presentation about first order differential equation

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First-Order Differential Equation

First-Order Differential Equation: First-order differential equations are defined by an equation dy/dx =f (x, y) where x and y are two variables and f(x, y) are two functions. It is defined as a region in the xy plane. These types of equations have only the first derivative dy/dx so that the equation is of the first order and no higher-order derivatives exist.

Differential equations of the first order are written as;

y’ = f (x,my) (d/dx)y = f (x, y)

Let’s learn more about First-order Differential Equations, types, and examples of First-order Differential equations in detail below.

Table of Content

What is a First-Order Differential Equation?

Example of first-order differential equation, types of first-order differential equations, first-order differential equation solution, properties of first-order differential equations, first-order differential equation formulas, applications of first-order differential equations, first order differential equation examples and solutions, first order differential equation questions.

A first-order differential equation is defined by an equation: dy/dx = f(x, y) involving two variables x and y, where the function f(x,y) is defined on a region in the xy-plane. For any linear expression in y, the first-order differential equation [Tex]y’ = f (x, y)[/Tex] is linear. Nonlinear differential equations are those that aren’t linear.

Some examples of first-order differential equation

• dy/dx = x + 11
• dy/dx = 4x – 5

This equation represents a first-order ordinary differential equation where the derivative of y concerning x is equal to 2x.

First-order differential Equations are classified into several forms, each having its characteristics. Types of the First-Order Differential Equations:

• Homogeneous Equations
• Exact Equations
• Separable Equations
• Integrating Factor

A linear differential equation consists of a variable, its derivative, and additional functions. It’s expressed in the standard form as:

dy/dx + P(x)y = Q(x)

• P(x) and Q(x) may be functions of x or numerical constants.

Homogenous First Order Differential Equation

A  homogeneous differential equation  is a function of the form (f(x,y) \frac{dy}{dx} = g(x,y)), where the degree of (f) and (g) is the same. A function (F(x,y)) can be considered homogeneous if it satisfies the condition: (F(\lambda x, \lambda y) = \lambda^n F(x,y)) for any nonzero constant (\lambda).

• In simpler terms, a differential equation is homogeneous if it involves a function and its derivatives in a way that maintains a consistent degree.

Example of Homogenous First Order Differential Equation

Consider the differential equation: (\frac{dy}{dx} = \frac{x^2 – y^2}{xy}). This equation is homogeneous because both the numerator and denominator have the same degree (1).

The formula Q (x,y) dy + P (x,y) dx = 0 is considered to be an exact differential equation if a function f of two variables, x and y, exists that has continuous partial derivatives and can be divided into the following categories.

The general solution of the equation is:

u(x, y) = C since ux(x, y) = p(x, y) and uy (x, y) = Q(x, y)

• C is Constant of Integration

Separable differential equations are a special type of differential equations where the variables involved can be separated to find the solution of the equation. Separable differential equations can be written in the form of:

dy/dx = f(x) × g(y) where x and y are the variables and are explicitly separated from each other.

Once the variables have been separated, it is simple to find the differential equation’s solution by integrating both sides of the equation. After the variables are separated, the separable differential equation

dy/dx = f(x) × g(y) is expressed as dy/g(y) = f(x) × dx

Integrating Factor Homogenous Differential Equation

The  integrating factor  is a function used to solve first-order differential equations. It is most commonly applied to  ordinary linear differential equations of the first order . If a linear differential equation is written in the standard form y’ + a(x)y = 0

Then, the integrating factor (μ) is defined as: (\mu = e^{\int P(x)dx})

First-Order Differential Equation is generally solved and simplified using two methods mentioned below.

• Using Integrating Factor
• Method of variation of constant

Solution using Integration Factor

Using integrating factor can be used to simplify and facilitate the solution of linear differential equations. The entire equation becomes exact when the integrating factor, which is a function of x, is multiplied.

For any linear differential equation is written in the standard form as:

y’ + a(x)y = 0

Then, the integrating factor is defined as:

u(x) = e (∫a(x)dx)

Multiplication of integrating factor u(x) to the left side of the equation converts the left side into the derivative of the product y(x).u(x). General solution of the differential equation is:

y = {∫u(x).f(x)dx + c}/u(x)

where C is an arbitrary constant.

