MODELLING OF SECOND ORDER DIFFERENTIAL EQUATION AND ITS APPLICATION

1. MODELING OF SECOND ORDER
DIFFERENTIAL EQUATION
And Applications of Second
Order Differential Equations:-

2. GROUP MEMBERS
 AYESHA JAVED(30)
 SAFEENA AFAQ(26)
 RABIA AZIZ(40)
 SHAMAIN FATIMA(50)
 UMAIRA ZIA(35)

4. MODELING
 A mathematical model is a description of a system using
mathematical concepts and language. The process of developing a
mathematical model is termed as Mathematical modeling.
Mathematical models are used in the
 Natural sciences
(such as physics, biology, earth science, chemistry)
 Engineering disciplines
(such as computer science, electrical engineering)
 Social sciences
(such as economics, psychology, sociology, political science).

5. EXPLANATION
 A model may help to explain a system and to study the
effects of different components, and to make predictions
about behavior.

7. SECOND ORDER DIFFERENTIAL EQUATION
 A second order differential equation is an equation
involving the unknown function y, its derivatives y'
and y'', and the variable x. We will consider explicit
differential equations of the form:
Explicit solution is a solution where the dependent
variable can be separated.
 Where a, b, and c are constants.

8. HOMOGENOUS EQUATION
The differential equation:
Is a second order, constant coefficient, linear, homogenous
differential equation. Its solution is found from the solutions
to the auxiliary equation:
aD2+bD+c=0
These are:

9. AUXILIARY EQUATION
The characteristic equation (or auxiliary equation) is
an algebraic equation of degree n upon which depends
the solution of a given nth-order differential equation
or difference equation. The characteristic equation can
only be formed when the differential or difference
equation is linear and homogeneous, and has
constant coefficients. There are three types of roots:
 Real and distinct roots
 Real and repeater roots
 Complex roots

10. REAL AND DISTINCT:
If the Auxiliary equation:
am2+bm+c=0
With solution:
Where:
m1and m2 are real and m1≠ m2
Then the solution is:

11. REAL AND REPEATED:
If the Auxiliary equation:
am2+bm+c=0
With solution:
Where:
m1and m2 are real and m1=m2
Then the solution is:

12. COMPLEX ROOTS
If the Auxiliary equation:
am2+bm+c=0
With solution:
Where:
m1and m2 are complex
Then the solution to differential equation whose roots are
complex can be written as:

14. NON-HOMOGENOUS EQUATION
The second order, constant coefficient, linear, non-
homogenous differential equation is an equation of the
type:
The solution is in two parts yc+ yp:
a) Part1, yc is the solution to the homogenous equation
and is called complementary function which is the
solution to the homogenous equation
b) Part 2, yp is called the particular integral.

15. f(t) Choice for yp(t)
a xn cnxn+cn-1xn-1+…+c1x1+c0
a xneat eat (cnxn+cn-1xn-1+…+c1x1+c0 )
K sin(at) c1 sin(at)+c2 cos(at)
K cos(at) c1 sin(at)+c2 cos(at)
K eat sin(bt) eat (c1 sin(bt)+c2 cos(bt))
K eat cos(bt) eat (c1 sin(bt)+c2 cos(bt))
TABLE

16. EXAMPLE
Q: To solve:
 Complementary function: (part1)
Auxiliary equation:
D2 -5D+6=0
D2 -3D -2D +6 =0
(D -3)(D -2)=0
D=2, 3
Complementary function:
yc = c1e2x +c2e3x

17.  Particular integer: (part 2)
Assume a form for yp as yp=Ax2 +Bx +C,
dyp/dx =2Ax +B
d2yp/dx2 =2A
then substitution in:
We get: 2A -10Ax -5B +6Ax2 +6Bx +6C = x2
6Ax2 +(6B-10A)x +(6C+2A-5B)=x2 +0x +0
comparing coefficients of x2, x and constant:
A=1/6; B=5/18; C=19/18
So that:
yp =(1/6)x2 +(5/18)x +(19/18)

18.  The Complete solution of:
Consist of:
Complementary function+ particular function
That is:
y= yc+yp = c1e2x +c2e3x +(1/6)x2 +(5/18)x +(19/18)

20. APPLICATIONS OF SECOND ORDER DIFFERENTIAL
EQUATION:
Second-order linear differential equations have a variety of
applications in science and engineering. In this section we
explore two of them:
1) The vibration of springs
2) Electric current circuits.

