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Linear Differential Equation
By
Nofal Umair
Introduction to Differential Equations
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 dyx y
dxdx
y y x t
xt 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
ydx
dy
dx
yd
is non - linear because in 2nd term is not of degree one.
.0932
2
ydx
dy
dx
ydExample:
is linear.
1.
2.
( , )y f x y
Linear Non-linear
Integrating Factor
SeparableHomogeneous 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 LDE
1. 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 EquationSolution 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 dyyg
ygdx
xf
xf
3. Integrating to find the solution:
Cdyyg
ygdx
xf
xf
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
yxf
yxx
yxxyxf
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
Cdvvg
vgdx
xf
xf
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
x
N
y
M
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 Equation
3. 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 get
ln N(t)=kt+ln C
or
or N(t)=Cekt
C can be determined if N(t) is given at certain time.
)()(
tkNdt
tdN
kdttN
tdN
)(
)(
Carbon dating
Let 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 Mwhere 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 obtainM(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
)( SDkdt
dP