ENGI 3424
2.
2 - Second Order Linear ODEs
Page 2-01
Second Order Linear Ordinary Differential Equations
The general second order linear ordinary differential equation is of the form
d2y
dy
 P  x
 Q  x y  R  x
2
dx
dx
Of the second (and higher) order ordinary differential equations, the linear equations with
constant coefficients will command most of our attention in this chapter:
d2y
dy
 P
 Q y  R  x
2
dx
dx
Contents:
2.1
2.2
2.3
2.4
2.5.
Complementary Function
Particular Solution (Variation of Parameters)
Particular Solution (Undetermined Coefficients)
Higher Order Linear Ordinary Differential Equations
Numerical Methods
ENGI 3424
2.1
2.1 – Complementary Function
Page 2-02
Complementary Function
d2y
dy
The homogeneous equation associated with the ODE
 P
 Q y  R  x  is
2
dx
dx
d2y
dy
 P
 Qy  0
2
dx
dx
The principle of superposition of solutions of the homogeneous equation is valid because
it is linear. That is, if y = u(x) and y = v(x) are both solutions of the homogeneous
ODE, then so also is any linear combination of these functions: y = c1 u(x) + c2 v(x) ,
where c1 and c2 are any constants.
Adding any solution of the homogeneous ODE to a particular solution of the original
ODE generates another solution of the original ODE.
Thus the general solution (abbreviated as G.S.) of
d2y
dy
 P
 Q y  Rx 
2
dx
dx
can be partitioned into two parts:
the complementary function (C.F., which is the general solution of the associated
homogeneous ODE) and a particular solution (P.S.).
If y  e x then
Substituting y  e x into the homogeneous ODE:
from which the auxiliary equation (A.E.) follows:
2 + P  + Q = 0
[The choice of y  e x as a trial solution to the homogeneous ODE is justified later, on
page 2-09, when a more general method for finding the complementary function is
introduced.]
ENGI 3424
2.1 – Complementary Function
The solution of the auxiliary equation  2 + P  + Q = 0 is
Distinct roots (1  2)  the complementary function is
[The case of equal roots will be dealt with later, on page 2.08 .]
Example 2.1.1
Solve the differential equation
y" + 3y'  4y = 0
The auxiliary equation is
The complementary function (which is also the general solution) is
Checking the solution:
Page 2-03
2.1 – Complementary Function
ENGI 3424
Page 2-04
Example 2.1.2
Solve
y"  2y' + 2y = 0
In general, when the roots of the auxiliary equation are a complex conjugate pair of
values,  = a ± bj , then the complementary function is


yC  x   eax c1 e jbx  c2 e jbx  eax  c3 cos bx  c4 sin bx 
(where the arbitrary constants are related by c3 = c1 + c2 and c4 = j (c2
or
yC(x) = A eax cos(bx  )

c3
c4
 where A  c32  c4 2 , cos  
and sin  
2
2

c3  c4
c32  c4 2

or
yC(x) = A eax sin(bx  )

c3
c4
 where A  c32  c4 2 , sin  
and cos  

c32  c4 2
c32  c4 2

 c1) )








Note that, for an auxiliary equation of this type, with real coefficients, where the solution
is constrained to be real, the arbitrary constants c3 and c4 are both real, but c1 and c2
often are not. For this reason, the forms involving the trigonometric functions are
usually preferred to the complex exponential form.
ENGI 3424
2.1 – Complementary Function
Page 2-05
Example 2.1.3
A spring, that is not at its natural length, experiences a restoring force R that is
proportional to the extension s beyond the natural length and is directed towards the
equilibrium position. In the absence of friction, this would lead to undamped simple
harmonic motion.
Let us suppose that there is also a friction force D that is
proportional to the speed and acts in the opposite direction to the velocity.
Restoring force proportional to displacement 
Friction (drag) proportional to speed 
Newton’s second law of motion:
Therefore the ODE governing the motion of the spring is
ENGI 3424
2.1 – Complementary Function
Page 2-06
Example 2.1.3 (continued)
Suppose that m = 1 kg, b = 6 kg s−1 , c = 25 kg s−2 and that the spring begins at its
equilibrium position, but moving at 2 m s−1 to the right, so that s(0) = 0 and v(0) = 2, then
the ODE becomes
Note that if b = 0 (no friction at all), then the system is totally undamped and exhibits
simple harmonic motion:
s(t) = A sin (kt  )
c
where k 
.
m
2.1 – Complementary Function
ENGI 3424
Page 2-07
The General Spring Problem
d 2s
b ds
c