Method of Variation of a Constant

Method of Variation of a Constant is a similar method to solve first order differential equation. In first step, we need to do y’ + a(x)y = 0 . In this method of solving first order differential equation, homogeneous equation always contains a constant of integration C.

• In solving non-homogeneous linear differential equations, we replace the constant (C) with an unknown function (C(x)).
• By substituting this solution into the non-homogeneous equation, we can determine the specific form of the function (C(x)).
• This approach allows us to find a particular solution that satisfies the given non-homogeneous equation.
• Interestingly, both the method of variation of constants and the method of integrating factors lead to the same solution.

Remember, this technique helps us handle non-homogeneous differential equations by introducing a function that varies with the independent variable!

Various properties of linear first-order differential equation are:

• No transcendental functions like trigonometric functions and logarithmic functions are used in linear first-order differential equation.
• In linear first-order differential equation, products of y and any of its derivatives are not present.

[Tex]\begin{array}{|c|c|c|} \hline \textbf{Type of Equation} & \textbf{General Form} & \textbf{Solution Method} \\ \hline \text{Linear Differential Equations} & \frac{dy}{dx} + p(x)y = q(x) & \text{Use an integrating factor, } \mu(x) = e^{\int p(x) \, dx}, \text{ then solve } \frac{d}{dx}(\mu(x)y) = \mu(x)q(x). \\ \hline \text{Homogeneous Equations} & \frac{dy}{dx} = f\left(\frac{y}{x}\right) & \text{Make the substitution } v = \frac{y}{x}, \text{ then solve the resulting separable differential equation for } v. \\ \hline \text{Exact Equations} & M(x,y) + N(x,y)\frac{dy}{dx} = 0 & \text{Find a potential function } \Psi \text{ such that } d\Psi = Mdx + Ndy, \text{ where } \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \\ \hline \text{Separable Equations} & \frac{dy}{dx} = g(x)h(y) & \text{Separate variables and integrate: } \int \frac{1}{h(y)} dy = \int g(x) dx + C. \\ \hline \text{Integrating Factor} & \text{Often used for non-exact linear equations} & \text{Find an integrating factor } \mu(x) \text{ that makes the equation exact, then proceed as with exact equations.} \\ \hline \end{array} [/Tex]

Numerous disciplines, including physics, engineering, biology, economics, and more, first-order differential equations are used. Among other things, they are used to simulate phenomena like fluid dynamics, electrical circuits, population dynamics, and chemical reactions. Various applications of the first-order differential equation are:

• Used in Newton’s Law of Cooling
• Used in Orthogonal Trajectories
• Used in Falling Body Problems
• Used in Dilution Problems

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Below are the example of problems on First Order Differential Equation.

Example 1: Solve the following separable differential equation:dy/dx = x/y 2

First, we separate the variables: y 2. dy = x.dx Integrating both sides: ∫y 2. dy = ∫x.dx y 3 /3 = x 2 /2 + C

Example 2: Solve the following linear differential equation: dy/dx + 2xy – x = 0

Equation in the standard form: dy/dx + 2xy – x = 0 Now, we can use an integrating factor to solve it: f(x) = e ∫2xdx f(x) = [Tex]e^{x^2}[/Tex] Multiplying both sides by the integrating factor: [Tex]e^{x^2} \frac{dy}{dx} + 2xye^{x^2} – xe^{x^2} = 0[/Tex] [Tex]\frac{d}{dx} (ye^{x^2}) – xe^{x^2} = 0[/Tex] Integrating both sides: [Tex]ye^{x^2} = \frac{x^2}{2} + C[/Tex]

Example 3: Solve the first-order differential equation x 3 y’ = x + 2

Solution: 

x 3 y’ = x + 2 ⇒ y’ = (x + 2)/x 3 Integrating both sides w.r.t. x ⇒ ∫(dy/dx) dx = ∫ {(x + 2)/x 3 } dx ⇒ y = -1/x – 1/x 2 + C

Q1: For differential equation dy/dx + yx 2 = sin x find integrating factor.

Q2: Find the solution of differential equation dy/dx = y 2 (x 2 +1).

Q3: Solve the differential equation dy/dx + 2x 3 y = x.

FAQs on First-Order Differential Equation

What is first order differential equation.

A first order differential equation is a differential equation where the maximum order of a derivative is one and no other higher-order derivative can appear in this equation. A first-order differential equation is generally of the form [Tex]dy/dx =f (x,y)[/Tex] .

