1. KNF1023
Engineering
Mathematics II
Second Order ODEs
Prepared By
Annie ak Joseph
Prepared By
Annie ak Joseph Session 200782009
2. Learning Objectives
Explain about variation of parameters
method in the inhomogeneous ODEs
Explain about Spring Mass System and
Electric Circuits
3. Method of Variation of Parameters
There is a more systematic approach for
finding a particular solution of the 2nd order linear
inhomogeneous ODE.
y "+ f ( x) y '+ g ( x) y = r ( x)
The approach known as the method of variation
of parameters makes use of two linearly
independent solutions of the corresponding
homogeneous ODE.
4. Method of Variation of Parameters
The solution of the second order inhomogeneous
ODEs is
y= y p ( x) + Y ( x)
{
{
a particular solution general solution of the
of in hom ogeneous ODE corresponding hom ogeneous ODE
y p = u ( x)Y1 ( x) + v( x)Y2 ( x)
5. Method of Variation of Parameters
Thus, the variation parameters will involved
two equations which is
r (x )y 2 (x )
u(x ) = −∫ ' '
dx
( y1 y 2 − y1 y 2 )
and
r ( x ) y1 ( x )
v( x ) = ∫ dx
( ' '
y1 y 2 − y1 y 2)
6. Example 1
Consider the inhomogeneous ODE
y − y = 3 exp(x )
''
Notice that y '' − y = 0 has two linearly
independent solutions given by
y1 = exp(− x ) and y 2 = exp( x )
Hence
y1 y2 − y1' y2 = exp ( − x ) exp ( x ) − ( − exp ( − x ) ) exp ( x ) = 2
'
7. Continue…
If we look for a particular solution of the
inhomogeneous ODE in the form y p = u ( x )e − x + v ( x )e x
then u and v are given by
3exp( x) exp( x) 3
u ( x) = − ∫ dx = − exp(2 x)
2 4
and
3 exp( x ) exp( − x ) 3
v( x) = ∫ dx = x
2 2
8. Continue…
Notice that we can ignore the constants of
integration, since we are only interested in getting
a particular solution.
So, a particular solution of the inhomogeneous
ODE y "− y = 3exp( x) is
y p = u ( x) exp(− x) + v( x) exp( x)
3 3
= − exp(2 x) exp(− x) + x exp( x)
4 2
3 3
= − exp( x) + x exp( x)
4 2
9. Continue…
The required general solution of the
inhomogeneous ODE is
3 3
y = − exp( x) + x exp( x) + C exp(− x) + D exp( x)
4 2
Where C and D are arbitrary constants.
This can again be rewritten as
3 3
y = x exp( x) + C exp(− x) + D − exp( x)
2 4
3
= x exp( x) + A exp(− x) + B exp( x) (as before)
2
3
Where , A = C and B = D − ,are arbitrary constants
4
10. APPLICATIONS OF SECOND-ORDER EQUATIONS -
Spring –Mass System
Simple harmonic motion may be defined as
Motion in a straight line for which the acceleration
Is proportional to the displacement and in the
opposite direction. Example of this type of motion
are a weight on a spring.
Damped System
If we considered a mass-spring system and
modeled it by the homogeneous linear ODE,
my"+cy '+ ky = 0
11. APPLICATIONS OF SECOND-ORDER EQUATIONS -
Spring –Mass System
Here y(t) as a function of time t is the
displacement of the body of mass m from rest.
Forces within this system are inertia my”, the
damping force cy’ (if c>0) and the spring force ky
acting as a restoring forces.
The corresponding characteristics equation is
2 c k
λ + λ+ =0
m m
The roots are
c 1
λ1, 2 =− ± c 2 − 4mk
2m 2m
12. APPLICATIONS OF SECOND-ORDER EQUATIONS -
Spring –Mass System
Using the abbreviated notations
c 1
α= and β= c 2 − 4mk
2m 2m
We can write
λ1 = −α + β
λ2 = −α − β
13. APPLICATIONS OF SECOND-ORDER EQUATIONS -
Spring –Mass System
The form of the solution will depend on the
damping, and,
Case I. c2> 4mk. Distinct real roots λ1, λ2
(Overdamping)
Case II. c2< 4mk. Complex conjugate
(Underdamping)
Case III. c2 = 4mk. A real double root
(Critical damping)
15. APPLICATIONS OF SECOND-ORDER EQUATIONS -
Spring –Mass System
Case 1. Overdamping.
When the damping constant c is so large that c2 >
4mk, then λ1,λ2 are distinct real roots, the general
solution is
− (α − β ) t − (α + β ) t
y (t ) = c1e + c2e
Case II.Underdamping.
If the damping constant c
is so small that c2< 4mk then β is pure imaginary.