22. NEWTON SECOND LAW OF MOTION

23. VIBRATING SPRINGS
AND
SIMPLE HARMONIC
MOTION
• We consider the motion of an
object with mass m at the end
of a spring that is either vertical
or horizontal on a level surface.

24. The force exerted by a spring is given by Hooke's
Law; this states that if a spring is stretched or
compressed a distance x from its natural length,
then it exerts a force given by the equation
HOOKE’S LAW

25. VIBRATING SPRINGS
According to Hooke’s Law, which says
that if the spring is stretched (or
compressed) units from its natural
length, then it exerts a force that is
proportional to :
restoring force= -kx

26. EXPLANATION:
Where k is a positive constant (called the spring constant) and By
Newton’s Second Law, we have:
F = ma = -kx
Acceleration is defined as the rate of change of velocity and
velocity of an object is the rate of change of its position with
respect to time t.
a= dv/dt ;
v= dx/dt ;
a= d2x/dt2
By substituting the value, we have:
m(d2x/dt2)= -kx

27. CONTINUE…
We get:
This is second-order linear differential equation. Its characteristic
equation is mD2 +k =0
D2 =-k/m;
D=√-k/m;
Its roots are: We can combine the constants k and m by making the
substitution: √ k/m= ω
D = ±ωi

28. This type of motion is called simple harmonic motion.
As r=d2x/dt2 The solution of Eq. is a function of time
D = ±ωi
a ±bi ; a=0 and b= ω The solution of Eq. (1) is a function of time
x(t) =eat (c1 cos(b)+ c2 sin(b))
Thus, the general solution is:
x(t) =e0t (c1 cos(ωt)+ c2 sin(ωt))
x(t) =c1 cos(ωt)+ c2 sin(ωt))
AS ω = √ k/m

29. EXAMPLE
A spring with a mass of 2 kg has natural length 0.5 m. A force of
25.6 N is required to maintain it stretched to a length of 0.7 m. If
the spring is stretched to a length of 0.7 m and then released with
initial velocity 0, find the position of the mass at any time t.
Solution:
From Hooke’s Law, the force required to stretch the spring is
k (0.2)=25.6
k =25.5/0.2 =128
So, Using this value of the spring constant , together with m=2,
we have

30. As in the earlier general discussion, the solution of this equation
is
2D2 +128 =0
D2 =-128/2 = -64
D = ± 8i
Thus, the general solution is:
x(t) =c1 cos(8t)+ c2 sin(8t)
We are given the initial condition that x(0)= 0.2. But from above
equation, we have:
x(0) =c1 cos(0)+ c2 sin(0) = c1
Comparing both we get:
c1 = 0.2

31. Differentiate equation, we get:
x’(t) = -8c1 sin(8t)+ 8c2 cos(8t)
Since the initial velocity is given as x’(0)=0, and
x’(0) = -8c1 sin(0)+ 8c2 cos(0)
x’(0) = 8c2
Compare both equations, we get:
c2 = 0
so the solution is:
x(t) =(1/5) cos (8t)
Or
x(t) =(0.2) cos (8t)

33. ELECTRIC CURRENT CIRCUIT
The simple electrical circuit consists of a resistor R in ohms; a
capacitor C in farads; an inductor L in henries ; and an
electromotive force (emf) E(t) in volts , usually a battery or a
generator, all connected in series. The current I flowing
through the circuit is measured in amperes and the charge Q on
the capacitor is measured in coulombs.

