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Ordinary Differential Equations

Last Updated : 09 Apr, 2024
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Ordinary Differential Equations(ODE) is the mathematical equation that describe how a function’s rate of change relates to its current state. It involves a single independent variable and its derivatives.

Ordinary-Differential-Equations

Ordinary Differential Equations

Let’s know more about Ordinary Differential Equations, it’s types, order and degree of Ordinary differential equation in detail below.

Ordinary Differential Equation

An ordinary differential equation (ODE) is a type of equation that involves ordinary derivatives, not partial derivatives. It typically includes variables and a derivative of the dependent variable with respect to the independent variable. Such equations contain at least one derivative of an unknown function, which can be either an ordinary derivative or a partial derivative.

Ordinary differential equations specifically involve ordinary derivatives, and they are commonly referred to simply as “differential equations.”

Definition of Ordinary Differential Equation

An ordinary differential equation is a mathematical equation that involves the derivatives of an unknown function with respect to a single independent variable. It describes the relationship between function and its derivatives, commonly used to model various dynamic systems in physics, engineering, and other scientific fields.

General Form of Ordinary Differential Equations

The general form of an ordinary differential equation (ODE) is represented as

F(x, y, y’, y”, …) = 0

Where (x) is the independent variable, (y) is the dependent variable, and (y’), (y”), etc., denote the first, second, and higher order derivatives of (y) with respect to (x) respectively.

Examples of Ordinary Differential Equation

Some examples of ODE are:

  • (dy/dx = 2x): This equation represents the first-order ordinary differential equation where the derivative of ( y ) with respect to ( x ) is equal to ( 2x ).
  • [Tex]\bold{\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 0}[/Tex]: This is a second-order ordinary differential equation where the second derivative of ( y ) with respect to ( x ), plus three times the first derivative of ( y ) with respect to (x), plus two times (y), equals zero.
  • [Tex]\bold{\frac{d^2y}{dx^2} + y\frac{dy}{dx} = 0}[/Tex]: Another example of a second-order ordinary differential equation, where the second derivative of ( y ) with respect to ( x ), plus (y) times the first derivative of (y) with respect to (x), equals zero.

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Order of Ordinary Differential Equation

The order of an ordinary differential equation (ODE) refers to the highest derivative present in the equation. It signifies the complexity of the equation and determines the number of initial conditions needed for a unique solution.

An example of an ordinary differential equation (ODE) of first order is:

dy/dx = x + y

And an example of an ODE of second order is:

d2y/dx2 + 2dy/dx + y = 0

In the first example, the highest derivative present is the first derivative, hence it’s a first-order ODE. In the second example, the highest derivative present is the second derivative, making it a second-order ODE.

Degree of ODE

The degree of an Ordinary Differential Equation (ODE) is defined as the highest power to which the derivative of the dependent variable appears in the equation. It represents the order of the highest derivative involved in the equation. The degree provides insight into the complexity of the ODE and influences the methods used for its solution.

For example: [Tex]\bold{\frac{d^3y}{dx^3} + 2 \frac{d^2y}{dx^2} – 3 \frac{dy}{dx} + 4y = 0 }[/Tex]

In this equation, the highest derivative of ( y ) is ( d3y/dx3), which appears with a coefficient of 1.

∴ Degree of this ODE is 3, indicating that it is a third-order differential equation.

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Types of Ordinary Differential Equation

There are four types of ordinary differential equations namely:

  • Homogeneous and Non-homogeneous Differential Equations
  • Linear and Non-linear Differential Equations
  • Autonomous and Non-autonomous Differential Equations

Difference between Homogeneous and Non-homogeneous Differential Equation

Below is the difference between Homogeneous and Non-homogeneous Differential Equation.

