MA108 ODE: Second Order Linear ODE’s

Lecture 9

Niranjan Balachandran
IIT Bombay

Niranjan Balachandran, IITB MA108

Example: Umdamped Simple Harmonic Vibrations
Consider a cart of mass M attached to a wall by means of a spring.
At the equilibrium postion x = 0, the spring exerts no force. If the
center of the cart is at a distance x from the spring, the spring
exerts a force
Fs = −kx
where k > 0 is a constant. By Newton’s second law of motion,
d 2x
M= −kx.
dt 2
Set r
k
a= .
M
Then the characteristic equation of the above differential equation
is m2 + a2 = 0,
i.e., m = ±ıa. Hence, the general solution of the above DE is
x = c1 sin at + c2 cos at
where c1 , c1 ∈ R.
Niranjan Balachandran, IITB MA108
Example continued

Suppose the cart is pulled aside to the initial position x = x0 with

initial velocity v = dx
dt = 0 at t = 0. Then

c2 = x0 and c1 = 0.

Hence,
x = x0 cos at.
where x0 is the amplitude of the simple harmonic vibration and T
is the time required for one complete cycle. Then

aT = 2π,

i.e., r
2π M
T = = 2π .
a K

Niranjan Balachandran, IITB MA108

Example: Damped Vibrations
Suppose, in the previous example, the cart of mass M experiences
a damping force Fd , due to the viscosity of the medium through
which it moves, such a sair or water. Assume that
dx
Fd = −c ,
dt
where c is a positive constant that measures the resistance of the
medium. Then d 2x
M 2 = Fs + Fd
dt
i.e.,
d 2x c dx k
+ + x = 0.
dt 2 M dt M
Set r
c k
2b = and a = .
M M
Then the characteristic equation of the above DE is
m2 + 2bm + a2 = 0.
√
The roots m1 , m2 are given by −b ± b 2 − a2 .
Niranjan Balachandran, IITB MA108
Example continued
Case I: b > a. Then m1 and m2 are distinct negative real numbers.
The general soultion of the DE is
x = c1 e m1 t + c2 e m2 t .
The initial conditions are
dx
x(0) = 0 and (0) = 0.
dt
Since dx
= m1 c1 e m1 t + m2 c2 e m2 t ,
dt
it follows that
c1 + c2 = x0
m1 c1 + m2 c2 = 0.
Solving the above two equations for c1 and c2 ,
m2 x0 m1 x0
c1 = and c2 = .
m2 − m1 m1 − m2
Hence
x0
m1 e m2 t − m2 e m1 t


x(t) =
m1 − m2
Niranjan Balachandran, IITB MA108
Example continued

Case II: b = a. Then m1 = m2 = −b = −a.

Hence
x(t) = c1 e −at + c2 te −at
is the general solution to the given DE. The initial conditions are
dx
x(0) = x0 and (0) = 0.
dt
It follows that

x0 = c1 ,
0 = −ac1 + c2 .

Hence,
x(t) = x0 e −at (1 + at).

Niranjan Balachandran, IITB MA108

Example continued
Case III: b < a. Then m1 , m2 are complex conjugates of each other
and are given by p
−b ± b 2 − a2 .
√
Set α = a2 − b 2 . Then m1 , m2 are given by −b ± ıα. The
general solution of the given DE is
x(t) = e −bt (c1 cos αt + c2 sin αt).
Recall the initial conditions
dx
x(0) = x0 and (0) = 0.
dt
It follows that x0 = c1 and
dx
= −be −bt (c1 cos αt + c2 sin αt) + e −bt (c2 α cos αt − c1 α sin αt)
dt
so that at t = 0, we get
bc1 bx0
c2 = = .
α α
Hence
bx0
x(t) = e −bt (x0 cos αt + sin αt).
α
Niranjan Balachandran, IITB MA108
Example continued
Recall that
bx0
x(t) = e −bt (x0 cos αt + sin αt).
α
Let tan θ = αb . Then
√
x0 e −bt b 2 + α2
x(t) = cos(αt − θ).
α
This function oscillates with an amplitude which decreases
exponentially. The graph passes the equilibrium position x = 0 at
regular intervals. Let T be the time required for one complete
cycle. Then
αT = 2π,
i.e.,
2π 2π 2π
T = =√ =q .
α 2
a −b 2 k
− c2
M 4M 2

Niranjan Balachandran, IITB MA108

Cauchy-Euler Equations
The equation
x 2 y 00 + axy 0 + by = 0
where a, b ∈ R is called a Cauchy-Euler equation. Assume x > 0.
Suppose y = x m is a solution to this DE. Then,

x 2 m(m − 1)x m−2 + axmx m−1 + bx m = 0.

