methods for solving various types of equations
Understanding algebraic equations is fundamental before tackling differential equations. Here are some common types:
- Linear Equations. Form: y = ax+b. Solution: This is a straight line where a is the slope (it measures the steepness) and b is the y-intercept (the point where the line crosses the y-axis). To find the point where the line intersects the x-axis, set y=0: 0 = ax +b⇒ $x = \frac{-b}{a}$.
- Quadratic Equations. Form: $y = ax^2 +bx +c$. Solution: This is a parabola. The x-intercepts (roots) can be found using the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$. Δ = b2 -4ac. If Δ > 0: Two real and distinct roots; If Δ = 0: One real root (repeated); If Δ < 0: Two complex conjugate roots.
- Cubic Equations. Form: $y = ax^3 +bx^2 +cx +d$. The general solution involves Cardano’s method for finding real roots.
- Exponential Equations. Form $y = ae^{bx}↭\frac{y}{a} = e^{bx}↭ ln(\frac{y}{a}) = bx ↭ x = \frac{ln(\frac{y}{a})}{b}$
Differential Equations
Differential equations describe how quantities change and are essential for modeling dynamic systems where variables depend on one another.
Definition. A differential equation is an equation that involves one or more functions and their derivatives. It relates the function itself (dependent variable), its derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$
Key Components (e.g., $\frac{dy}{dx} = 3x +5y$):
- Dependent variables: Variable(s) (y) that depend on one or more other variables. The function y (or y(x)) we are trying to solve for.
- Independent variables: Variable(s) (x) upon which the dependent variables depend.
- Derivatives: Expressions like $\frac{dy}{dx}$ or y’ that represent the rate at which the dependent variable (y) change with respect to the independent variable (x).
Famous Examples of Differential Equations
Differential equations serve as models for numerous phenomena in science, engineering, and everyday life. Here are some famous examples that illustrate the diversity and application of differential equations:
- Exponential Growth and Decay: $\frac{dy}{dt} = ky$. Solution: $y(t)=y_0e^{kt}$. When k > 0, we observe exponential growth (e.g., population growth or compound interest). When k < 0, we have exponential decay (e.g., radioactive decay or cooling).
- Newton’s Law of Cooling: $\frac{dT}{dt} = -k(T-T_{env})$. Solution: $T(t) = T_{env} + (T_0 -T_{env})e^{-kt}$. This equation describes how an object’s temperature T changes over time, approaching the surrounding environmental temperature Tenv. This model applies to cooling or heating processes.
- Simple Harmonic Motion: $\frac{d^2y}{dt^2}+w^2y = 0$. Solution: y(t) = Acos(ωt) + Bsin(ωt). This describes oscillating systems, such as a pendulum or a mass on a spring. The solution represents periodic motion with angular frequency
ω, where A and B depend on initial conditions.

Classification of Differential Equations
Differential equations are classified based on several criteria, each providing insight into the structure and solutions of the equation. Here is an expanded guide to these classifications:
- Order: The order of a differential equation is the order of the highest derivative present in the equation. First-order differential equation: Involves only the first derivative ($\frac{dy}{dx}$, e.g. y′=2(25−y), y’ + 2x = sin(y), y’ -exy = 0). Second-order differential equation: Involves up to the second derivative ($\frac{d^2y}{dx^2}$, e.g., y’ + y’’ = 2x, $\frac{d^2y}{dx^2} + 4\frac{dy}{dx} +3y = 0$.)
- Degree. The degree of a differential equation refers to the exponent of the highest derivative, assuming the equation has been made free from radicals and fractions involving derivatives, e.g., (y’’)2 + y’ = ln(x). Here, the highest derivative y′′ is squared, giving the equation a degree of 2.
- Linearity: A differential equation is linear if it can be written in the form: $a_n(x)\frac{d^ny}{dx^n}+ a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}} + ··· + a_0(x)y= g(x)$ where an(x), an-1(x), ···, a0(x) and g(x) are function of the independent variable x, and the dependent variables y and its derivatives appear linearly (i.e., no powers or products of y and its derivatives) e.g., y’ + sin(x)y = ex, y’’ + 3y’ + 2y = x2. It is nonlinear if it involves nonlinear terms of the dependent variable or its derivatives, e.g., y·y’ + 2x = 8, (y’’)2 + y’ = ln(x), sin(y) + y’ = 3.