s  0
2
dt
m dt
m
2
Case 1:
b
c
   4 
m
m
λ =
2
Case 2:
b
c
   4 
m
m
λ =
or
or
2
Case 3:
λ =
b
c
   4 
m
m
2.1 – Complementary Function
ENGI 3424
Page 2-08
Complementary Function when the Auxiliary Equation has Equal Roots
λ 1 = λ 2 (= λ)
 the ODE becomes
y" − 2λ y' + λ 2 y = 0
One solution to this equation is C1 eλx
We require another solution that is independent of this one (so that there will be two
distinct arbitrary constants of integration in the complementary function).
Try f (x) = C2 x eλx
[This second form arises naturally from the operator method, on page 2.09.]
Then f '(x) =
and
f "(x) =
Example 2.1.4
Solve
y"  6y' + 9y = 0
ENGI 3424
2.1 – Complementary Function
Page 2-09
The Operator Method
The homogeneous ordinary differential equation with constant coefficients,
d2y
dy
 P
 Qy  0
2
dx
dx
can also be written, using differential operators, in the form
d
 d

  k1    k 2  y  0
 dx
  dx

Justification:
The second order ODE can therefore be re-written as a linked pair of first order linear
ordinary differential equations [the method of reduction of order, section 1.4]:
d

(A)
  k1    0 ,
 dx

where
dy
 
 k2 y .
(B)
dx
Solution:
ENGI 3424
2.1 – Complementary Function
Operator Method (continued)
Page 2-10
2.1 – Complementary Function
ENGI 3424
Page 2-11
Operator Method (continued)
Summary for the Complementary Function:
ODE:
y" + P y' + Q y = 0
A.E.:
λ2 + P λ + Q = 0
1x
λ real and distinct

yC  A e
 Be
λ real and equal

yC   Ax  B  e x
2x
λ complex conjugate pair
 yC  ea x  C cos bx  D sin bx  , where a  Re    , b  Im   
2.2 – Variation of Parameters
ENGI 3424
2.2
Page 2-12
Particular Solution (Variation of Parameters)
The method of variation of parameters is a general method for finding the particular
solution of a linear ODE.
If the complementary function for the ODE
d2y
dy
 P
 Q y  Rx 
2
dx
dx
is yC(x) = C1 y1(x) + C2 y2(x) , (so that { y1(x), y2(x) } is a basis for the space of all
solutions to the homogeneous ODE), then the particular solution is
yP(x) = u(x) y1(x) + v(x) y2(x) ,
where the functions u(x) and v(x) need to be determined.
We need two constraints in order to pin down the functional forms of u(x) and v(x).
An obvious constraint is that u(x) y1(x) + v(x) y2(x) be a particular solution of the ODE.
We have considerable freedom as to what the other constraint will be.
yP = u y1 + v y2
Impose our “free” constraint,
2.2 – Variation of Parameters
ENGI 3424
Page 2-13
Define the Wronskian function W(x) to be
y
Wy1, y2  x   det  1
 y1
y2 
y2 
together with the associated determinants
 0 y2 
y 0
W1  det 
  y2 R and W2  det  1

   y1R
 R y2 
 y1 R 
then Cramer’s rule yields solutions for u' and v' :
W1
W
and v  2 .
W
W
Therefore a particular solution is yP = u y1 + v y2 , where
u 
u  x  

