**SILMULTANEOUS DIFFERENTIAL EQUATION AND ITS APPLICATION**

**TABLE OF CONTENT**

CHAPTER ONE

INTRODUCTION

1.1 DEFINITION OF TERMS

1.2 SOLUTIONS OF LINEAR EQUATIONS

CHAPTER TWO

SIMULTAENOUS LINEAR DIFFERENTIAL EQUATION WITH CONSTRAINTS COEFFICIENTS.

2.1 LINEAR OPERATOR

CHAPTER THREE

APPLICATION OF SIMULTAENOUS DIFFERENTIAL EQUATIONS AND EXAMPLES

3.2 APPLICATION TO ASTRONOMY

CHAPTER FOUR

SUMMARY

4.1 CONCLUSION

4.2 RECOMMENDATION

REFERENCES

CHAPTER 1

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 Lebnize.

Sir Isaac Newton (1642-1727) and Goltfied wihelm 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 elacticity, 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) bernailli of basel 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 firt 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 brachristochrone problem. The determination of the curve of fastest decent.

The brachistrochone 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.

DEFINATION 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

Ii 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 characterized 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.

While we are not in a position to prove theorem 4.1 in its full generality a demonstration of the uniqueness of the solution in the special case.

A2y’’ + a1y’ + a0y = g(x)

Y(0) = y0

Y’(0) = y’0

Where a2,a1and a0 are positive contants and g(x) is continuous for all x.

The requiremets in the theorem 4.1 that ai(x), i=1,2…n be continuous and an(x)≠ 0 for every x in I are both important specifically, if an(x)=0 for some x in the interval then the solution of linear initial –value problem may not be unique or even exist.

Another type of problem consists of solving differential equation of order two or greater in which the dependent variable y(or its derivatives) is specified at two different points

A problem such as

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

Subject to : y(a)=y0, y(b) = y, is called a two-point boundary-value problem or simply a boundary-value problem.

4.1.2 linear dependence and linear independence: the next two concepts are basic to the study of linear differential equations.

DEFINATION: 4.1 a set of function F1(x), F2(x),……, Fn(x) is said to be linearly dependent on an interval I if there exist constatnts C1, C2, ……, Cn , not all zero, such that

C1 F1 (x) + C1, F2(x) + ------ +CnFn(x) =0

For every x in the interval.

DEFINATION 4.2 A set of function F1(x), F2 (x),…. Fn(x) is used to be linearly independent on an interval I if it is not linearly dependent on the interval.

In other words, a set of functions is linearly independent on an interval if the only constant for which

C1 F1 (x) + C1, F2(x) + ------ CnFn(x) =0

For every x in the interval, are C1 = C2 = ---- = Cn = 0.

It is easy to understand these definitions in the case of two functions F1(x) and F2(x). if the functions are linearly dependent on an interval then there exist constant C1 and C2

Which are not both zero such that for every X in the interval

C1 F1 (x) + C2F2(x) =0

Therefore, if we assume that C1≠ 0, it follows that F1 (x) =-c_2/c_1 F2(x) that is, if two fucntions are linearly dependent then one is simply a constant multiple of the other. Conversely, if F1(x) = C1F2(x), for some constant C2, then

(-1). F1(x) + C2 F2 (x) =0

For every X on some interval. Hence the functions are linear dependent since at least one of the constants (namely, C1=-1) is not zero. We conclude that two functions are linear independent when neither is a constant multiple of the other on an interval.

THEOREM 4.2

Suppose F1(x), F2(x), ………….Fn(x) posses at least n-1 derivatives. If

[?(f_1&f_2&?&f_n@?f'?_1&?f'?_2&…&?f'?_n@?&?&?&?@f_3^((m-1))&f_2^((m-1))&…&f_n^((m-1)) )]≠0

For at least one point in the interval I. then, the function F1(x), F2(x),……..,Fn(x) are linearly independent on the interval.

The determinant in the above theorem is denoted by

W (f1 (x), f2 (x), ------, fn(x)) and is called the Wronskian of the functions.

PROOF: we prove 4.2 by contradiction for the case when n=2. Assume that

W(F1(x0), F2(x0) ) ≠ 0 for a fixed x0 in the interval I and that F1(x) and F2(x) are linearly dependent on the interval. The fact that the function are linearly dependent means there exist constant C1 and C2 not both zero for which C1F1 (x) + C2 F2 (x) =0 for every x in I. differentiating this combination then gives

C1 F’’1 (x) + C1 F’’2 (x) =0

Thus we obtain the system of linear equation

? ?(C_1 F_1 (x)+ C_2 F_2 (x)=0@C_2 ?F'?_1 (x)+ C_2 ?F'?_2 (x)=0)}----------(2)

But the linear dependence of F1 and F2 , implies that (2) possesses a non trivial solution for each x in the interval. Hence

W (F1 (x), F2 (x) ) = [?(f_1 (x)&f_2 (x)@?f'?_1 (x)&?f'?_2 (x))]= 0

For every x in I. this contradicts the assumption that W(F1 (xo), F2 (xo)) ≠ 0. We conclude that F1 and F2 are linearly independent.

COROLLARY if F1 (x), F2 (x),…. Fn (x) possess at least n-1 derivative and are linear dependent on I then the interval that is W(F1 (x), F2 (x),…….., Fn(x)) ≡ 0 on the interval.