Homogeneous Differential EquationsNon-homogeneous Differential Equations
A differential equation in which all terms are of the same degreeA differential equation in which terms of different degrees are present
Example: d2y​/dx2 + 2dy/dx + y = 0, y” – 3y’ + 2y = 0Example: d2y​/dx2 + 3dy/dx + y = x2, y” – 3y’ + 2y = ex

Linear and Non-linear Differential Equations

Below is the difference between Linear and Non-linear Differential Equations.

LinearNon-linear
A differential equation in which the dependent variable and its derivatives appear linearlyA differential equation in which the dependent variable and its derivatives appear non-linearly
Example: d2y​/dx2 + 2dy/dx + 3y = 0, y’ + 2xy = 0Example: d2y​/dx2 + sin(y) = 0, y” + y2 = 0

Difference between Autonomous and Non-autonomous Differential Equations

Below is the difference between Autonomous and Non-autonomous Differential Equation.

Autonomous Differential EquationsNon-autonomous Differential Equations
A differential equation where the independent variable doesn’t explicitly appearA differential equation where the independent variable explicitly appears
Example: dy/dt​ = y, y’ = y(1 – y)Example: dy/dt ​= ty, y’ = y – x

Solution of Ordinary Differential Equations

Solutions of ordinary differential equations refer to functions that satisfy the given equation, making it true for all values of the independent variable within a specified domain. These solutions can be found analytically or numerically, depending on the complexity of the equation and available techniques.

Analytical methods involve finding closed-form expressions for the solutions, while numerical methods use iterative algorithms to approximate solutions. The solution to a differential equation typically includes arbitrary constants determined by initial conditions or boundary conditions, allowing for a family of solutions that fulfill the equation.

General and Particular Solutions

A general solution of a differential equation is a solution that includes all possible solutions to the equation, often expressed with arbitrary constants. It represents a family of solutions that satisfy the given equation but may vary depending on the values of these constants.

A particular solution, on the other hand, is a specific solution obtained by assigning values to the arbitrary constants in the general solution. It corresponds to a unique solution that meets additional conditions, such as initial conditions or boundary conditions, specified for the problem.

For example, the first-order ordinary differential equation ( dy/dx = x + C ), where ( C ) is an arbitrary constant.

  • The general solution of this equation is ( y = x2/2 + C ), where ( C ) represents any constant.
  • A particular solution could be ( y = x2/2 + 3 ), obtained by assigning a specific value, say ( C = 3 ), to the constant (C).

Initial Value Problems

Initial value problems (IVPs) involve finding a particular solution to a differential equation that satisfies both the equation itself and specified initial conditions. These conditions typically include the value(s) of the dependent variable and its derivative(s) at a specific point or within a specific interval.

ProblemEquationInitial Conditions
First-Order ODE IVPdy/dx = 2xy(0) = 1
Second-Order ODE IVP

[Tex]\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 0[/Tex]

y'(0) = 0
Systems of Differential Equations IVPdx/dt = -2x + y and dy/dt = x + 3yy(0) = 2

Boundary Value Problems

Boundary value problems (BVPs) involve finding a solution to a differential equation subject to conditions specified at different points in the domain of the problem, typically at the boundaries of the domain. Unlike initial value problems, which specify conditions at a single point or within an interval, boundary value problems involve conditions at multiple points.

ProblemEquationBoundary Conditions
Dirichlet Boundary Value Problem

d2y/dx2 + y = 0

y(0) = 0, y(1) = 1

Neumann Boundary Value Problem

d2y/dx2 = x

dy/dx (0) = 0, y(1) = 1

Mixed Boundary Value Problem

d2y/dx2 = ƛ y

y(0) = 0, dy/dx(1) = 0

Difference between Ordinary Differential Equation and Partial Differential Equations

The difference between ODE and PDE is mentioned below in the table based.

Aspect

Ordinary Differential Equations (ODEs)

Partial Differential Equations (PDEs)

Definition

Equations involving one independent variableEquations involving multiple independent variables.