The roots of the characteristics equation are
complex conjugate,
16. APPLICATIONS OF SECOND-ORDER EQUATIONS -
Spring –Mass System
λ1 = −α + jω
λ2 = −α − jω
The general solution is
y (t ) = e −α t
( A cos ωt + B sin ωt )
17. APPLICATIONS OF SECOND-ORDER EQUATIONS -
Spring –Mass System
Case III. Critical damping.
If c2 = 4mk, then β=0, λ1 = λ2 = − α , and the
general solution is
y (t ) = (c1t + c2 )e −αt
Undamped System
If the damping of the system is so small and can be
disregarded, then
my"+ky = 0
The general solution is
y (t ) = A cos ω0t + B sin ω0t
18. Example 2
In testing the characteristics of a particular
type of spring, it is found that a weight of 4.90 N
stretches the spring 0.49 m when the weight and
spring are placed in a fluid which resists the motion
with a force equal to twice the velocity. If the
weight is brought to rest and then given a velocity
of 12 m/s, find the equation of motion.
19. Continue…
Using Newton’s second law, we have
my"+cy '+ ky = 0
my"+2.00 y '+ ky = 0
my" = −2.00 y '− ky
The weight is 4.90 N and the acceleration due to
gravity is 9.80 m/s2. Thus, the mass, m is
m=4.90/9.80 = 0.500 kg
The constant k is found from the fact that the
spring stretches 0.49 m for a force of 4.9 N, Thus,
using Hooke’s Law,
20. Continue…
4.90 = k (0.490)
k = 10.0 N/m
This means that the differential equation to be
solved is
0.500 y"+2.00 y '+10 y = 0
1.00 y"+4.00 y '+20.0 y = 0
Solving this equation, we have
21. Continue…
2
1 . 00 m + 4 . 00 m + 20 . 0 = 0
− 4 . 00 ± 16 . 0 − 4 ( 20 . 0 )( 1 . 00 )
m =
2 . 00
m = − 2 . 00 ± 4 . 00 j
− 2 . 00 t
y = e ( c 1 cos 4 . 00 t + c 2 sin 4 . 00 t )
22. Continue…
Since the weight started from the equilibrium
position with a velocity of 12.0 m/s, we know that
y= 0 and y’ = 12.0 for t = 0. Thus,
0
0 = e (c1 + 0c2 )
c1 = 0
Thus, since c1=0, we have
−2.00t
y = c2e sin 4.00t
y' = c2e− 2.00t (cos 4.00t )(4.00) + c2 sin 4.00t (e− 2.00t )(−2.00)
12.0 = c2e0 (1)(4.00) + c2 (0)(e0 )(−2.00)
c2 = 3.00
24. Application on Electric Circuit
Given the RLC circuit as below,
(a) Show that the circuit can be modeled as
d 2i di i
L 2 + R + = E ' (t )
dt dt C
()
(b) Given that E t = −0.01e − t Solve the differential
equation for the initial conditions, when
and t = 0,i = 0 di = 0
dt
25. Solution
(a) KVL : VL + VR + VC = E ( t )
di 1
L + iR + ∫ idt = E ( t ) differential with respect
dt C to t.
2
d i di i
L 2 + R + = E ' (t )
dt dt C
(b) Given E ( t ) = −0.01e , hence
−t
E ' ( t ) = 0.01e − t
d 2i di i
0.001 2 + 0.005 + = 0.01e − t
dt dt 250
d 2i di
+ 5 + 4i = 10e − t
dt 2 dt
26. Continue…
'' '
i + 5i + 4i = 0
2
λ + 5λ + 4 = 0
λ = −4@− 1
−4 t −t
I = Ae + Be
For Particular Solution
i p = α te − t
i p = α e − t − α te −t
'
i = −α e − (α e − α te
''
p
−t −t −t
) = −2α e −t
+ α te −t
27. Continue…
i '' + 5i ' + 4i = ( −2α e − t + α te− t ) + 5 (α e − t − α te − t ) + 4 (α te− t )
−t
= 3α e
Compare to 10e − t
3α = 10
10
α=
3
10 − t
Thus i p = te
3
28. Continue…
Thus, the general solution is
−4 t −t 10 −t
i = i p + I = Ae + Be + te
3
Given i ( 0 ) = 0 and i ' ( 0 ) = 0
−4( 0 ) −0 10 −0
0 = Ae + Be + 0e , A + B = 0 − − − (1)
3
' −4 t −t 10 − t 10 − t
i = −4 Ae − Be + e − te
3 3
−4( 0 ) 10 −0 10 10
0 = −4 Ae − Be + e − ( 0 ) e , 4 A + B = − − − (2)
−0 −0
3 3 3
29. Continue…
10 10
A= ,B = −
9 9
10 −4t 10 −t 10 − t
i ( t ) = e − e + te
9 9 3
Try to use variation parameter to
solve this application question and
check whether the answer is the
same or not…
30. Prepared By
Annie ak Joseph
Prepared By
Annie ak Joseph Session 2007/2008