What are the types of First Order Differential Equations?

First Order Differential Equations are classified into three categories: (i) Separable Differential Equation, (ii) Linear Differential Equation and (iii) Exact Differential Equation

Give one example of First Order Differential Equation.

An Example of First Order Differential Equations: (dy/dx = 2x)

What are application of First Order Differential Equations?

First Order Differential Equation are used in various fields like physics, engineering, biology, applied mathematics, etc.

What is Homogeneous First Order Differential Equation?

A first order differential equation M(x, y) dx + N(x, y) dy = 0 is said to be homogeneous if both M(x, y) and N(x, y) are homogeneous. We can write a homogeneous linear first-order differential equation is of the form y’ + p(x)y = 0.

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Unit 1: First order differential equations

About this unit.

Differential equations relate a function to its derivative. That means the solution set is one or more functions, not a value or set of values. Lots of phenomena change based on their current value, including population sizes, the balance remaining on a loan, and the temperature of a cooling object.

Intro to differential equations

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Exact equations and integrating factors

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1. Second order; linear 2. Third order; nonlinear because of (dy/dx) 4 3. Fourth order; linear 4. Second order; nonlinear because of cos(r + u) 5. Second order; nonlinear because of (dy/dx) 2 or 1 + (dy/dx) 2 6. Second order; nonlinear because of R 2 7. Third order; linear 8. Second order; nonlinear because of ˙ x 2 9. Writing the differential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2. However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x. 10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ue u we see that it is linear in v. However, writing it in the form (v + uv − ue u)(du/dv) + u = 0, we see that it is nonlinear in u. 11. From y = e −x/2 we obtain y = − 1 2 e −x/2. Then 2y + y = −e −x/2 + e −x/2 = 0. 12. From y = 6 5 − 6 5 e −20t we obtain dy/dt = 24e −20t , so that dy dt + 20y = 24e −20t + 20 6 5 − 6 5 e −20t = 24. 13. From y = e 3x cos 2x we obtain y = 3e 3x cos 2x − 2e 3x sin 2x and y = 5e 3x cos 2x − 12e 3x sin 2x, so that y − 6y + 13y = 0. 14. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and y = tan x + cos x ln(sec x + tan x). Then y + y = tan x. 15. The domain of the function, found by solving x + 2 ≥ 0, is [−2, ∞). From y = 1 + 2(x + 2) −1/2 we have (y − x)y = (y − x)[1 + (2(x + 2) −1/2 ] = y − x + 2(y − x)(x + 2) −1/2 = y − x + 2[x + 4(x + 2) 1/2 − x](x + 2) −1/2 = y − x + 8(x + 2) 1/2 (x + 2) −1/2 = y − x + 8.

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We discuss types of differential equations, examples and solutions.

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First Order Linear Differential Equations

You might like to read about Differential Equations and Separation of Variables first!

A Differential Equation is an equation with a function and one or more of its derivatives :

Here we will look at solving a special class of Differential Equations called First Order Linear Differential Equations

First Order

They are "First Order" when there is only dy dx , not d 2 y dx 2 or d 3 y dx 3 etc

A first order differential equation is linear when it can be made to look like this:

dy dx + P(x)y = Q(x)

Where P(x) and Q(x) are functions of x.

To solve it there is a special method:

• We invent two new functions of x, call them u and v , and say that y=uv .
• We then solve to find u , and then find v , and tidy up and we are done!

And we also use the derivative of y=uv (see Derivative Rules (Product Rule) ):

dy dx = u dv dx + v du dx

Here is a step-by-step method for solving them:

• 2. Factor the parts involving v
• 3. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)
• 4. Solve using separation of variables to find u
• 5. Substitute u back into the equation we got at step 2
• 6. Solve that to find v
• 7. Finally, substitute u and v into y = uv to get our solution!

Let's try an example to see:

  dy dx − y x = 1

First, is this linear? Yes, as it is in the form

dy dx + P(x)y = Q(x) where P(x) = − 1 x and Q(x) = 1

So let's follow the steps:

Step 1: Substitute y = uv , and   dy dx = u dv dx + v du dx

Step 2: Factor the parts involving v

Step 3: Put the v term equal to zero

Step 4: Solve using separation of variables to find u

Step 5: Substitute u back into the equation at Step 2

Step 6: Solve this to find v

Step 7: Substitute into y = uv to find the solution to the original equation.