y2  x  R  x 
dx , v  x   
W  x

y
y1  x  R  x 
dx , W  x   1
y1
W  x
y2
y2
Note that we can ignore the arbitrary constants of integration in both integrals, because
A y1 and B y2 are both solutions of the homogeneous ODE and can therefore be
absorbed into the complementary function.
Example 2.2.1
Find the general solution of the ODE
y" + 2y' − 3y = x2 + e2x .
ENGI 3424
2.2 – Variation of Parameters
Example 2.2.1 (continued)
Particular Solution by Variation of Parameters:
Page 2-14
ENGI 3424
Example 2.2.1 (continued)
2.2 – Variation of Parameters
Page 2-15
ENGI 3424
2.2 – Variation of Parameters
Example 2.2.2
Find the general solution of:
y" + y = tan x
Page 2-16
2.2 – Variation of Parameters
ENGI 3424
Example 2.2.2
(continued)
Page 2-17
ENGI 3424
2.2 – Variation of Parameters
Page 2-18
Example 2.2.3
Use the variation of parameters method to find the particular solution, then find the
complete solution of the initial value problem
y" − 2y' + y = ex,
y(0) = 0, y'(0) = 1
2.2 – Variation of Parameters
ENGI 3424
Example 2.2.3
Page 2-19
(continued)
Note that a complete solution requires additional information (often in the form of
initial conditions). Two pieces of information are needed in order to evaluate both
arbitrary constants of integration. However, do not substitute these conditions into the
complementary function; wait until the general solution has been obtained.
2.2 – Variation of Parameters
ENGI 3424
Page 2-20
The method of variation of parameters may also be used to find the particular solution of
a linear second order ODE whose coefficients are not constant. However, general
methods for finding the complementary functions in these cases are beyond the scope of
this course.
Example 2.2.4
Verify that the complementary function for the ODE
d2y
dy
x2 2  5x
 8 y  3x
dx
dx
is yC  A x 4  B x 2
and hence find the general solution of the ODE.
yC  A x 4  B x 2

yC 
The particular solution is
yP  x   u  x  y1  x   v  x  y2  x  where
y1  x 4 and y2  x 2
ENGI 3424
2.2 – Variation of Parameters
Page 2-21
Example 2.2.4 (continued)
[This example appeared in the journal note “A seldom used formula for ODEs”,
George, G.H., Mathematical Gazette, vol. 92, #524, pp. 344-348]
2.3 – Undetermined Coefficients
ENGI 3424
2.3
Page 2-22
Particular Solution (Undetermined Coefficients)
The general solution to
d2y
dy
 P
 Q y  Rx 
2
dx
dx
is the sum of the complementary function and any one solution (the particular solution)
that we can find to the original inhomogeneous ODE.
The method of variation of parameters to find the particular solution is powerful, but it
can involve an unnecessary level of effort. In some cases, an alternative method
(undetermined coefficients) is available that is often faster to use.
If the function R(x) does not contain any part of the complementary function, then
assume that the particular solution yP(x) is of the same form as R(x).
Example 2.3.1
(same as Example 2.2.1)
Find the general solution of the ODE
y" + 2y' − 3y = x2 + e2x .
A.E.:
C.F.:
P.S.:
R(x) contains neither e−3x nor ex .
R(x) is the sum of a quadratic function and e2x .
Therefore try the sum of a quadratic function and a multiple of e2x ,
where all four coefficients are to be determined.
2.3 – Undetermined Coefficients
ENGI 3424
Example 2.3.1
(continued)
yP = a x2 + b x + c + d e2x
Matching coefficients:
G.S.: y(x) = yC(x) + yP(x)
Therefore
Page 2-23
2.3 – Undetermined Coefficients
ENGI 3424
Page 2-24
General Method:
d2y
dy
 P
 Q y  Rx 
2
dx
dx
If the function R(x) does not contain any part of the complementary function, then
assume that the particular solution yP(x) is of the same form as R(x).
If R(x) = ekx ,
then try yP = c ekx , with c to be determined.
If R(x) = (a polynomial of degree n),
then try yP = (a polynomial of degree n), with all (n + 1) coefficients to be determined.
If R(x) = (a multiple of cos kx and/or sin kx),
then try yP = c cos kx + d sin kx , with c and d to be determined.
This method can be extended to cases where
R(x) = (a sum and/or product of the functions above).
But: if part (or all) of yP is included in the complementary function yC , then
multiply yP by x.
As seen above in Example 2.3.1, the method of undetermined coefficients can be used in


