CHAPTER ONE

INTRODUCTION

In mathematics, the history of differential equation traces the development of “D.E” from calculus, with itself was independently inverted by English physicist Isaac Newton and German mathematician Gottfried Lebanese.

Sir Isaac Newton (1642-1727) and Golfed whelp Leibniz (1646-1716) were the two prominent scientists who independently discovered the fundamental ideas of calculus. This provides a tool for solving problems involving motion and other physical phenomena such as elasticity, analysis of the bending of beams and the shape a string will form under various conditions.

Generally, first-order and higher-order differential equations problems analytically.

Langrange said of Euler’s work in mechanics identified the condition for exactness of first order differential equation in (1734-1735) developed the theory of integrating factors and gave the general solution of homogeneous.

The brothers jakob (1654-1705) and Johann (1667-1748) Bernoulli of basil did much to develop methods of solving differential equations, with the aid of calculus they formulated as differential equations and solved a number of problems in mechanics. In the same paper (in 1690) he first used the term “integral” in the modern sense. In 1694 Johann Bernoulli was able to solve the equation

d_y/d_x = y⁄ax

Even though it was not yet known that      d (In x)  =dx⁄x

Which led to much friction between them, was the brachistochrone problem. The determination of the curve of fastest decent.

The brachistochrone was also solved by Leibniz and newton in addition to Bernoulli brother.

The study of differential equation originated in the beginnings of the calculus. Newton did relatively work in differential equations, his development of the calculus and elucidation of the basic principles of mechanics provided a basis for their application in the eighteenth century, most notably by Euler. Newton classified first order differential equation which is written in the form  d_y/d_x  = f(x)

d_y/d_x  = f (y) and d_y/d_x  = f(x,y)

By the end of the eighteenth century many elementary methods of solving ordinary differential equations had been discovered. In the nineteenth century interest turned toward the investigation of the theoretical equations of existence and uniqueness and to the development of less elementary methods such as those based on power series method. Partial differential equations also began to be studied intensively, as their crucial role in mathematical physics became clear.

In 1693, Leibniz solved his first differential equation and that same year newton published the results of previous D.E solution methods a year that is said to mark the inception for D.E as a distinct field in mathematics.

DEFINITION OF TERMS

This section focuses mainly on the basis terminologies associated with the study of first-order ordinary differential equations.

A differential equation is an equation that contains the derivatives or differentials of one or more dependent variables, with respect to one or more independent variables.

 

Types of differential equations

 

An ordinary differential equation

 

A partial differential equation

 

An ordinary differential equation is an equation which contains only ordinary derivatives of one or more independent variables, with respect to a single independent variables.

 

For example

 

dy/dx – y = 2

 

(x-y)dx-4ydy=0

 

(d^2 y)/〖dx〗^2  – dy/dx+y=0

 

Are ordinary differential equations

 

A partial differential equation is an equation is an equation involving the partial  derivatives of one or more dependent variables with respect to two or more independent variables.

 

For example,

 

∂u/(∂y )= (-∂v)/(    ∂x )

 

x ∂u/(∂y )+y ∂y/∂x=u

 

∂u/(∂x∂y  )=x-y

 

Are partial differential equations

 

Order of differential equations is the order of the highest derivative in a differential equation. For example

 

((d^2 y)/〖dx〗^2  )^5 + (dy/dx )^2-y=x

 

Is a second order differential equation

 

dy/dx + y = 0

 

Is a first order differential equation

 

K2(∂^3 y)/〖dx〗^3 +∂y/∂t=0

 

Is a third-order partial differential equation.

 

Degree of Differential Equation;- is the degree of the highest order of the differential equation. For example

 

( (∂^3 y)/〖dx〗^3  )^3+(∂y/∂x )^6+y=x^3

 

Is of degree 3

 

K2(∂^4 u)/〖dx〗^4 +(d^2 u)/(∂t^2 )=0 is of degree 1

 

Linear and non-linear differential equations

 

A differential equations is said to be linear if it has the form

 

a_n (x)  (d^n y)/(dn^n )+ a_(n-1) (x)  (d^(n-1) y)/(dn^(n-1) )+⋯ a_1 (x)  dy/dx+a_0 (x)y=g(x)

 

It should be observed that linear differential equation are characterised by two properties.

 

The dependent variable y and its derivatives are of the first degree; and

 

Each coefficient depends only on the independent variable x. an equation that is not linear is said to be non-linear.

 

the equation x dy + y dx = 0

 

y’’ – y’ + y = 0 and x^3  (d^3 y)/(dx^3 )+3x dy/dx – y = x^3

 

Are linear first, second and third order differential equations, respectively on the other hand   dy/dx= 〖xy〗^(1/2),yy^”-y^’=x+4

 

And

 

(d^3 y)/( dx^3 )+ y^2=0are non-linear first, second and third

 

Order ordinary differential equations respectively.