And it produces this nice family of curves:

What is the meaning of those curves?

They are the solution to the equation   dy dx − y x = 1

In other words:

Anywhere on any of those curves the slope minus y x equals 1

Let's check a few points on the c=0.6 curve:

Estmating off the graph (to 1 decimal place):

Why not test a few points yourself? You can plot the curve here .

Perhaps another example to help you? Maybe a little harder?

  dy dx − 3y x = x

dy dx + P(x)y = Q(x) where P(x) = − 3 x and Q(x) = x

And one more example, this time even harder :

  dy dx + 2xy= −2x 3

dy dx + P(x)y = Q(x) where P(x) = 2x and Q(x) = −2x 3

Let's see ... we can integrate by parts ... which says:

∫ RS dx = R ∫ S dx − ∫ R' ( ∫ S dx ) dx

(Side Note: we use R and S here, using u and v could be confusing as they already mean something else.)

Choosing R and S is very important, this is the best choice we found:

• R = −x 2 and
• S = 2x e x 2

So let's go:

Put in R = −x 2 and S = 2x e x 2

And also R' = −2x and ∫ S dx = e x 2

And we get this nice family of curves:

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17.2: First Order Homogeneous Linear Equations

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A simple, but important and useful, type of separable equation is the first order homogeneous linear equation :

Definition: first order homogeneous linear differential equation

A first order homogeneous linear differential equation is one of the form

\[\dot y + p(t)y=0\]

or equivalently

\[\dot y = -p(t)y.\]

"Linear'' in this definition indicates that both \(\dot y\) and \(y\) occur to the first power; "homogeneous'' refers to the zero on the right hand side of the first form of the equation.

Example \(\PageIndex{2}\)

The equation \(\dot y = 2t(25-y)\) can be written \(\dot y + 2ty= 50t\). This is linear, but not homogeneous. The equation \(\dot y=ky\), or \(\dot y-ky=0\) is linear and homogeneous, with a particularly simple \(p(t)=-k\).

Because first order homogeneous linear equations are separable, we can solve them in the usual way:

\[\eqalign{ \dot y &= -p(t)y\cr \int {1\over y}\,dy &= \int -p(t)\,dt\cr \ln|y| &= P(t)+C\cr y&=\pm\,e^{P(t)}\cr y&=Ae^{P(t)},\cr} \]

where \(P(t)\) is an anti-derivative of \(-p(t)\). As in previous examples, if we allow \(A=0\) we get the constant solution \(y=0\).

Example \(\PageIndex{3}\)

Solve the initial value problems \(\dot y + y\cos t =0\), \(y(0)=1/2\) and \(y(2)=1/2\).

We start with

\[P(t)=\int -\cos t\,dt = -\sin t,\]

so the general solution to the differential equation is

\[y=Ae^{-\sin t}.\]

To compute \(A\) we substitute:

\[ {1\over 2} = Ae^{-\sin 0} = A,\]

so the solutions is

\[ y = {1\over 2} e^{-\sin t}.\]

For the second problem,

\[ \eqalign{{1\over 2} &= Ae^{-\sin 2}\cr A &= {1\over 2}e^{\sin 2}\cr}\]

so the solution is

\[ y = {1\over 2}e^{\sin 2}e^{-\sin t}.\]

Example \(\PageIndex{4}\)

Solve the initial value problem \(y\dot y+3y=0\), \(y(1)=2\), assuming \(t>0\).

We write the equation in standard form: \(\dot y+3y/t=0\). Then

\[P(t)=\int -{3\over t}\,dt=-3\ln t\]

\[ y=Ae^{-3\ln t}=At^{-3}.\]

Substituting to find \(A\): \(2=A(1)^{-3}=A\), so the solution is \(y=2t^{-3}\).