We get:
m(m − 1) + am + b = 0.
i.e.,
m2 + (a − 1)m + b = 0.
This is called the auxiliary equation of the given Cauchy-Euler
equation. The roots are
p
(1 − a) ± (a − 1)2 − 4b
m1 , m2 = .
2
Niranjan Balachandran, IITB MA108
Cauchy-Euler Equations

Case I: Distinct real roots.

Are x m1 and x m2 linearly independent? Yes. Hence the general
solution is given by

y = c1 x m1 + c2 x m2 ,

for c1 , c2 ∈ R.

Niranjan Balachandran, IITB MA108

Cauchy-Euler Equations
Case II: Equal real roots.
i.e.,
1−a
m1 = m2 = .
2
Hence
1−a
y = f (x) = x 2

is a solution. To get a solution g linearly independent from f , set

g (x) = v (x)f (x).
Now,
g 0 = v 0 f + vf 0 ,
and
g 00 = v 00 f + 2v 0 f 0 + f 00 v .
Thus,
x 2 (v 00 f + 2v 0 f 0 + f 00 v ) + ax(v 0 f + vf 0 ) + bvf = 0;
Niranjan Balachandran, IITB MA108
Cauchy-Euler Equations
x 2 (v 00 f + 2v 0 f 0 + f 00 v ) + ax(v 0 f + vf 0 ) + bvf = 0;
i.e.,
v (x 2 f 00 + axf 0 + bf ) + x 2 (v 00 f + 2v 0 f 0 ) + axv 0 f = 0.
Thus,
x 2 v 00 f + v 0 x(2f 0 x + af ) = 0.
1−a
Now, for f (x) = x 2 , we have:
1 − a − 1 (a+1) 1−a
2f 0 x + af = 2x · x 2 + ax 2
2
1−a 1−a 1−a
= (1 − a)x 2 + ax 2 =x 2 = f (x).
Hence, we get:
x 2 v 00 f + v 0 xf = 0.
i.e.,
v 00 1
0
=− .
v x
Niranjan Balachandran, IITB MA108
Cauchy-Euler Equations

Hence,  
1
ln |v 0 | = − ln x = ln .
x
Take,
1
v0 = .
x
Set
v (x) = ln x.
Hence,
1−a
g (x) = (ln x)x 2 .
Thus the general solution is given by
1−a 1−a
y = c1 x 2 + c2 x 2 ln x,

c1 , c2 ∈ R.

Niranjan Balachandran, IITB MA108

Cauchy-Euler Equations
Case III: Complex roots.
Roots are m1 = µ + ıν, m2 = µ − ıν.

ıν
x m1 = x µ x ıν = x µ e ln x = x µ e ıν ln x = x µ (cos(ν ln x) + ı sin(ν ln x)),
Similarily,
x m2 = x µ e −ıν ln x = x µ (cos(ν ln x) − ı sin(ν ln x)),
It follows that
x m1 + x m2
= x µ cos(ν ln x),
2
x m1 − x m2
= x µ sin(ν ln x)
2ı
These are linearly independent solutions of the given DE and hence
the general solution is given by
y = x µ (c1 cos(ν ln x) + c2 sin(ν ln x)),
c1 , c2 ∈ R.
Niranjan Balachandran, IITB MA108
Non-homogeneous Second Order Linear ODE’s

Consider the non-homogeneous DE

y 00 + p(t)y 0 + q(t)y = r (t)

where p(t), q(t), r (t) are continuous functions on an interval I .

The associated homogeneous DE is

y 00 + p(t)y 0 + q(t)y = 0.

Can we relate the solutions of the above two DE’s?

Niranjan Balachandran, IITB MA108

Non-homogeneous Second Order Linear ODE’s
Theorem
Let f (t) be any solution of

y 00 + p(t)y 0 + q(t)y = r (t)

and y1 (t), y2 (t) be a basis of the solution space of the

corresponding homogeneous DE. Then the set of solutions of the
non-homogeneous DE is

{c1 y1 (t) + c2 y2 (t) + f (t) | c1 , c2 ∈ R}.

