Transcript Second-Order Linear Differential Equations

SECOND-ORDER LINEAR
DIFFERENTIAL EQUATIONS
AP Calculus BC
ORDER OF A DIFFERENTIAL EQUATION
The order of a differential equation is the order of the highest
derivative.
yʹʹʹ – 2xyʹ + y = sin x is a 3rd order differential equation.
SECOND ORDER DIFFERENTIAL EQUATIONS
Second order linear differential equations have the form:
2
d y
dy
P( x) 2  Q( x)  R( x) y  G ( x)
dx
dx
If G(x) = 0, it is a homogeneous differential equation.
If G(x) ≠ 0, it is a non-homogeneous differential equation.
USEFUL FACT #1
If y1 and y2 are both solutions to a linear homogeneous equation,
and c1 and c2 are any constants, then the function
y(x) = c1y1(x) + c2y2(x)
is also a solution. (In other words, the linear combination of y1 and
y2 is also a solution.)
PROOF OF USEFUL FACT #1
If y1 and y2 are both solutions to the equation, then
P( x) y1  Q( x) y1  R ( x) y1  0
P( x) y2  Q( x) y2  R ( x) y2  0
Now substitute c1y1(x) + c2y2(x) for y in the original equation to show that it
is a solution:
P( x)  c1 y1  c2 y2   Q ( x)  c1 y1  c2 y2   R ( x )  c1 y1  c2 y2 




 P ( x) c1 y1  c2 y2  Q ( x ) c1 y1  c2 y2  R ( x )  c1 y1  c2 y2 
 c1  P ( x) y1  Q ( x) y1  R ( x) y1   c2  P ( x ) y2  Q ( x ) y2  R ( x ) y2 




 y  c1 y1  c2 y2 is a solution.
 c1  0  c2  0  0
USEFUL FACT #2
y1 and y2 are linearly independent if neither is a constant
multiple of the other.
y1 = 5x and y2 = x  not linearly independent
y1 = ex and y2 = xex  linearly independent
USEFUL FACT #3
If y1 and y2 are linearly independent solutions to a homogeneous
differential equation, and P(x) ≠ 0, then the general solution is
given by y(x) = c1y1(x) + c2y2(x), where c1 and c2 are arbitrary
constants.
The general solution to the differential equation is a linear
combination of two linearly independent solutions. This means if
we know two linearly independent solutions, we know every
solution.
SOLVING
For now, we will assume that P, Q, and R are constant functions,
and we will call them a, b, and c, respectively, which gives us:
ay  by  cy  0
We are looking for a function whose second derivative times a
constant plus its first derivative times another constant plus
another constant times y is equal to 0.
A good candidate is y = erx (where r is a constant)
So now ar2erx + brerx + cerx = 0
Factor out erx  erx(ar2 + br + c) = 0
Therefore, ar2 + br + c = 0
This is called the characteristic equation (or auxiliary
equation) of the differential equation.
ROOTS OF THE CHARACTERISTIC EQUATION
This means that y = erx is a solution to the differential equation.
Find the roots of ar2 + br + c = 0 by factoring or using the quadratic
formula.
 There are three cases:
1.
b2
– 4ac > 0  Two real roots
2. b2 – 4ac = 0  One real root
r1 x
r2 x
y

c
e

c
e
Solution is
1
2
Solution is y  c1e rx  c2 xe rx
3. b2 – 4ac < 0  Two complex roots  r1 = α + iβ
General solution is y = eαx(c1 cos βx + c2 sin βx)
r2 = α – iβ
EXAMPLE 1
Solve the equation yʹʹ + yʹ – 6y = 0.
The auxiliary equation is r2 + r – 6 = 0
Factor  (r + 3)(r – 2) = 0
r = –3, r = 2
General solution is y = c1e–3x + c2e2x
You can check this by substituting back into the original equation.
EXAMPLE 2
2
d y dy
y0
Solve 3 2 
dx
dx
(use quadratic formula for this one)
Characteristic equation is 3r2 + r – 1 = 0
1  1  4(3)(1) 1  13

r
6
2(3)
General solution is
y  c1e
 1 13 

x
6


 c2e
 1 13 

x
6


EXAMPLE 3
Solve the equation 4yʹʹ + 12yʹ + 9y = 0
Auxiliary equation is 4r2 + 12r + 9 = 0
Factor  (2r + 3)2 = 0
Only root is r = –3/2, so general solution is:
y = c1e–3/2x + xc2e–3/2x
EXAMPLE 4
Solve yʹʹ – 6yʹ + 13y = 0
Auxiliary equation is r2 – 6r + 13 = 0
Can’t factor, non-real answer
6  16
r
 3  2i
2
General solution is y = e3x(c1 cos 2x + c2 sin 2x)
EXAMPLE 5 – INITIAL VALUE PROBLEM
Solve the equation yʹʹ + yʹ – 6y = 0 when y(0) = 1 and yʹ(0) = 0.
This is the same as Example 1, so the general solution is
y = c1e–3x + c2e2x
Substitute in values:
1 = c1 + c2
0 = –3c1 + 2c2
2
3
Solve as a system of equations  c1  , c2 
5
5
2 3 x 3 2 x
So the solution is y  e  e
5
5
EXAMPLE 6 – ANOTHER INITIAL VALUE PROBLEM
Solve the equation yʹʹ + y = 0 when y(0) = 2 and yʹ(0) = 3.
r2 + 1 = 0
r = ±i
So α = 0 and β = 1
General solution is y = c1 cos x + c2 sin x
Substitute in values:
2 = c1
3 = c2
So solution is y = 2 cos x + 3 sin x.