Example 2.2.1 R  x   x 2  e2x .
It can also be used in Example 2.2.3  R  x   e x  .
However, in Example 2.2.2, R(x) = tan x , which does not fall into any of these
categories. The method of undetermined coefficients cannot be used in this case to find
the particular solution (which was yP    cos x   ln sec x  tan x ).
ENGI 3424
2.3 – Undetermined Coefficients
Example 2.3.2
Consider a model of the simple series RLC
circuit, where the constants R, L, C are the
resistance,
inductance
and
capacitance
respectively, E(t) is the applied electromotive
force, t is the time and I(t) is the resulting
current.
Examine the voltage drops around the circuit:
R:
L:
C:
and note that i 
dQ
.
dt
Page 2-25
ENGI 3424
2.3 – Undetermined Coefficients
Example 2.3.2 (continued)
Particular solution
If E(t) = Eo (constant), then
Suppose that the e.m.f. is sinusoidal, so that E(t) = Eo sin ωt , then
E  cos  t
1 dE
 o
L dt
L
P.S.:
Try
iP = a sin ωt + b cos ωt

iP' = −bω sin ωt + aω cos ωt

iP" = −aω2 sin ωt − bω2 cos ωt
R
1
iP 
iP 
L
LC
b R
a 
a R
b 


2
2


 a 
 sin  t   b 
 cos  t
L
LC 
L
LC 


E
 0 sin  t  o cos  t
L
 iP 

b R
a

 a 2
L
LC
 b 
a
1   2 LC 
 RC
 a

Eo
a R
1 
 
1   2 LC    2 

 
L
LC 
L
  RC

 a 1   2 LC 2 
a R  RC C 
  Eo

 
  RC LC 
 RC LC
L


Eo C  RC 
E
 RC LC
 a  o 

2
2
2
2
L  RC   1   2 LC 
 RC   1   2 LC 

Eo C 1   2 LC 
Eo C  RC 
1   2 LC
 b 


2
2
2
2
 RC

RC

1


LC



  RC 2  1   2 LC 
Page 2-26
2.3 – Undetermined Coefficients
ENGI 3424
Page 2-27
Example 2.3.2 (continued)
Therefore the particular solution is
 iP 

Eo C  RC  sin  t  1   2 LC  cos  t
 RC 
2
 1   2 LC 

2
which is a steady-state sinusoidal response to the sinusoidal electromotive force, but with


 RC


a phase difference of   arccos 
.
2
2
  RC   1   2 LC  


The total current is then
 E C  RC  sin t  1   2 LC  cos t





 Rt 
D
D
i  t   e 2 L  A sin 
t   B cos 
t    
2
2


2

2
L
2
L

RC

1


LC











transient
 
steady  state
As a specific example, if E(t) = 17 sin 2t, R = 120 Ω, C = 1 mF and L = 10 H,
then it can be shown that
1
i  t   e6t  A sin 8t  B cos8t  
 sin 2t  4 cos 2t 
120
The transient current, iC  t   e 6t  A sin 8t  B cos8t  , dies away very quickly.
Its magnitude falls to under 1% of the total current permanently in less than a second.
The values of the two arbitrary constants can be found from the initial conditions, but,
given that the complementary function becomes negligible in a very short time, one often
does not try to evaluate them.
Example 2.3.3
Find the complete solution of the initial value problem
y" + 2 y' + y = ex , y(0) = y' (0) = 1


ENGI 3424
2.3 – Undetermined Coefficients
Example 2.3.3 (continued)
Page 2-28
ENGI 3424
2.4
2.4 - Higher Order ODEs
Page 2-29
Higher Order Linear Ordinary Differential Equations
The nth order ordinary differential equation
dny
d n 1 y
d n 2 y
d2y
dy

a

a



a
 an 1
 an y  Rx 
1
2
n 2
n
n 1
n 2
2
dx
dx
dx
dx
dx
(where the coefficients ai are all constant) can be solved as follows.
Form the auxiliary equation
n + a1n1 + ... + an22 + an11 + an = 0
Find all n values for .
Form the complementary function yC , which will be a linear combination of
e
1x
,e
2 x
,
,e
n x
 (except for repeated roots).
Complex conjugate pairs can be re-written in terms of sine and cosine functions.
Find a particular solution yP (by inspection, undetermined coefficients, or variation of
parameters, as extended to this higher order equation).
Write down the general solution y = yC + yP.
n initial and/or boundary conditions will be needed at this stage to evaluate all of the n
arbitrary constants of integration.
Example 2.4.1
Find the general solution of
5
4
d y
d y
d3y
d2y
dy

