Reduction of Order
Repeated Roots of the Characteristic Equation
So far…
•
We’ve Learned How To Solve Second Order Linear
Homogeneous ODEs with Constant Coefficients
•
Characteristic Equation
•
We’ve seen distinct real roots, complex
conjugate roots
Characteristic Equation
Ay 00 + By 0 + Cy = 0
Gives the Characteristic Equation
2
Ar + Br + C = 0
Case 1:
If roots are real and distinct, r1 and r2
General solution: y = C1 e
r1 t
+ C2 e
r2 t
Case 2:
If roots are complex, a + bi and a
at
bi
at
General solution: y = B1 e cos (bt) + B2 e sin (bt)
Characteristic Equation
Ay 00 + By 0 + Cy = 0
Gives the Characteristic Equation
2
Ar + Br + C = 0
What if there’s only one root,
p
B ± B 2 4AC
r=
2A
B
2
and
4AC = 0 ?
B
r1 =
2A
Characteristic Equation
Ay 00 + By 0 + Cy = 0
Gives the Characteristic Equation
2
Ar + Br + C = 0
What if there’s only one root,
B
r1 =
2A
Then we know one solution:
y=e
r1 t
How do we find another?
Reduction of Order
If we have a linear homogeneous second order equation
00
0
y + p (t)y + q (t)y = 0
and we know one solution
y1
How do we find another?
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
In general, finding solutions is difficult
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
0
0
(v (t)y1 ) = v (t) y1 + v (t) y1
00
0
00
=
(v (t)y1 )
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00
Insert guess into ODE
y 00 + p (t) y 0 + q (t) y = 0
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
0
0
(v (t)y1 ) = v (t) y1 + v (t) y1
0
00
00
=
(v (t)y1 )
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00
Insert guess into ODE
00
(v (t)y1 ) +
0
+
p (t)(v (t) y1 v (t) y1 ) + q (t) v (t) y1 = 0
0
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
0
0
(v (t)y1 ) = v (t) y1 + v (t) y1
0
00
00
=
(v (t)y1 )
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00
Insert guess into ODE
00
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00 +
0
+
p (t)(v (t) y1 v (t) y1 ) + q (t) v (t) y1 = 0
0
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
0
0
(v (t)y1 ) = v (t) y1 + v (t) y1
0
00
00
=
(v (t)y1 )
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00
Insert guess into ODE
00
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00 +
0
+
p (t) v (t) y1 p (t)v (t) y1 + q (t) v (t) y1 = 0
0
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
0
0
(v (t)y1 ) = v (t) y1 + v (t) y1
0
00
00
=
(v (t)y1 )
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00
Insert guess into ODE
00
v (t) y1 + 2 v 0 (t) y1 0 + p (t) v 0 (t) y1 +
00
0
v (t)( y1 + p (t) y1 + q (t) y1) = 0
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
0
0
(v (t)y1 ) = v (t) y1 + v (t) y1
0
00
00
=
(v (t)y1 )
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00
Insert guess into ODE
00
v (t) y1 + 2 v 0 (t) y1 0 + p (t) v 0 (t) y1 +
00
0
v (t)( y1 + p (t) y1 + q (t) y1) = 0
=0
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
0
0
(v (t)y1 ) = v (t) y1 + v (t) y1
0
00
00
=
(v (t)y1 )
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00
Insert guess into ODE
00
0
0
+
v (t) y1 2 v (t) y1 + p (t) v 0 (t) y1 = 0
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
0
0
(v (t)y1 ) = v (t) y1 + v (t) y1
0
00
00
=
(v (t)y1 )
v (t) y1 + 2 v 0 (t) y1 0 + v (t) y1 00
Insert guess into ODE
00
0
y1 v (t) + (2 y1 + p (t) y1) v 0 (t) = 0
Can be solved with a first order linear ODE
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
