Chapter 5
2nd Order Linear Differential Equations
2nd Order Equations

We already have one example position/velocity/acceleration

Also electrical circuits

And springs & pendulums
2nd Order Equations - Definition

Recall definition

General form –

The equation is linear if none of the y’s are multiplied
together or in the denominator.




G ( x, y, y ', y '')  0
General Form: A( x) y '' B ( x) y ' C ( x) y  F ( x)
Or, as long as A( x )  0, we can divide it out to get
y '' p( x) y ' q( x)  f ( x)
Finally, the equation is homogeneous if
Examples
x y '' yy ' 2 y  cos( x)
2
1
y '' 3 y '  0
y
xy '' e x y ' x 2 y  0
Solutions to 2nd order equations


As with 1st-order equations, we will after the first
step have a general solution – one containing
arbitrary constants.
Theorem 5.1. If p(x) and q(x) are continuous on an
open interval (a,b), and k0 and k1 are real
numbers, then
y '' p ( x) y ' q ( x)  0; y ( x0 )  k 0, y '( x0 )  k1

Has a unique solution valid for x in (a,b)
Homogeneous 2nd Order Equations


Solutions to homogeneous equations can be joined
linearly to form new ones.
Th 5.2 p 198. : If y1 , y2 are both solutions to
y '' p( x) y ' q( x)  0

Then for any constants c1 , c2another solution is given
by c1 y1  c2 y2

This says that any linear combination is also a
solution (principle of superposition)
Particular Solutions to 2nd order equations

As we have seen in the case of the x(t), v(t), a(t)
model, we need two initial conditions to get a
particular solution. So a typical 2nd order IVP looks
like
A( x) y '' B ( x) y ' C ( x)  F ( x); y (a)  k0 , y '(a )  k1
Example






y '' y ' 6 y  0;
Solutions are given by y1  e2 x ; y2  e3 x
Identify p(x) and q(x)
What is the interval of validity of the solution?
Verify y1 and y2:
Form the general solution:
Find a particular solution such that y (0)  7, y '(0)  1
Example 2
Solutions are given by y1  cos 2 x; y2  sin 2 x
Identify p(x) and q(x)
What is the interval of validity of the solution?
Verify y1 and y2:
 y '' 4 y  0;





Form the general solution:
Find a particular solution such that y (0)  2, y '(0)  1
Linear Independence; Wronskian; Th 5.1.5

Suppose we have two solutions y1 and y2 of a 2nd
order DE, and form the quantity
W ( y1 , y2 )  y1 y2 ' y1 ' y2


Then W is either always = 0, or never = 0 for x in
(a,b)
W is called the Wronskian of y1, y2, and is usually
written as the determinant
y1 y2
y1 '

y2 '
If W ( y1 , y2 )  0 , then the solutions are linearly
independent, and form a fundamental set
Example 2 Revisited

y '' 4 y  0; Solutions are given by y1  cos 2 x; y2  sin 2 x

W(y1,y2) = (Nspire)

Try p 203 #1, 2, 3, 5abc (Quiz), 9, 10a (collect)