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First Order Partial Differential Equations

Mar 09, 2012

1.49k likes | 4k Views

First Order Partial Differential Equations. Method of characteristics. Web Lecture WI2607-2008. H.M. Schuttelaars . Delft Institute of Applied Mathematics. Contents . Linear First Order Partial Differential Equations Derivation of the Characteristic Equation Examples (solved using Maple)

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• partial differential equations
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• characteristic curves
• first order
• its tangent vector

Presentation Transcript

First Order Partial Differential Equations Method of characteristics Web Lecture WI2607-2008 H.M. Schuttelaars Delft Institute of Applied Mathematics

Contents • Linear First Order Partial Differential Equations • Derivation of the Characteristic Equation • Examples (solved using Maple) • Quasi-Linear Partial Differential Equations • Nonlinear Partial Differential Equations • Derivation of Characteristic Equations • Example

Contents • Linear First Order Partial Differential Equations • Derivation of the Characteristic Equation • Examples (solved using Maple) After this lecture: • you can recognize a linear first order PDE • you can write down the corresponding characteristic equations • you can parameterize the initial condition and solve the characteristic equation using the initial condition, either analytically or using Maple

First Order Linear Partial Differential Equations Definition of a first order linear PDE:

First Order Linear Partial Differential Equations Definition of a first order linear PDE: This is the directional derivative of u in the direction <a,b>

First Order Linear Partial Differential Equations Example

First Order Linear Partial Differential Equations Plot the direction field:

First Order Linear Partial Differential Equations Plot the direction field: t x

First Order Linear Partial Differential Equations Direction field: Through every point, a curve exists that is tangent to <a,b> everywhere. t x

First Order Linear Partial Differential Equations Direction field: Through every point, a curve exists that is tangent to <a,b> everywhere: 1) Take points (0.5,0.5), (-0.1,0.5) and (0.2,0.01) X X X

First Order Linear Partial Differential Equations Direction field: Through every point, a curve exists that is tangent to <a,b>everywhere: 1) Take points (0.5,0.5), (-0.1,0.5) and (0.2,0.01) 2) Now draw the lines through those points that are tangent to <a,b> for all points on the lines.

First Order Linear Partial Differential Equations Zooming in on the line through (0.5,0.5), tangent to <a,b> for all x en t on the line: Direction field:

First Order Linear Partial Differential Equations Zooming in on the line through (0.5,0.5), tangent to <a,b> for all x en t on the line: Direction field: Parameterize these lines with a parameter s

First Order Linear Partial Differential Equations SHORT INTERMEZZO

First Order Linear Partial Differential Equations SHORT INTERMEZZO Parameterization of a line in 2 dimensions Parameter representation of a circle

First Order Linear Partial Differential Equations Or in 3 dimensions Parameter representation of a helix

First Order Linear Partial Differential Equations Or in 3 dimensions NOW BACK TO THE CHARACTERISTIC BASE CURVES Parameter representation of a helix

First Order Linear Partial Differential Equations • Parameterize these lines with a parameter s: Direction field: s=0.1 s=0.02 s=0 • For example: • s=0: • (x(0),t(0)) = (0.5,0.5) • changing s results in other points on this curve s=0.04 s=0.01

First Order Linear Partial Differential Equations • Parameterize these lines with a parameter s: • Its tangent vector is: Direction field: s=0.1 s=0.02 s=0 s=0.04 s=0.01

First Order Linear Partial Differential Equations • Parameterize these lines with a parameter s, x=x(s), t=t(s). • Its tangent vector is given by • On the curve:

First Order Linear Partial Differential Equations • Its tangent vector is given by • On the curve: IN WORDS: THE PDE REDUCES TO AN ODE ON THE CHARACTERISTIC CURVES

First Order Linear Partial Differential Equations • The PDE reduces to an ODE on the characteristic curves. • The characteristic equations (that define the characteristic curves) read:

First Order Linear Partial Differential Equations • The PDE reduces to an ODE on the characteristic curves. • The characteristic equations (that define the characteristic curves) read: One can solve for x(s) and t(s) without solving for u(s).

First Order Linear Partial Differential Equations • The PDE reduces to an ODE on the characteristic curves. • The characteristic equations (that define the characteristic curves) read: One can solve for x(s) and t(s) without solving for u(s). Gives the characteristic base curves

First Order Linear Partial Differential Equations The equations for the characteristic base were solved to get the base curves in the example: Solving

First Order Linear Partial Differential Equations The equations for the characteristic base were solved to get the base curves in the example: Solving gives

First Order Linear Partial Differential Equations This parameterisation, i.e., was plotted for (0.5,0.5) (x(0),t(0)) = (-0.1,0.5) (0.2,0.01) by varying s!

