MA266: LN 08
Nov. 14-18, 2021
Fall 2021
1
Section 4.2:
Reduction of Order
2
Reduction of Order
Suppose that 𝒚𝟏 denotes a nontrivial solution of
𝑎2 𝑥 𝑦 ′′ + 𝑎1 𝑥 𝑦 ′ + 𝑎0 𝑥 𝑦 = 0
(1)
and 𝑦1 is defined on an interval 𝐼. We seek a second solution 𝒚𝟐 so that the set consisting of 𝑦1
and 𝑦2 is linearly independent on 𝐼 in the form of
𝒚𝟐 𝒙 = 𝐮 𝒙 𝒚𝟏 𝒙
Method of Solution:
• Find the first and second derivatives of 𝑦2 𝑥
𝑦2′ 𝑥 = 𝑢′ 𝑥 𝑦1 𝑥) + 𝑢 𝑥 𝑦1′ (𝑥
𝑦2′′ 𝑥 = 𝑢′′ 𝑥 𝑦1 𝑥 + 2𝑢′ 𝑥 𝑦1′ 𝑥 + 𝑢 𝑥 𝑦1′′ (𝑥)
• Substitute 𝑦2 𝑥 , 𝑦2′ 𝑥 and 𝑦2′′ 𝑥 into Equation (1) and simplify it.
• Use the substitution 𝑤 = 𝑢′ to reduce the equation to a first order liner DE.
• Find 𝑢(𝑥) and then find 𝑦2 𝑥 = 𝑢 𝑥 𝑦1 𝑥 .
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Example 1: Page 132
The function 𝑦1 = 𝑒 𝑥 is a solution of 𝑦 ′′ − 𝑦 = 0 on the interval (-∞,∞), use reduction of order
to find a second solution 𝑦2 .
Solution:
If 𝑦 = 𝑢 𝑥 𝑦1 𝑥 = 𝑢𝑒 𝑥 . Then, 𝑦 ′ = 𝑒 𝑥 𝑢′ + 𝑢𝑒 𝑥 and
𝑦 ′′ = 𝑒 𝑥 𝑢′′ + 2𝑒 𝑥 𝑢′ + 𝑢𝑒 𝑥
Plug these into the given equation 𝑦 ′′ − 𝑦 = 0, then we see that
𝑦 ′′ − 𝑦 = 𝑒 𝑥 𝑢′′ + 2𝑒 𝑥 𝑢′ + 𝑢𝑒 𝑥 − 𝑢𝑒 𝑥 =𝑒 𝑥 𝑢′′ + 2𝑢′ = 0
Since 𝑒 𝑥 ≠0 , the last equation requires
𝑢′′ + 2𝑢′ = 0
If we make the substitution 𝑤 = 𝑢′ ⇒
𝑤 ′ = 𝑢′′ . Then, we have
𝑤 ′ + 2𝑤 = 0
which is a linear 1st order linear equation in 𝑤 with the integrating factor:
𝜇 𝑥 = 𝑒 ‫ ׬‬2𝑑𝑥 = 𝑒 2𝑥
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Example 1: Page 132 Continued …
The function 𝑦1 = 𝑒 𝑥 is a solution of 𝑦 ′′ − 𝑦 = 0 on the interval (-∞,∞), use reduction of order
to find a second solution 𝑦2 .
Solution: Continued …
𝑤 ′ + 2𝑤 = 0,
Then, we can write
𝑑 2𝑥
𝑒 𝑤 =0
𝑑𝑥
𝜇 𝑥 = 𝑒 2𝑥
⇒
𝑒 2𝑥 𝑤 = 𝑐1
⇒
𝑢′ = 𝑐1 𝑒 −2𝑥
⇒
⇒
𝑤 = 𝑐1 𝑒 −2𝑥
𝑐1 −2𝑥
𝑢=− 𝑒
+ 𝑐2
2
Thus
𝑐1 −𝑥
𝑦=𝑢 𝑥
= − 𝑒 + 𝑐2 𝑒 𝑥
2
By choosing 𝑐2 = 0 and 𝑐1 = −2, we obtain the desired second solution
𝑒𝑥
𝑦2 = 𝑒 −𝑥
Because 𝑊(𝑒 𝑥 , 𝑒 −𝑥 ) ≠ 0 for every 𝑥, the solutions are linearly independent on (−∞, ∞).