34. EXPLANATION
It is known that the voltage drops across a resistor, a capacitor,
and an inductor are respectively RI, (1/C)Q, and L(dI/dt)
where Q is the charge on the capacitor. The voltage drop
across an emf is E(t). Thus, from Kirchhoff's loop law, we
have:
1

35. SYMBOLS:

36. CONTINUE…
The relationship between I and Q is
I = dQ/dt, dI/dt = d2Q/ dt2
Substituting these values into the above equation, we
obtain
which is a second-order linear differential equation with
constant coefficients. If the charge Q and the current I
are known at time 0, then we have the initial condition

37. Thus
Q(0) = Q0 , (dQ/dt)|t=0 =I(0) = I0
To obtain a differential equation for the current, we differentiate
equation (1) with respect to t and then substitute it directly into
the resulting equation to obtain
The first initial condition is
I(0)= I0

38. The second initial condition is obtained from equation (1) by
solving for dI/dt and then setting t=0. Thus
Eq 1
L dI/dt=E(t)-RI-Q/C
(dI/dt)= (1/L)E(t) –(R/L)I –(1/LC )Q
NOW t=0 and I=I0 , Q= Q0
(dI/dt)|t=0 = (1/L)E(0) –(R/L)I0 –(1/LC )Q0

39. Characteristic equation
L D2 +RD+1/C=0
R=40Ὠ, L=1H, C= 16X 10-4F, E(t)= 100cos 10t,
e.g:-
Qc (t) =e-at(c1cos bt+c2sin bt)

41. EXAMPLE
Find the charge and current at time in the circuit if R=40Ὠ,
L= 1H, C= 16X 10-4F, E(t)= 100cos 10t, and the initial
charge and current are both 0.
Solution:
With the given values of L, R, C, and E(t), it becomes:
(Part 1) The auxiliary equation is r2+40r+625= 0 with
roots

42. SOLUTION:
so the solution of the complementary equation is
Qc (t) =e-20t(c1cos 15t+c2sin 15t)
(Part 2) For the method of undetermined coefficients we try the
particular solution
Qp (t)= Acos 10t+Bsin 10t
Qp'(t) = -10Asin10t +10Bcos 10t
Qp" (t) = -100Acos 10t -100Bsin 10t
Substituting it in first equation, we have
(-100Acos 10t -100Bsin 10t ) +40(-10Asin10t +10Bcos 10t)
+625(Acos 10t+Bsin 10t) = 100cos 10t

43. (525A + 400B) cos 10t +(-400A +525B)sin 10t = 100cos 10t
The solution of this system is
A = 84/697; B =64/697
so a particular solution is
Qp (t)= (84/697)cos 10t + (64/697)sin 10t)
and the general solution is
Q (t) =Qc (t) +Qp (t) =e-20t(c1cos 15t+c2sin 15t) + (84/697)cos 10t
+(64/697)sin 10t
Imposing the initial condition Q (0)=0 , we get
Q (0)= c1 +(84/697) =0
c1 = -84/697

44. To impose the other initial condition we first differentiate to find the
current:
I = Qp'(t)=e-20t(-15c1sin15t+15c2cos15t -20 e-20t(c1cos 15t+c2sin 15t) -
10(84/697)sin 10t +10(64/697)cos 10t
I = e-20t[-15c1sin15t+15c2cos15t -20c1cos 15t -20c2sin 15t]
+(40/697)[-21 sin 10t +16 cos 10t]
I = e-20t[(-20c1 +15c2)cos 15t +(-15c1 -20c2)sin 15t ] +(40/697)[-21 sin
10t +16 cos 10t]
I (0) = -20c1 +15c2 +(640/697) = 0
c2 = -(464/2091)

45. Thus, the formula for the charge Q is:
and the expression for the current I is