Dependent Variable

Functions of one variableFunctions of multiple variables

Order

Described by order (degree)Described by order and degree

Typical Form

F(x, y, y′, y′′,…)=0F(x, y, ∂y​/∂x, ∂2y​/∂x2,…)=0

Boundary/Initial Conditions

Requires initial conditions (for initial value problems) or boundary conditions (for boundary value problems)Requires boundary conditions

Examples

dy/dx = 2x, y” = 3y’ + 2y = 0∂t/∂u = α[∂2u/∂x2] , 2 ϕ = 0

Solution Methods

Often solved analytically or numerically using methods like separation of variables, variation of parameters, or numerical integrationSolutions often involve methods like separation of variables, Fourier transforms, Green’s functions, or numerical methods such as finite difference or finite element methods

Physical Applications

Modeling single-variable processes like population growth, radioactive decay, spring-mass systemsModeling multi-variable phenomena such as wave propagation, heat conduction, fluid dynamics

Complexity

Generally simpler to solve and understandOften more complex due to the involvement of multiple variables and derivatives

Applications of Ordinary Differential Equations

Some Applications of Ordinary Differential Equations are:

  • ODEs are used extensively in physics to model various physical phenomena like motion, heat conduction, wave propagation, and quantum mechanics.
  • ODEs are fundamental in engineering for analyzing and designing systems such as electrical circuits, control systems, mechanical vibrations, and fluid dynamics.
  • ODEs are utilized in economics to model economic systems, growth models, supply and demand dynamics, and financial derivatives.
  • ODEs play a crucial in every aspect where change can be observed.
  • In mathematical modeling, ODEs are very useful to model various real life scenarios.

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Sample Questions on ODE

Example 1: The population of a certain species grows at a rate proportional to the current population size. If the population doubles in 10 years, and the initial population is 1000, find the population as a function of time. Then, determine how long it will take for the population to reach 5000.

Using basic differential equation for exponential growth:

dP/dt = kP

Where:

  • ( P ) represents the population size.
  • ( k ) is the growth rate constant.

Given that the population doubles in 10 years, we can find ( k ) using this information. When the population doubles, it means that ( P(t) = 2P(0) ), where (P(0)) is the initial population. Using the exponential growth equation:

2P(0) = P(0)e10k

2 = e10k

Taking the natural logarithm of both sides:

ln(2) = 10k

k = ln(2)/10

Now that we have the growth rate constant ( k ), we can find the population as a function of time (P(t)). The general solution to the differential equation is:

P(t) = P(0)ekt

Given that (P(0) = 1000), we have:

P(t) = 1000[Tex]e^{\frac{\ln(2)}{10}t}[/Tex]

To find out how long it takes for the population to reach 5000, set P(t) = 5000 and solve for ( t ):

5000 = 1000[Tex]e^{\frac{\ln(2)}{10}t}[/Tex]

5 = [Tex]e^{\frac{\ln(2)}{10}t}[/Tex]

ln(5) = [Tex]\frac{\ln(2)}{10}t[/Tex]

t = 10ln(5) / ln(2)

t = [Tex]\frac{10\ln(5)}{\ln(2)}[/Tex]

t = [Tex]\frac{10 \times \ln(5)}{\ln(2)}[/Tex]

t ≈ [Tex]\frac{10 \times 1.609}{0.693}[/Tex]

t ≈ 16.09 / 0.693

t ≈ 23.22

So, it takes approximately 23.22 years for the population to reach 5000.

Example 2: A mass attached to a spring obeys the equation ([Tex]m\frac{d^2x}{dt^2}[/Tex] + kx = 0), where (m) is the mass, (k) is the spring constant, and (x) is the displacement from the equilibrium position. If (m = 2) kg, (k = 5) N/m, and the initial displacement is (x(0) = 3) m, and initial velocity is (v(0) = -2) m/s, find the position function (x(t)) and describe the motion of the mass.