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Exercise 3: Page 134
The function 𝑦1 = cos 4𝑥 is a solution of 𝑦 ′′ + 16𝑦 = 0, use reduction of order to find a second
solution 𝑦2 . Write down a general solution of DE
6
Exercise 9: Page 134
The function 𝑦1 = 𝑥 4 is a solution of 𝑥 2 𝑦 ′′ − 7𝑥𝑦 ′ + 16𝑦 = 0, use reduction of order to find a
second solution 𝑦2 . Write down a general solution of DE
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Section 4.3:
Homogeneous Linear Equations
with Constant Coefficients
8
Homogeneous Linear Equations with Constant Coefficients
The DE:
𝑎𝑛 𝑦
𝑛
+ 𝑎𝑛−1 𝑦
𝑛−1
+ ⋯ + 𝑎1 𝑦 ′ + 𝑎0 𝑦 = 0
• Linear, because every derivative in power one and coefficients are all constant.
• Homogeneous, because the right-hand side is zero.
• Constant coefficients, where 𝑎𝑛 , 𝑎𝑛−1 , … , 𝑎1 , 𝑎0 are real constants.
We can produce exponential solutions for homogeneous linear higher-order DEs.
For the previous DE, we try to find a solution of the form 𝑦 = 𝑒 𝑚𝑥
⇒ 𝑦 ′ = 𝑚𝑒 𝑚𝑥 ,
𝑦 ′′ = 𝑚2 𝑒 𝑚𝑥 ,
…,
𝑦
𝑛
= 𝑚𝑛 𝑒 𝑚𝑥
after substitution into the DE, we have
𝑎𝑛 𝑚𝑛 + 𝑎𝑛−1 𝑚𝑛−1 + ⋯ + 𝑎1 𝑚 + 𝑎0 𝑒 𝑚𝑥 = 0
Finally, we get the following equation, called the auxiliary equation:
𝑎𝑛 𝑚𝑛 + 𝑎𝑛−1 𝑚𝑛−1 + ⋯ + 𝑎1 𝑚 + 𝑎0 = 0
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Auxiliary Equation for 2nd-Order DE With Constant Coefficients
Consider second order equation
𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 0
where 𝑎, 𝑏 and 𝑐 are constants. We have the following auxiliary equation
𝑎𝑚2 + 𝑏𝑚 + 𝑐 = 0
We have three cases:
• Case I: Distinct Real Roots (𝑏 2 − 4𝑎𝑐 > 0): Roots are 𝑚1 and 𝑚2 , the general solution is
𝑦 = 𝑐1 𝑒 𝑚1𝑥 + 𝑐2 𝑒 𝑚2𝑥 ,
𝑦1 = 𝑒 𝑚1𝑥 ,
𝑦2 = 𝑒 𝑚2𝑥
• Case II: Repeated Real Roots (𝑏 2 − 4𝑎𝑐 = 0): Roots are 𝑚1 = 𝑚2 , the general solution is
𝑦 = 𝑐1 𝑒 𝑚1𝑥 + 𝑐2 𝑥𝑒 𝑚1𝑥 ,
𝑦1 = 𝑒 𝑚1𝑥 ,
𝑦2 = 𝑥𝑒 𝑚1𝑥
• Case III: Conjugate Complex Roots (𝑏 2 − 4𝑎𝑐 < 0):
Roots are 𝑚1 = 𝛼 + 𝑖𝛽 and 𝑚2 = 𝛼 − 𝑖𝛽, the general solution is
𝑦 = 𝑒 𝛼𝑥 𝑐1 cos 𝛽𝑥 + 𝑐2 sin 𝛽𝑥 ,
𝑦1 = 𝑒 𝛼𝑥 cos 𝛽𝑥 ,
𝑦2 = 𝑒 𝛼𝑥 sin 𝛽𝑥
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Example 1: Page 137
Solve the following differential equations.
a 2𝑦 ′′ − 5𝑦 ′ − 3𝑦 = 0,
b 𝑦 ′′ − 10𝑦 ′ + 25𝑦 = 0,
c 𝑦 ′′ + 4𝑦 ′ + 7𝑦 = 0
Solution:
a) The Auxiliary Equation:
1
− 5𝑚 − 3 = 0 ⇒ 2𝑚 + 1 𝑚 − 3 = 0 ⇒ 𝑚1 = − ,
2
Then, the general solution is (Case I: Distinct real roots)
2𝑚2
𝑦 = 𝑐1 𝑦1 + 𝑐2 𝑦2 = 𝑐1 𝑒
1
−2 𝑥
𝑚2 = 3
+ 𝑐2 𝑒 3𝑥
b) The Auxiliary Equation:
𝑚2 − 10𝑚 + 25 = 0 ⇒
𝑚−5
2
=0
⇒ 𝑚1 = 𝑚2 = 5
Then, the general solution is (Case II: Repeated real roots)
𝑦 = 𝑐1 𝑦1 + 𝑐2 𝑦2 = 𝑐1 𝑒 5𝑥 + 𝑐2 𝑥𝑒 5𝑥
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Example 1: Page 137 Continued…
Solve the following differential equations.