 

Linear equation;- we defined the general form of linear D.E of order ‘n’ to be

 

a_n (x)  (d^n y)/(dx^n )+ a_(n-1)  (d^(n-1) y)/(dn^(n-1) )+⋯+ a_1  dy/dx+a_0 (x)  y=f(x)

 

The coefficients are functions of x only, and that y all its derivatives are raised to the first power, now if n=1 we obtain the linear first order equation

 

a_1 (x)  dy/dx+ a_0 (x)  y=f(x)

 

Dividing all by a1(x), we have

 

dy/dx+p(x)  y=Q(x)——————————————-(1)

 

Where P(x) and Q(x) are continuous function. It follows that (1) has the following solution. Equation (1) can be written as

 

-dy + [p (x) y – Q(x) ]dx = 0—————————————–(2)

 

Linear equation posses the pleasant property that x function u(x) can always be found such that, the multiple of (2)

 

μ(x)dy+ μ(x)[p(x)y-Q (x) ]dx=0……………………………………..………(3)

 

In an exact D.E using exact

 

∂/∂x  μ(x)= ∂/∂x  μ(x)[p(x)  y -Q(x)]………………..………………………….……………..(4)

 

∂μ/∂x = μp(x)

 

Using variable seperable, to determine u(x)

 

∫〖∂u/μ= 〗 ∫〖p(x)dx〗

 

 

 

In / μ / = ∫p(x)dx

 

μ (x) = l^∫〖p(x)dx〗——————————————–(5)

 

The function u(x) defined in (5) is called integrating factor for the linear equation.

 

From equation (4), we solve

 

l^∫〖p(x)dx〗 dy + l^∫〖p(x)dx〗 [p(x)y-Q (x) ]dx

 

l^∫〖p(x)dx〗 dy + l^∫〖p(x)dx〗  p(x)ydx

 

Are exact differentials we now write (3) in the form

 

l^∫ p(x)dx dy + l^∫ p(x)dx  p(x)ydx= l^∫ p(x)dx Q(x)dx

 

⇒d [l^∫ p(x)dx  y] = l^∫ p(x)dx  f(x)dx

 

Integrating both sides, gives

 

l^∫▒p(x)dx  y = ∫▒〖l^∫▒p(x)dx  Q(x)dx〗

 

Y = l^(-∫▒p(x)dx) ∫▒〖l^∫▒p(x)dx  Q(x)dx〗—————–(6)

 

The equation (1) has a solution it must be of from (6)

 

VARIABLE SEPARABLE;- if g(x) is a given continuous function then the first –order equation

 

dy/dx=g(x)—————————————–(1)

 

Can be solve by integrating. the solution (1) is

 

Y =∫▒〖g(x)dx+c〗

 

OR

 

A D.E of the form

 

dy/dx= (g(x))/(h(y))

 

Is said to be separable

 

hydy/dx=g(x)…………………………………………….(2)

 

Now if y = f(x) denotes a solution of (2) we now have

 

h[f(x)) f’(x) = g(x) therefore

 

∫▒〖h(f(x) ) f^’ (x)  dx〗= ∫▒g(x)dx+c …………………………………..(3)

 

But dy =f1 (x)dx so (3), becomes

 

∫▒〖h(y)dy=∫▒〖g(x)+c〗〗………………………………………………………(4)

 

Equation (4) is the required solution.

 

Initial value and boundary- value problems: for a linear nth-order differential equation the problem solve

 

a_n (x)  (d^n y)/(dx^n )+ a_(n-1)  (d^(n-1) y)/(dx^(n-1) )+⋯+ a_1 (x)dy/dx+a_0 (x)  y=g(x)………………..(1)

 

Subject to :■(y(x_0 )=&y_0@y'(x_0 )=&〖y’〗_0@⋮&⋮@⋮&⋮@y^((n-1) ) (x_0 )=&y_o^((n-1)) )

 

Where y0, y01—————y0(n-1)  are arbitrary constant, is called an initial-value problem. we seek a solution on some interval I containing the point x=x0

 

In the important case of a linear second-order equation, a solution of

 

a_2 (x)  (d^2 y)/(dx^2 )+ a_1 (x) dy/dx+a_0 (x)  y=g(x)

 

Y(x0) = yo

 

Y’(xo) = yo’

 

Is a function defined on I whose graph passes through (X0,yo) such that the slope of the curve at the point is the number Y10.

 

The next theorem give sufficient conditions for the existence of a unique solution (1)

 

THEOREM 4.1

Let an(x), an-1(x)——–, a1(x), a0(x) and  g(x) be continuous on an interval I and let  an(x)≠0 for every x in this interval. If x=x0 is any point in this interval, then a solution y(x) of the initial-value problem (1) exists on the interval and is unique.