Given the equation 2d2x/dt2 + 5x = 0), the general solution is:

x(t) = A cos(ωt) + B sin(ω t)

Differentiating (x(t)) twice, we get:

dx/dt = -Aω sin(ω t) + Bω cos(ω t)

d2x/dt2 = -Aω2 cos(ω t) – Bω2 sin(ω t)

Substituting these into the differential equation, we get:

2(-Aω2 cos(ω t) – Bω2 sin(ω t)) + 5(A cos(ω t) + B sin(ω t)) = 0

Solving for (A), (B), and (ω) using the initial conditions (x(0) = 3) and (v(0) = -2), we find:

A = 3

B = -2

ω = [Tex]\sqrt{\frac{5}{2}}[/Tex]

Thus, the solution to the differential equation with the given initial conditions is: [Tex]x(t) = 3\cos\left(\sqrt{\frac{5}{2}}t\right) – 2\sin\left(\sqrt{\frac{5}{2}}t\right)[/Tex]

Example 3: The temperature (T) of a cup of coffee in a room obeys Newton’s Law of Cooling: ([Tex]\frac{dT}{dt} = -k(T – T_a)[/Tex]), where (k) is a constant, (T) is the temperature of the coffee, and (Ta) is the ambient temperature. If (T(0) = 90° C), (Ta = 20° C), and (k = 0.05), find the temperature of the coffee after 10 minutes and determine how long it will take for the coffee to cool to (40° C).

Given that (T(0) = 90° C), (Ta = 20° C), and (k = 0.05), let’s solve for (T(t)).

Separating variables and integrating, we get:

dT/[T – Ta] = -kdt

Integrating both sides:

[Tex]\int \frac{1}{T – T_a} dT = -\int k dt[/Tex]

ln|T – Ta| = -kt + C1

T – Ta = Ce-kt

Given the initial condition (T(0) = 90° C), we can solve for (C):

90 – 20 = C ⇒ C = 70

So, the particular solution is:

T(t) = 70e-0.05t + 20

Now, to find the time it takes for the coffee to cool to (40° C), we set (T(t) = 40) and solve for (t):

40 = 70e-0.05t + 20

20 = 70e-0.05t

e-0.05t

= 20/70

= 2/7

[Tex]t = -\frac{\ln{\frac{2}{7}}}{0.05} \approx 15.98[/Tex]

So, it takes approximately (15.98) minutes for the coffee to cool to (40° C).

Ordinary Differential Equations – FAQs

What is ordinary differential equation?

An ordinary differential equation (ODE) is an equation that involves functions and their derivatives with respect to a single independent variable. It describes how a function changes concerning its input variable.

What are types of ordinary differential equations?

Ordinary differential equations can be classified based on various factors, including linearity, order, and degree. Common types include linear ODEs, nonlinear ODEs, first-order ODEs, second-order ODEs, and higher-order ODEs.

What is the order of an ordinary differential equation?

Order of an ordinary differential equation is determined by the highest derivative present in the equation. For example, a first-order ODE involves only the first derivative of the unknown function, while a second-order ODE involves the second derivative.

How are ordinary differential equations different from other types of differential equations?

Ordinary differential equations specifically deal with functions of one variable and their derivatives. In contrast, partial differential equations involve functions of multiple variables and their partial derivatives with respect to those variables.

Write two examples of ordinary differential equations.

Examples of ordinary differential equations include the simple first-order linear ODE dy/dx ​= 2x and the classic second-order linear ODE d2y/dx2​ + 3dy/dx​+2y = 0.

What is Stability Analysis?

Stability analysis is crucial in the study of ordinary differential equations to determine the behavior of solutions over time and assess their long-term stability.

What is Stiff Equations?

Stiff equations refer to a specific type of ordinary differential equations (ODEs) that involve multiple time scales, where some components of the solution change much more rapidly than others.

Define Higher-order ODEs.

Higher-order ordinary differential equations (ODEs) are differential equations that involve derivatives of a function up to a certain order.




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