4.1.3 SOLUTIONS OF LINEAR EQUATIONS: HOMOGENOUS EQUATIONS;-

A linear nth – order differential equation of the form

a_n (x) (d^n y)/(dx^n )+ a_(n-1) (d^(n-1) y)/(dx^(n-1) )+?+ a_1 dy/dx+a_0 (x) y=0……………………………..(3)

Is said to be homogenous where as

a_n (x) (d^n y)/(dx^n )+ a_(n-1) (x) (d^(n-1) y)/(dx^(n-1) )+?+ a_1 (x)dy/dx+a_0 (x) y=g(x)…………………….(4)

g(x) not identically zero, is said to be non-homogenous.

Note: to avoid needles repetition throughout the remainder of this text we shall as a matter of course make the following important assumptions when given definitions and proving theorems about the linear equation (3) and (4) on some common interval I.

The coefficient ai (x), i=0,1,…, n are continuos

The right-hand member g(x) is continuous

And an(x)≠0 for every x in the interval.

The following theorem is known as the superposition principle

THEOREM 4.3

Let y1, y2 ………….., yk be solutions of the homogenous linear nth-order differential equation (3) on an interval I. then the linear combination

Y = C1Y1(x) + C2Y2(x) + ----------------+ CkYk (x)-----------------------------------(5)

Where the Ci, I = 1, 2, -----k are arbitrary constants, is also a solution on the interval.

Proof:

We prove the case when n=k =2. Let y1(x) and y2(x) be solution of

A2(x) y’’ + a1 (x) y’ + a0 (x) y =0 if we defined

Y = C1Y1(x) + C2Y2 (x) then

A2(x) [C1Y’’1 + C2Y’’2 ] + a1(x) [C1Y’1 + C2Y’1] + a0(x) [C1Y1 + C2Y2]

C1 [a2 (x) y’’1 + a1(x) Y’1 + ao (x) y1] + C2 [a2 (x) y’’2 + a1(x) Y’’2 + a0 (x) Y2 ]

= C1.0 + C2.0

=0

DEFINATION

Let yp be a given solution of the homogenous linear nth-order differential equation (4) on an interval I and let

Yc = c1y1 (x) + c2y2 (x) +-----------------+Cnyn (x)

Denote the general solution of the associated homogenous (3) on the interval. The general solution of the non-homogenous equation, on the interval is defined to be

y= c1y1 (x) + c2y2 (x) +-----------------+Cnyn (x) + Yp (x) = yc + yp

From the above definition, the general solution of (3), is called the complementary function for equation (4). In other words, the general solution of a form non homogenous linear differential equation is

y= complementary function + any particular solution.

Homogenous Linear Equation with constant coefficient:-

We have seen that the linear first-order equations dy/dx + ay = 0, where a is a constant, has the experimental solution y= C1e-ax on -∞ < x < ∞. Therefore it is natural to seek to determine whether exponential solutions exist on -∞ < x < ∞ for higher order equation such as

a_n y^((n)) + a_(n-1) y^((n-1)) +?+ a_2 y^''+a_1 y^'+ a_0 y=0-----------------------------(1)

Where ai, I = 0, 1, 2 ----, n are constants. The solutions of (1) are exponential function or constructed out of exponential functions. We shall begin by considering the special case of the second order equation.

ay’’ + by’ + cy = 0----------------------------------------------------------------------------------- (2)

If we try a solution of the form y=emx , then y’= memx and y’’= m2emx so that equation (2) becomes am2lmx + bmlmx + cmlmx = 0

Or

lmx(am2 + bm + c) = 0

Because lmxis never zero for real value of x, it is apparent that the only way that this exponential function can satisfy the differential equation is to choose in so that it is root of the quadratic equation

am2 + bm + c = 0--------------------------------------(3)

Where equation (3) is called the auxillary equation or characteristics equation of the differential equation (2). We shall consider three cases namely, the solutions of the auxillary equations corresponding to real distinct roots, real but equal roots, and complex conjugate roots.

CASE 1. The auxillary equation (3) has two unequal roots, M1 and M2 we find two solution y1= lm1xand y2= lm2xsince these functions are linearly independent on -∞ < x < ∞

It follows the general solution of (2) on this interval is

Y = c1em1x + c2em2x------------------------------------------(4)

CASE II when m1= m2 we obtain only one exponential solution Y1=c1em1x

However, the second solution is

Y2=em1x∫?l^(-(b/a)x)/l^2m1x dx---------------------------------------(5)

But from quadratic formular we have

M = -b ± √((b^2-4ac)/4a= ) (-b)/2a

Since m1=m2 to have b2-4ac = 0. Implies

2m1 = (-b)/a …………………………..(5) becomes

Y2 = em1x∫?l^2m1x/l^2m1x dx=em1x∫?dx

?=xl?^m1x

The general solution of (2) is then

Y=C1em1x + C2xem1x………………………..(6)

CASE III: if M1& M2 are complex then we can write m1=α + βi and M2 = α - βi where α and βare real and i2 =-1. Where

Y=c1l^(α+iβ)x+c2l^(α-iβ)x-----------------------------------------(7)

Using Euler’s formular

l^2θ=cos?θ+i sinθ, whereθis any real number from this result

We can write

l^2B=cos??βx+i sin?βx and l^(-iβx)=cos??βx-i sin?βx ? ?

Thus (7) becomes

Y =l^αx [c1l^iβx+c2l^(-iβx)]

=l^αx [c_1 {cos??βx+i sin?βx ? }+c_2 {cos??βx-i sin?βx ? } ]

=l^αx [(c_1+ c_2 ) cos??βx+i (c_1- c_(2 ) ) sin?βx ?]

Sincel^αx cosβx and l^αx sin?βxthemselves form a fundamental set of solutions of the given differential equation on -∞ < x < ∞ we can

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