Linear differential equations project

Linear Differential Equations

BYMOHAMMAD SAEED KHAWWAM

201401379

Instructor:  Rana Hani KanjSection: MATH-283-3-CRN:10260

Differential Equations An equation which involves unknown function of one or several variables that

relates the values of the function itself and its derivatives of various orders.

ordinary differential equation (ode) : not involve partial derivatives partial differential equation (pde) : involves partial derivatives order of the differential equation is the order of the highest derivatives

Examples:

second order ordinary differential equation

first order partial differential equation

2

23 sin

d y dy x ydxdx

y y x txt x x t

Terminologies In Differential Equation

• Existence: Does a differential equation have a solution?• Uniqueness: Does a differential equation have more

than one solution? If yes, how can we find a solution which satisfies particular conditions?

• A problem in which we are looking for the unknown function of a differential equation where the values of the unknown function and its derivatives at some point are known is called an initial value problem (in short IVP).

• If no initial conditions are given, we call the description of all solutions to the differential equation the general solution.

Differential EquationsSome Application of Differential Equation in Engineering

Linear Differential Equation A differential equation is linear, if

1. dependent variable and its derivatives are of degree one,2. coefficients of a term does not depend upon dependent

variable.

Example:

364

3

3

ydxdy

dxyd

is non - linear because in 2nd term is not of degree one.

.0932

2

ydxdy

dxydExample:

is linear.

1.

2.

( , )y f x y

Linear Non-linear

Integrating Factor

Separable Homogeneous Exact

IntegratingFactor

Transform to ExactTransform to separable

First Order Linear Equations• A linear first order equation is an equation

that can be expressed in the form

Where P and Q are functions of x

HistoryYEAR PROBLEM DESCRIPTION MATHAMATICIAN

1690 Problem of the Isochrones

Finding a curve along which a body will fall with uniform

vertical velocity

James Bernoulli

1728 Problem of Reducing 2nd Order

Equations to 1st Order

Finding an integrating factor

Leonhard Euler

1743 Problem of determining

integrating factor for the general linear

equation

Concept of the ad-Joint of a differential

equation

Joseph Lagrange

1762 Problem of Linear Equation with

Constant Coefficients

Conditions under which the order of a

linear differential equation could be

lowered

Jean d’Alembert

Methods Solving LDE1. Separable variable

M(x)dx + N(y)dy = 0 2. Homogenous

M(x,y)dx+N(x,y)dy=0, where M & N are nth degree 3. Exact

M(x,y)dx + N(x,y)dy=0, where M/ðy=0, where ðM/ðy = ðN/ðx 

Solution of Differential Equation

1st Order DE - Separable EquationsThe differential equation M(x,y)dx + N(x,y)dy = 0 is separable if the equation can be written in the form:

02211 dyygxfdxygxf

Solution :

1. Multiply the equation by integrating factor: ygxf 12

1

2. The variable are separated :

0

1

2

2

1 dyygygdx

xfxf

3. Integrating to find the solution:

Cdyygygdx

xfxf

1

2

2

1

1st Order DE - Homogeneous EquationsHomogeneous Function

f (x,y) is called homogenous of degree n if : y,xfy,xf n Examples:

yxxy,xf 34 homogeneous of degree 4

yxfyxx

yxxyxf,

,4344

34

yxxyxf cos sin, 2 non-homogeneous

yxfyxxyxxyxf

n ,

cos sin

cos sin,22

2

1st Order DE - Homogeneous EquationsThe differential equation M(x,y)dx + N(x,y)dy = 0 is homogeneous if M(x,y) and N(x,y) are homogeneous and of the same degree

Solution :

1. Use the transformation to : dvxdxvdyvxy 2. The equation become separable equation:

0,, dvvxQdxvxP

3. Use solution method for separable equation

Cdvvgvgdx

xfxf

1

2

2

1

4. After integrating, v is replaced by y/x

1st Order DE – Exact EquationThe differential equation M(x,y)dx + N(x,y)dy = 0 is an exact equation if :

Solution :

The solutions are given by the implicit equation xN

yM

CyxF ,

1. Integrate either M(x,y) with respect to x or N(x,y) to y. Assume integrating M(x,y), then :

where : F/ x = M(x,y) and F/ y = N(x,y)

ydxyxMyxF ,,

2. Now : yxNydxyxMyy

F ,',

or :

dxyxMy

yxNy ,,'

1st Order DE – Exact Equation3. Integrate ’(y) to get (y) and write down the result F(x,y) = C

Examples:

1. Solve : 013 32 3 dyyxdxyxAnswer:

Newton's Law of Cooling• It is a model that describes, mathematically, the change in temperature of

an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature T of the object and the temperature Te of the environment surrounding the object.

d T / d t = - k (T - Te) Let x = T - Te

so that dx / dt = dT / dt d x / d t = - k x

The solution to the above differential equation is given by x = A e - k t

substitute x by T – Te T - Te = A e - k t

Assume that at t = 0 the temperature T = To 

T0 - Te = A e o

 which gives A = To-Te

The final expression for T(t) is given by T(t) = Te + (To- Te) e - k t

This last expression shows how the temperature T of the object changes with time.

Growth And Decay• The initial value problem

where N(t) denotes population at time t and k is a constant of proportionality, serves as a model for population growth and decay of insects, animals and

human population at certain places and duration.

Integrating both sides we getln N(t)=kt+ln C

or or N(t)=Cekt

C can be determined if N(t) is given at certain time.

)()( tkNdttdN

kdttNtdN

)()(

Carbon datingLet M(t) be the amount of a product that decreases withtime t and

the rate of decrease is proportional to the amount M as follows d M / d t = - k M

where d M / d t is the first derivative of M, k > 0 and t is the time.Solve the above first order differential equation to obtain

M(t) = Ae-kt

 where A is non zero constant. It we assume that M = Mo at t = 0, then

M= Ae0

which gives  A = MoThe solution may be written as follows

M(t) = Mo e-kt

Economics and Finance

• The problems regarding supply, demand and compounding interest can be calculated by this equation

is a separable differential equation of first-order. We can write it as

dP=k(D-S) dt.

Integrating both sides, we get P(t)=k(D-S)t+A

where A is a constant of integration. 

Similarly  S(t)=S(0) ert ,Where S(0) is the initial money in the account

)( SDkdtdP