Insert guess into ODE
0
00
y1 v (t) + (2 y1 + p (t) y1) v 0 (t) = 0
Make a clever “renaming”
0
w (t) = v (t)
0
00
w (t) = v (t)
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
Insert guess into ODE
0
0
y1 w (t) + (2 y1 + p (t) y1) w (t) = 0
Make a clever “renaming”
0
w (t) = v (t)
0
00
w (t) = v (t)
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
Insert guess into ODE
0
0
y1 w (t) + (2 y1 + p (t) y1) w (t) = 0
First
Order
Linear
ODE
Z
Z
0
w (t)dt = v (t)dt = v (t)
Reduction of Order
y 00 + p (t)y 0 + q (t)y = 0
One solution y1
Guess second solution has the form: v (t) y1
Insert guess into ODE
0
0
y1 w (t) + (2 y1 + p (t) y1) w (t) = 0
FirstZ Order Linear ODE
w (t)dt = v (t)
Second solution is
y2 = v (t) y1
Use Wronskian to establish a Fundamental Set of Solutions
Important Example
y
00
0
2y + y = 0
Characteristic Equation:
2
r
2r + 1 = 0
2
(r 1) = 0
One Solution
t
y1 = e
Important Example
y
00
0
2y + y = 0
00
0
p (t) =
2
One Solution
y1 = e t
y + p (t)y + q (t)y = 0
0
q (t) = 1
0
y1 w (t) + (2 y1 + p (t) y1) w (t) = 0
Important Example
y
00
0
2y + y = 0
00
0
p (t) =
2
One Solution
y1 = e t
y + p (t)y + q (t)y = 0
e
t w0
q (t) = 1
t
(t) + (2 e +
2 et ) w (t) = 0
Important Example
y
00
0
2y + y = 0
e
t w0
One Solution
y1 = e t
t +
t
+
2 e ) w (t) = 0
(t) (2 e
=0
Important Example
y
00
0
2y + y = 0
e
t w0
One Solution
(t) = 0
y1 = e t
Important Example
y
00
0
2y + y = 0
One Solution
y1 = e t
w0 (t) = 0
w (t) = C
Z
Z
w (t)dt = Cdt = Ct
v (t) =
Second Solution
v (t)y1 = Ctet
Important Example
y
00
0
2y + y = 0
t
y
=
e
One Solution: 1
t
y
=
te
Second Solution: 2
Wronskian:
t
t
t
e
e
+
te
Wronskian:
0
0
y1 y2 y 1 y 2
0
t
y1 = e
y20 = et + tet
et tet = e2t > 0
Form a Fundamental Set of Solutions
t
General Solution: y = C1 e + C2 te
t
This Holds in General
Ay 00 + By 0 + Cy = 0
Gives the Characteristic Equation
2
Ar + Br + C = 0
Only one root,
p
B ± B 2 4AC
r=
2A
B
2
and
4AC = 0
B
r1 =
2A
This Holds in General
B 0 C
y + y + y=0
B2
A
A
B
r1 =
2A
B
p (t) =
A
One solution:
r1 t
y1 = e
00
4AC = 0
Determine Reduction of Order Equation
0
y
+
y1 w (t) (2 1 + p (t) y1) w (t) = 0
0
This Holds in General
B 0 C
y + y + y=0
B2
A
A
B
r1 =
2A
B
p (t) =
A
One solution:
r1 t
y1 = e
00
4AC = 0
Determine Reduction of Order Equation
✓
◆
B r1 t
r1 t 0
r1 t
e w (t) + 2r1 e + e
w (t) = 0
A
This Holds in General
B 0 C
y + y + y=0
B2
A
A
B
r1 =
2A
B
p (t) =
A
One solution:
r1 t
y1 = e
00
4AC = 0
Determine Reduction of Order Equation
✓
◆
B r1 t B r1 t
r1 t 0
e w (t) + 2
e + e
w (t) = 0
2A
A
=0
This Holds in General
B 0 C
y + y + y=0
B2
A
A
B
r1 =
2A
B
p (t) =
A
One solution:
r1 t
y1 = e
00
4AC = 0
Determine Reduction of Order Equation
er1 t w0 (t) = 0
This Holds in General
B 0 C
y + y + y=0
B2
A
A
B
r1 =
2A
B
p (t) =
A
One solution:
r1 t
y1 = e
00
4AC = 0
Determine Reduction of Order Equation
0
w (t) = 0
w (t) = C
v (t) = Ct
r1 t
v
(t)y
=
Cte
Second Solution:
1
This Holds in General
Ay 00 + By 0 + Cy = 0
If Characteristic Equation
2
Ar + Br + C = 0
Has only one root,
B
r1 =
2A
General Solution takes the form:
y = C1 e
r1 t
+ C2 te
r1 t
Summary
•
We now know how to solve Second Order Linear
Homogeneous with Constant Coefficients:
•
Characteristic Equation
•
•
Distinct Real Roots, Complex Conjugate Roots,
One (Repeated) Root
If we have a Second Order Linear Homogeneous
Equation and one solution, use Reduction of Order
to find second solution.
Questions?