Proof: Let φ(t) be any solution of

L(y ) = y 00 + p(t)y 0 + q(t)y = r (t).
Then,
L(φ(t) − f (t))

Niranjan Balachandran, IITB MA108

Non-homogeneous Second Order Linear ODE’s
Theorem
Let f (t) be any solution of

y 00 + p(t)y 0 + q(t)y = r (t)

and y1 (t), y2 (t) be a basis of the solution space of the

corresponding homogeneous DE. Then the set of solutions of the
non-homogeneous DE is

{c1 y1 (t) + c2 y2 (t) + f (t) | c1 , c2 ∈ R}.

Proof: Let φ(t) be any solution of

L(y ) = y 00 + p(t)y 0 + q(t)y = r (t).
Then,
L(φ(t) − f (t)) = L(φ(t)) − L(f (t)) = r (t) − r (t) = 0.

Niranjan Balachandran, IITB MA108

Non-homogeneous Second Order Linear ODE’s

Hence, φ(t) − f (t) is a solution of the homogeneous DE. Thus,

φ(t) − f (t) = c1 y1 (t) + c2 y2 (t),

for c1 , c2 ∈ R. Hence,

φ(t) = c1 y1 (t) + c2 y2 (t) + f (t).

Summary: In order to find the general solution of a

non-homogeneous DE, we need to
get one particular solution of the non-homogeneous DE
get the general solution of the corresponding homogeneous
DE.

Niranjan Balachandran, IITB MA108

Method of Undetermined Coefficients

Suppose the non-homogeneous ODE has constant coefficients. In

this case, we know how to write down the general solution of the
corresponding homogeneous ODE. So we need to find one solution
of the non-homogeneous DE. One way to do this is called the
method of undetermined coefficients. Thus, we have:

y 00 + py 0 + qy = r (t),

with p, q ∈ R, and r (t) - a continuous function on I . The method

of undetermined coefficients does not work for any r (t), but only
when we know more about r (t). We’ll use this method only if r (t)
involves e at , sin at, cos at or polynomials in t.

Niranjan Balachandran, IITB MA108

Method of Undetermined Coefficients
Example: Find a particular solution of the DE:
y 00 − 3y 0 − 4y = 3e 2t .
We’ll search for a solution of the form ae 2t , where a is a constant.
So put y = ae 2t . We get:
(ae 2t )00 − 3(ae 2t )0 − 4ae 2t = 3e 2t .
Thus,
4ae 2t − 6ae 2t − 4ae 2t = 3e 2t .
Thus,
1
a=− .
2
Hence − 21 e 2t is a particular solution of the DE.
How do you get the general solution? Analyse roots of
m2 − 3m − 4 = 0. So general solution is
1
y = c1 e 4t + c2 e −t − e 2t .
2
Niranjan Balachandran, IITB MA108
Method of Undetermined Coefficients

If
r (t) = r1 (t) + r2 (t) + . . . + rn (t),
where ri (t) are e at or sin at or cos at or polynomials in t, consider
the n subproblems

y 00 + py 0 + qy = ri (t).

If yi (t) is a particular solution of this problem, then,

y (t) = y1 (t) + y2 (t) + . . . + yn (t)

is a particular solution of

y 00 + py 0 + qy = r (t).

Niranjan Balachandran, IITB MA108

Method of Undetermined Coefficients
Example: Find a particular solution of
y 00 − 3y 0 − 4y = 2 sin t.
Make a guess as to functions of which form we’ll search for as a
solution. a sin t? No. a sin t + b cos t? Yes. So set
y (t) = a sin t + b cos t.
Thus,
y 0 = a cos t − b sin t; y 00 = −a sin t − b cos t.
Substituting, we get:
(−5a + 3b − 2) sin t + (−3a − 5b) cos t = 0.
Thus,
−5a + 3b = 2; 3a + 5b = 0
5 3
(Why?). Thus, a = − 17 , b= 17 , and a particular solution is
5 3
y (t) = − sin t + cos t.
17 17
Niranjan Balachandran, IITB MA108
Method of Undetermined Coefficients
Example: Find a particular solution of
y 00 − 3y 0 − 4y = 4t 2 − 1.
Set
y (t) = at 2 + bt + c.
Substituting, we get:
−4at 2 + (−6a − 4b)t + (2a − 3b − 4c) = 4t 2 − 1.
Thus,
−4a = 4, − 6a − 4b = 0, 2a − 3b − 4c = −1.
Thus,
3 11
a = −1, b = , c = − .
2 8
Thus, a particular solution is
3 11
y (t) = −t 2 + t − .
2 8
Niranjan Balachandran, IITB MA108
Method of Undetermined Coefficients
Example: Find a particular solution of
y 00 − 3y 0 − 4y = −8e t cos 2t.
We should search for a solution of the form
y (t) = ae t cos 2t + be t sin 2t.
Then,
y 0 (t) = (a + 2b)e t cos 2t + (−2a + b)e t sin 2t,
and
y 00 = (−3a + 4b)e t cos 2t + (−4a − 3b)e t sin 2t.
Substituting, we get:
−10a − 2b = −8, 2a − 10b = 0.
Thus, a particular solution is
10 2
y (t) = e t cos 2t + e t sin 2t.
13 13
Niranjan Balachandran, IITB MA108
Method of Undetermined Coefficients

