Second Order Ordinary Differential Equation
Definition of a 2nd order differential equation:
A second order ODE is called linear if it can be written as
y  p( x) y  q( x) y  r ( x)      (1)
d2y
dy

p
(
x
)
 q ( x) y  r ( x)
dx
dx 2
If r(x) =0, then equation (1) can be written as
y  p( x) y  q( x) y  0 ------------------------(1)
It is a homogeneous 2nd ODE.
The general solution of the above differential equation is given by
y( x)  c1 y1( x)  c2 y2 ( x)
where c1 and c2 are arbitrary constants.
Solution of the differential equation (1) where p(x) and q(x) are
constants
Let y=emx

dy
 me mx
dx
d2y
 2  m 2 e mx
dx
Now substitute the above values in the differential equation of the form
ay  by  cy  0
d2y
dy
a 2 b
 cy  0
dx
dx
am2emx  bmemx  ce mx  0
emx (am2  bm  c)  0
As emx  0 for every x we get the auxiliary or the characteristics equation
am2  bm  c  0
 b  b 2  4ac
m
2a
Case 1
Discriminant Real Roots
Discriminant > 0
b2  4ac  0
If m1 and m2 are distinct real roots then the general solution is given by
y( x)  c1e
m1 x
 c2e
m2 x
Case 2
Equal Real Roots
Discriminant = 0
b 2  4ac  0
If m1 and m2 are equal real roots then the general solution is given by
y( x)  c1e
m1 x
 c2 xe
m1 x
Case 3
Imaginary Roots
Discriminant < 0
b 2  4ac  0
If m1    i and m2    i are imaginary roots then the general
solution is given by
y( x)  ex (c1 cos x  c2 sin x)
Example 1: Find the general solution of y  y  6 y  0
d 2 y dy
  6y  0
dx 2 dx
 m2  m  6  0
 (m  3)(m  2)  0
 m  3, (2)
y( x)  c1e
m1 x
 c2e
m2 x
y( x)  c1e2 x  c2e3 x
Example 1:
Solve the equation y  y  6 y  0
Example 2:
Solve the equation 4 y  12 y  9 y  0
Example3:
y  y  y  0
Find the general solution of
Solution: We have,
m2  m  1  0
p 2  4q  0  (1) 2  4(1)  (3)  0
 b  b 2  4ac
m
2a
m
1 1 4
2
m
1  3
2
m
1 i 3
2
z
1
3
i
2
2
Here, a   1
2
and b  3
2
y( x)  eax (c1 cosbx  c2 sin bx)
 y ( x)  e
x
3
1 cos 2
2 (c
Exercise:
Find the general solution of
(i)
(ii)
y  5 y  6 y  0
y  8 y  16 y  0
x  c2 sin
3
2
x)
(iii)
(iv)
(v)
(vi)
y  16 y  0
y  6 y  9 y  0
y  y  0
y  2 y  y  0
(vii) y  3 y  0
(viii) 8 y  2 y  y  0
Initial Value Problem:
An initial value problem consists of finding a solution y of the differential equation
that also satisfies initial condition of the form
y( x0 )  y1
y( x0 )  y0
Where
y0
and
y1 are the given constants.
Example:
Solve the initial value problem y  y  0 where
Solution: We have
y   y  0
m2 1  0
(m  1)(m  1)  0
m  (1),1
y( x)  c1e
m1 x
 c2e
m2 x
y( x)  c1e x  c2e x
Now,
y (0)  c1e 0  c2 e 0
1  c1  c2        (1)
y(0)  1, y(0)  0
Also, y( x)  c1e x  c2e x
y(0)  c1e 0  c2e0
0  c1  c2
c1  c2      (2)
Substituting equation (2) in equation (1), we get
c1  c2 
1
2
1
1
 y ( x)  e  x  e x
2
2
Exercise:
Solve the following Initial Value Problems:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
y  16 y  0; y(0)  3; y(0)  8
y  25 y  0; y(0)  0.8; y(0)  6.5
y  2 y  2 y  0; y(0)  1; y(0)  1
y  6 y  9 y  0; y(0)  1.4; y(0)  4.6
y   y   4 y  0; y(1)  11; y (1)  (6)
y  2 y  3 y  0; y(0)  1; y(0)  9
(vii) y  6 y  13 y  0; y(0)  (2); y(0)  0
(viii) y  4 y  y  0; y(0)  5; y(0)  4