2

3

4
4
 8x
5
4
3
2
dx
dx
dx
dx
dx
Auxiliary equation:
 5 + 2 4  3 3  4 2 + 4 = 0
ENGI 3424
Example 2.4.1 (continued)
2.4 - Higher Order ODEs
Page 2-30
ENGI 3424
2.4 - Higher Order ODEs
Page 2-31
Example 2.4.2
A thin uniform beam of length L experiences a transverse deflection y(x) at location x
(0 < x < L) due to a transverse force per unit length W(x) . From the elementary theory
of beams, the governing differential equation is
EI
d4y
 W  x
dx 4
where EI is the flexural rigidity of the beam,
E is Young’s modulus and I is the moment
of inertia of the beam about its central axis.
If the beam has uniform elastic properties and a uniform cross section along its length,
then its flexural rigidity EI is constant.
The initial conditions are:
y  0  a  the deflection at x = 0,
y  0  b  the slope at x = 0,
EI y  0  c  the bending moment at x = 0 and
EI y  0  d  the shear at x = 0.
Find the deflection for a cantilever beam with uniform load W(x) = W (constant).
For a cantilever beam, the deflection and slope are zero at the fixed end (x = 0), while the
bending moment and the shear are zero at the free end (x = L ).
ENGI 3424
Example 2.4.2 (continued)
2.4 - Higher Order ODEs
Page 2-32
ENGI 3424
2.5.
Numerical Methods
2.5 - Numerical Methods
Page 2-33
[not examinable]
An initial value problem with an nth order ordinary differential equation can be re-written
as a linked set of n first order initial value problems. Where analytical solutions are
difficult or impossible, extensions of numerical methods, such as the Euler and RK4
methods, can be applied to obtain approximate solutions to these initial value problems at
particular values of the independent variable.
A more general case for initial value problems involving second order ODEs,
d2y
dy
 P  x, y 
 Q  x, y  y  R  x  , y  0   a , y   0   b ,
2
dx
dx
transforms into the linked pair of first order initial value problems
dy
dz
 z , y  0   a and
 R  x   Q  x, y  y  P  x, y  z , z  0   b
dx
dx
This pair of problems can be solved sequentially (as in the operator method on pages 2.09
to 2.11) if the coefficients Q  x, y   0 and P  x, y   f  x  .
Otherwise analytical
solutions are mostly beyond the scope of this course. Symbolic algebra software, such
as Maple®, may be able to provide exact solutions or numerical approximations.
Example 2.5.1
Use software to estimate the value of y(0.5) given that
d2y
dy
 y2
 y  sin x , y  0   0 , y  0   1 .
2
dx
dx
Using Maple software, the commands
> with(DEtools):
> DEplot(diff(y(x),x$2)+y(x)*y(x)*diff(y(x),x)+y(x)-sin(x),y(x),
x=-3..12, [[y(0)=0,D(y)(0)=1]],
y=-6..3, stepsize=.05, linecolour=[blue]);
produce the plot
ENGI 3424
2.5 - Numerical Methods
Page 2-34
Example 2.5.1 (continued)
which does not resemble any standard function.
Zooming in (to x=0..0.5, y=0..0.5 and enhancing slightly),
we can see that our estimate is
y  0.5  0.495
The Maple worksheet is on the course web
site, at
"www.engr.mun.ca/~ggeorge/3424/
demos/ex251.mws"
The further sequence of Maple commands
>
>
>
>
ode1 := diff(y(x),x$2)+y(x)*y(x)*diff(y(x),x)+y(x)-sin(x);
ic1 := y(0) = 0, D(y)(0) = 1;
dsol1 := dsolve({ode1, ic1}, numeric, range=0..1);
dsol1(0.5);
produces y(0.5) = .494 671...
ENGI 3424
2.5 - Numerical Methods
Page 2-35
Example 2.5.2
Find the general solution of the ordinary differential equation
d2y
dy
x2 2  6 x
 12 y  24
dx
dx
and find the complete solution given the additional information y 1  2,
y 1  1 .
This is an example of a “dimensionally homogeneous” ODE (also known as a CauchyEuler ODE). A change of variables x  et will convert the ODE into a form with
constant coefficients, which can be solved exactly using the methods in the earlier
sections of this chapter.
The sequence of Maple commands
>
>
>
>
with(DEtools):
ode1 := x^2*diff(y(x),x$2) - 6*x*diff(y(x),x) + 12*y(x) - 24;
ic1 := y(1) = 2, D(y)(1) = 1;
dsolve(ode1);
produces the general solution
y(x) = x^3 _C2 + x^4 _C1 + 2
The additional command
> dsolve({ode1, ic1});
produces the complete solution
y(x) = –x^3 + x^4 + 2
The Maple worksheet is available at
"www.engr.mun.ca/~ggeorge/3424/demos/ex252.mws"
One can easily verify that this complete solution is correct:
ENGI 3424
2.5 - Numerical Methods
Example 2.5.2
Page 2-36
(continued)
The exact solution (not examinable except possibly as a bonus question) is obtained as
follows:
dx
 et  x
dt
dy dy dt dy dx
dy
1 dy
  