Example: Find a particular solution of

y 00 + 4y = 3 cos 2t.

Since r (t) = 3 cos 2t, you would look for solutions of the form

y (t) = a cos 2t + b sin 2t.

Thus,
y 0 (t) = −2a sin 2t + 2b cos 2t,
y 00 (t) = −4a cos 2t − 4b sin 2t.
Substituting in the given DE, we get:

(−4a cos 2t − 4b sin 2t) + 4(a cos 2t + b sin 2t) = 3 cos 2t.

But the lhs is 0! So can’t solve for a and b.

Niranjan Balachandran, IITB MA108

Method of Undetermined Coefficients
Why this ...? The problem was that sin 2t and cos 2t are also
solutions of the associated homogeneous ODE: y 00 + 4y = 0. So
lesson learned? When we search for solutions of a particular form,
we need to make sure that it’s not a solution of the associated
homogeneous equation.
We now modify the proposed solution as:

y (t) = at cos 2t + bt sin 2t.

Then,
y 0 (t) = (b − 2at) sin 2t + (a + 2bt) cos 2t,
y 00 (t) = −4at cos 2t − 4bt sin 2t − 4a sin 2t + 4b cos 2t.
Substituting, we get:

−4a sin 2t + 4b cos 2t = 3 cos 2t.

Thus, a = 0, b = 43 , and a particular solution is y (t) = 34 t sin 2t.

Niranjan Balachandran, IITB MA108
Method of Undetermined Coefficients

If the obvious candidate for a solution, say y (t) = f (t), as well as

this one multiplied by t, y (t) = tf (t), turn out to be solutions of
the associated homogeneous ODE, then what to do? Modify the
proposed solution by multiplying it with t 2 ; i.e., set

y (t) = t 2 f (t).

Can this too be a solution of the homogeneous ODE? No, since

the solution space is two dimensional.

Niranjan Balachandran, IITB MA108

Method of Undetermined Coefficients
Example: Find a particular solution of

y 00 − 3y 0 − 4y = 3e 2t + 2 sin t + 4t 2 − 1 − 8e t cos 2t.

Here,
r (t) = r1 (t) + r2 (t) + r3 (t) + r4 (t).
We need to solve
y 00 − 3y 0 − 4y = ri (t),
get a particular solution yi (t), and then

y (t) = y1 (t) + y2 (t) + y3 (t) + y4 (t)

is a particular solution of the given problem. Thus, a particular

solution is
1 5 3 3 11 10 2
y (t) = − e 2t − sin t+ cos t−t 2 + t− + e t cos 2t+ e t sin 2t.
2 17 17 2 8 13 13
Niranjan Balachandran, IITB MA108
Method of Undetermined Coefficients

Example: Find a particular solution of

y 00 + y = x 3 sin x.

What’s your candidate? Presence of sin x indicates both sin x and

cos x in the answer. Presence of x 3 indicates a generic cubic
polynomial. Thus,

(a1 x 3 + b1 x 2 + c1 x + d1 ) cos x + (a2 x 3 + b2 x 2 + c2 x + d2 ) sin x?

This wouldn’t do since sin x and cos x are already solutions of the
homogeneous part. So work with

x(a1 x 3 + b1 x 2 + c1 x + d1 ) cos x + x(a2 x 3 + b2 x 2 + c2 x + d2 ) sin x.

Niranjan Balachandran, IITB MA108