First Order Linear Partial Differential Equations To solve the original PDE, u(x,t) has to be prescribed at a certain curve C =C (x,t).

First Order Linear Partial Differential Equations To solve the original PDE, u(x,t) has to be prescribed at a certain curve C =C (x,t). The corresponding system of ODE’s has to be solved such that u(x,t) has the prescribed value at this curve C .

First Order Linear Partial Differential Equations The corresponding system of ODE’s has to be solved such that u(x,t) has the prescribed value at this curve C . • As a first step, parameterize the initial curve C with the parameter τ: x=x(τ), t=t(τ) and u=u(τ).

First Order Linear Partial Differential Equations The corresponding system of ODE’s has to be solved such that u(x,t) has the prescribed value at this curve C . • As a first step, parameterize the initial curve C with the parameter τ: x=x(τ), t=t(τ) and u=u(τ). • Next, the family of characteristic curves, determined by the points on C , may be parameterized by x=x(s, τ), t=t(x, τ) and u=u(s, τ), with the initial conditions prescribed for s=0.

First Order Linear Partial Differential Equations • As a first step, parameterize the initial curve C with the parameter τ: x=x(τ), t=t(τ) and u=u(τ). • Next, the family of characteristic curves, determined by the points on C , may be parameterized by x=x(s, τ), t=t(x, τ) and u=u(s, τ), with the initial conditions prescribed for s=0. This gives the solution surface

First Order Linear Partial Differential Equations EXAMPLE 1

First Order Linear Partial Differential Equations • Consider with The corresponding PDE reads:

First Order Linear Partial Differential Equations • Consider with • Parameterize this initial curve with parameter l:

First Order Linear Partial Differential Equations • Consider with • Parameterize this initial curve with parameter l: • Solve the characteristic equations with these initial conditions.

First Order Linear Partial Differential Equations • Consider with • The (parameterized) solution reads:

First Order Linear Partial Differential Equations Visualize the solution for various values of l:

First Order Linear Partial Differential Equations When all values of l and s are considered, we get the solution surface:

First Order Linear Partial Differential Equations EXAMPLE 2

First Order Linear Partial Differential Equations PDE: Initial condition:

First Order Linear Partial Differential Equations PDE: Initial condition: Char eqns:

First Order Linear Partial Differential Equations PDE: Initial condition: Char eqns: Parameterised initial condition:

First Order Linear Partial Differential Equations Char eqns: Parameterized initial condition: Parameterized solution:

First Order Linear Partial Differential Equations Char eqns: Initial condition:

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Sufficient criteria for oscillation of even-order neutral differential equations with distributed deviating arguments

• Shaimaa Elsaeed 1 , 
• Osama Moaaz 1,2 ,  ,  , 
• Kottakkaran S. Nisar 3 , 
• Mohammed Zakarya 4 , 
• Elmetwally M. Elabbasy 1
• 1. Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, Egypt
• 2. Department of Mathematics, College of Science, Qassim University, P.O. Box 6644, Buraydah 51452, Saudi Arabia
• 3. Department of Mathematics, College of Science and Humanities in Alkharj, Prince Sattam Bin Abdulaziz University, Alkharj 11942, Saudi Arabia
• 4. Department of Mathematics, College of Science, King Khalid University, P.O. Box 9004, Abha 61413, Saudi Arabia
• Received: 25 January 2024 Revised: 14 April 2024 Accepted: 23 April 2024 Published: 07 May 2024

MSC : 34C10, 34K11

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This paper presents novel criteria for investigating the oscillatory behavior of even-order neutral differential equations. By employing a comparative approach, we established the oscillation properties of the studied equation through comparisons with well-understood first-order equations with known oscillatory behavior. The findings of this study introduce fresh perspectives and enrich the existing body of oscillation criteria found in the literature. To illustrate the practical application of our results, we provide an illustrative example.

• even-order ,
• oscillatory behavior ,
• neutral differential equations ,
• distributed deviating arguments

Citation: Shaimaa Elsaeed, Osama Moaaz, Kottakkaran S. Nisar, Mohammed Zakarya, Elmetwally M. Elabbasy. Sufficient criteria for oscillation of even-order neutral differential equations with distributed deviating arguments[J]. AIMS Mathematics, 2024, 9(6): 15996-16014. doi: 10.3934/math.2024775

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