a 2𝑦 ′′ − 5𝑦 ′ − 3𝑦 = 0,
b 𝑦 ′′ − 10𝑦 ′ + 25𝑦 = 0,
c 𝑦 ′′ + 4𝑦 ′ + 7𝑦 = 0
Solution: Continued…
c) The Auxiliary Equation:
𝑚2 + 4𝑚 + 7 = 0 ⇒ 𝑚1 = −2 + 3𝑖,
⇒ 𝛼 = −2,
𝑚2 = −2 − 3𝑖
𝛽= 3
Then, the general solution is (Case III: Conjugate complex roots)
𝑦 = 𝑒 𝛼𝑥 𝑐1 cos(𝛽𝑥) + 𝑐2 sin(𝛽𝑥)
𝑦 = 𝑒 −2𝑥 𝑐1 cos
3𝑥 + 𝑐2 sin
3𝑥
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Example 2: Page 137
Solve the IVP: 4𝑦’’ + 4𝑦’ + 17𝑦 = 0,
𝑦 0 = −1, 𝑦’ 0 = 2
Solution: The Auxiliary Equation:
1
4𝑚2 + 4𝑚 + 17 = 0 ⇒ 𝑚1 = − + 2𝑖,
2
Then, the general solution is
𝑦=𝑒
1
1
𝑚2 = − − 2𝑖 ⇒ 𝛼 = − ,
2
2
𝛽=2
1
−2 𝑥
𝑐1 cos 2𝑥 + 𝑐2 sin 2𝑥
𝑥
1 −𝑥
−
′
⇒ 𝑦 = − 𝑒 2 𝑐1 cos(2𝑥) + 𝑐2 sin(2𝑥) +𝑒 2 −2 𝑐1 sin(2𝑥) + 2𝑐2 cos(2𝑥)
2
Applying the given conditions, we have
−1 = 𝑒 0 𝑐1 cos 0 + 𝑐2 sin 0 = 𝑐1 ⇒ 𝑐1 = −1.
1 0
3
′
0
𝑦 0 = 2 ⇒ 2 = − 𝑒 𝑐1 cos 0 + 𝑐2 sin 0 +𝑒 −2 𝑐1 sin 0 + 2𝑐2 cos 0 ⇒ 𝑐2 =
2
4
Hence the solution of the IVP is:
𝑥
3
−2
𝑦=𝑒
− cos(2𝑥) + sin(2𝑥)
4
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𝑦 0 = −1 ⇒
Example 3: Page 138
Solve
𝑦 ′′′ + 3𝑦 ′′ − 4𝑦 = 0
Solution:
The Auxiliary Equation:
𝑚3 + 3𝑚2 − 4 = 0
⇒ 𝑚1 = 𝑚2 = −2,
⇒
𝑚+2
𝑚3 = 1
2
𝑚−1 =0
(Case 1 and Case 2)
Thus, the general solutions of the DE is
𝑦 = 𝑐1 𝑦1 + 𝑐2 𝑦2 + 𝑐3 𝑦3 = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑥𝑒 −2𝑥 + 𝑐3 𝑒 𝑥
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Example 4: Page 138
Solve
𝑑4 𝑦
𝑑𝑦 2
+2 2+𝑦 =0
𝑑𝑥 4
𝑑𝑥
Solution:
The Auxiliary Equation:
𝑚4 + 2𝑚2 + 1 = 0
⇒ 𝑚1 = 𝑚3 = 𝑖,
⇒
𝑚2 + 1
2
𝑚2 = 𝑚4 = −𝑖
= 0 ⇒ 𝑚2 = −1
⇒
𝛼 = 0,
⇒
𝛽=1
𝑚 = ±𝑖
(Case 2 and 3)
Thus, the general solutions of the DE is
𝑦 = 𝑒 𝛼𝑥 𝑐1 cos(𝛽𝑥) + 𝑐2 sin(𝛽𝑥) + 𝑥𝑒 𝛼𝑥 𝑐3 cos(𝛽𝑥) + 𝑐4 sin(𝛽𝑥)
= 𝑐1 cos 𝑥 + 𝑐2 sin 𝑥 + 𝑐3 𝑥 cos 𝑥 + 𝑐4 𝑥 sin 𝑥
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Exercises 2, 6 & 9: Page 140
Solve the following differential equations.
2 𝑦 ′′ − 36𝑦 = 0,
6 𝑦 ′′ + 10𝑦 ′ + 25𝑦 = 0,
9 𝑦 ′′ + 9𝑦 = 0
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Exercise 13: Page 140
Find the general solution of the following second-order DE:
3𝑦 ′′ + 2𝑦 ′ + 𝑦 = 0
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Practice Problem for the Students
Exercise 29 (Page 140): Solve the IVP
𝑦 ′′ + 16𝑦 = 0,
𝑦 0 = 2,
𝑦 ′ 0 = −2
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