- Ordinary Differential Equations (ODEs) vs. Partial Differential Equations (PDEs)
- Ordinary differential equations or (ODEs) are equations where the derivatives are taken with respect to a single independent variable, e.g., $\frac{d^2y}{dx^2}+\frac{dy}{dx} = 3xsin(y)$. This equation involves derivatives with respect to
x only, so it’s an ODE.
- Partial differential equations or (PDEs) are equations that depend on partial derivatives with respect to multiple independent variables, e.g., $\frac{∂y}{∂t}+z\frac{∂y}{∂z} = \frac{x-t}{x+t}$. This equation involves partial derivatives with respect to both t and z, making it a PDE.
A partial differential equation (PDE) is a type of mathematical equation that involves a function of several variables and its partial derivatives with respect to those variables, e.g., $\frac{∂f}{∂t} = \frac{∂^2f}{∂x^2}, \frac{∂w}{∂t}-\frac{∂^2w}{∂x^2} = 0, etc.$ Partial differential equations are incredibly important in many fields of science and engineering.
Solving First-Order Ordinary Differential Equations
Consider the differential equation: $\frac{dy}{dx}=f(x)$.
This type of equation represents a first-order ordinary differential equation (ODE) because it involves only the first derivative of y with respect to x, and the rate of change of y with respect to x depends only on x. The goal is to find the function y(x) that satisfies this equation.
Steps to Solve the Equation
- Isolate all terms involving y on one side and all terms involving x on the other side by multiplying both sides by dx: dy = f(x)dx. Now, all y terms are on the left, and all x terms are on the right.
- Integrate both sides with respect to their respective variables: $\int dy = \int f(x)dx ↭[\text{This yields}] y + C_1 = F(x) + C_2$ where C1 and C2 are constants of integration. Combining them into a single constant C = C2 -C1, simplifies the expression.
- Write down the general solution. The general solution for y(x) is y = $\int f(x)dx + C = F(x) + C$ where F(x) is the antiderivate of f(x) and C represents the constant of integration, accounting for all possible vertical shifts of the solution curve, giving a family of solutions.
Examples of Solving First-Order ODEs
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y’ = x2ex⇒[Integrate both sides] $y = \int x^2e^xdx + C$ [Integration by parts ∫udv=uv−∫vdu, u = x2 ⇒ du = 2xdx; dv = exdx ⇒ v = ex] = $x^2e^x -\int 2xe^xdx$ [Integration by parts ∫udv=uv−∫vdu, u = 2x ⇒ du = 2dx; dv = exdx ⇒ v = ex] = $x^2e^x -(2xe^x-\int 2e^xdx) = x^2e^x-2xe^x +2e^x + C$ where C has been renamed conveniently for simplicity.
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$y’ = \frac{x^3}{\sqrt{1-x^2}} ⇒[\text{Integrate Both Sides:}] y = \int \frac{x^3}{\sqrt{1-x^2}}dx$ [We choose the trigonometric substitution x = sin(θ), by the Pythagorean Theorem sin2(θ)+cos2(θ) = 1 ⇒ 1 -sin2(θ) = cos2(θ) ⇒ $\sqrt{1-x^2}=cos(θ)$. x = sin(θ), then differentiating both sides gives: dx = cos(θ)dθ] $\int \frac{sin^3(θ)cos(θ)dθ}{cos(θ)}dx = \int sin^3(θ)dθ = \int sin^2(θ)sin(θ)dθ = \int (1-cos^2(θ))sin(θ)dθ$ [Substitution u = cos(θ), du = -sin(θ)dθ] $\int (u^2-1)du = \frac{1}{3}u^3-u + C = \frac{1}{3}cos^3(θ)-cos(θ)+C = \frac{1}{3}(1-x^2)^{\frac{3}{2}}-\sqrt{1-x^2}+C$
The solution to a first-order ODE generally represents a family of curves, each differing by a constant of integration.
Bibliography
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