 e t

By the chain rule,
dx dt dx dt dt
dt
x dt
Let x  et


x
dy dy

dx dt
d 2 y d  dy  d  dy  dx
1 d  t dy  1  t d 2 y
dy 




 e
 et

 
 
  e
2
2
dx
dx  dx  dt  dx  dt
x dt 
dt  x 
dt
dt 
2
1  d 2 y dy 
d 2 y dy
2 d y
 2 2  
 x
 2 
x  dt
dt 
dx 2
dt
dt
d2y
dy
 bx
 cy  r  x  ,
2
dx
dx
(where b and c are constants) therefore transforms into the ODE
 d 2 y dy 
dy
d2y
dy


b

cy

r
x

  b  1  cy  r  x 


 2

2
dt 
dt
dt
dt
 dt
In this case, b = –6, c = 12 and r(x) = 24.
d2y
dy
The equivalent ODE is
 7  12 y  24
2
dt
dt
2
A.E.:   7  12  0     3   4  0    3, 4
Any Cauchy-Euler ODE of the type x 2
 
 
C.F.:
yC  t   Ae3t  Be4t  A et
P.S.:
r  x  is a constant. Therefore try yP  x   c  yP  x   0  yP  x   0
3
 B et
Substituting into the ODE: 0  0  12c  24
4

yC  x   Ax 3  Bx 4
 c2
 yP  x   2
Therefore the general solution is y  x   Ax3  Bx 4  2
Applying the conditions y 1  2,
y 1  1 quickly leads to A   1, B  1 .
The complete solution is therefore y  x   x4  x3  2 .
ENGI 3424
2.5 - Numerical Methods
Page 2-37
Example 2.5.3 (same as Example 5.11.2)
Find the general solution (as a power series about x = 0) to the ordinary differential
equation
d2y
 x2 y  0
2
dx
The sequence of Maple commands
> with(DEtools):
> ode := diff(y(x),x$2)
> dsolve(ode);
+ x*x*y(x);
produces the exact general solution
 1 x2 
 1 x2 
y  x   _ C1 x BesselJ  ,   _ C 2 x BesselY  , 
4 2 
4 2 
where the Bessel functions J  x  and Y  x  are beyond the scope of this course.
Converting the exact solution into a power series,
> Order := 15;
> dsolve(ode, y(x), series);
produces
1
1
1
1
y  0 x4 
D  y  0  x5 
y  0  x8 
D  y  0  x9
12
20
672
1440
1
1

y  0  x12 
D  y  0  x13  O x15
88704
224640
Letting A  y  0  and B  y  0  this solution can be re-written as
y  x   y  0   D  y  0  x 
 
1
1 8
1

y  x   A 1  x 4 
x 
x12 
12
672
88704

1 5
1 9
1

 B x 
x 
x 
x13
20
1440
224640








ENGI 3424
2.5 - Numerical Methods
[Space for Additional Notes]
END OF CHAPTER 2
Page 2-38