Solving Homogeneous Linear Differential Equations with Constant Coefficients, Lecture notes of Differential Equations

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HOMOGENEOUS LINEAR

DIFFERENTIAL EQUATIONS

WITH CONSTANT

COEFFICIENTS

INTENDED LEARNING OUTCOMES:

• Solve for the solutions of homogeneous

linear D.E. with constant coefficients.

Homogeneous linear differential equations of order n with constant coefficient are of the form, 𝑎 0

𝑛 𝑦 𝑑𝑥 𝑛

𝑛− 1 𝑦 𝑑𝑥 𝑛− 1

OPERATOR FORM:

𝑛

• 𝑎 1 𝐷 𝑛− 1
• … + 𝑎𝑛− 1 𝐷 + 𝑎𝑛)𝑦 = 0 or ∅ 𝐷 𝑦 = 0 where ∅ 𝐷 = 𝑎 0 𝐷 𝑛
• 𝑎 1 𝐷 𝑛− 1
• … + 𝑎𝑛− 1 𝐷 + 𝑎𝑛 HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS

AUXILIARY EQUATION, ∅ 𝒎

Associated with 𝑎 0 𝐷 𝑛

• 𝑎 1 𝐷 𝑛− 1
• … + 𝑎𝑛− 1 𝐷 + 𝑎𝑛 𝑦 = 0 is an auxiliary equation written in the following form: ∅ 𝑚 = 𝑎 0 𝑚 𝑛
• 𝑎 1 𝑚 𝑛− 1
• … + 𝑎𝑛− 1 𝑚 + 𝑎𝑛 = 0 If the roots of this auxiliary equation are r 1 , r 2 , …, r n then ∅ 𝑚 = 𝑎 0 𝑚 − 𝑟 1 𝑚 − 𝑟 2 … 𝑚 − 𝑟𝑛 = 0 Relative to this we can write (𝑎 0 𝐷 𝑛
• 𝑎 1 𝐷 𝑛− 1
• … + 𝑎𝑛− 1 𝐷 + 𝑎𝑛)𝑦 = 0 in the factored form: ∅ 𝐷 𝑦 = 𝑎 0 𝐷 − 𝑟 1 𝐷 − 𝑟 2 … 𝐷 − 𝑟𝑛 𝑦 = 0

SOLUTIONS OF HOMOGENEOUS LINEAR D.E.

When using the operator D, it can be shown that if y = 𝑒 𝑚𝑥 𝐷 𝑒 𝑚𝑥 = 𝑚𝑒 𝑚𝑥 𝐷 2 𝑒 𝑚𝑥 = 𝑚 2 𝑒 𝑚𝑥 ∴ 𝐷 𝑛 𝑒 𝑚𝑥 = 𝑚 𝑛 𝑒 𝑚𝑥 (𝑛 = 1 , 2 , … ) So, if ∅ 𝐷 is to be operated on 𝑒 𝑚𝑥 , ∅ 𝐷 𝑒 𝑚𝑥 = (𝑎 0 𝐷 𝑛

• 𝑎 1 𝐷 𝑛− 1
• … + 𝑎𝑛− 1 𝐷 + 𝑎𝑛)𝑒 𝑚𝑥 Hence, if 𝑚 = 𝑟 is a root of the auxiliary equation ∅ 𝑚 = 0 , then ∅ 𝐷 𝑦 = ∅ 𝐷 𝑒 𝑚𝑥 = ∅ 𝐷 𝑒 𝑟𝑥 = 0 ∴ y = 𝑒 𝑚𝑥 is a solution of ∅ 𝐷 𝑦 = 0. In general, it could be written as y = 𝑐𝑒 𝑚𝑥 where c is a constant.

SOLUTIONS OF HOMOGENEOUS LINEAR D.E. WITH CONSTANT COEFFICIENTS “Any linear combination of solutions of a homogeneous linear differential equation is also a solution.”

• If 𝑦 1 , 𝑦 2 , …, 𝑦𝑛 are solutions to the homogeneous linear differential equation with constant coefficients, then 𝑦 = 𝑐 1 𝑦 1 + 𝑐 2 𝑦 2 + … + 𝑐𝑛𝑦𝑛 is a solution, where 𝑐 1 , 𝑐 2 , … , 𝑐𝑛 are constants.

SOLUTIONS OF HOMOGENEOUS LINEAR D.E. WITH CONSTANT COEFFICIENTS CASE 1: Distinct Real Roots Example : Solve the D.E. 𝑦" + 𝑦′ − 2𝑦 = 0. The D.E. in operator form: 𝐷 2

• 𝐷 − 2 𝑦 = 0 ∅ 𝑚 = 𝑚 2
• 𝑚 − 2 = 0 ∅ 𝑚 = 𝑚 + 2 𝑚 − 1 = 0 The roots of the auxiliary equation are: 𝑚 1 = − 2 & 𝑚 2 = 1 Hence, the solution of the given D.E.: 𝒚 = 𝒄𝟏𝒆 𝒙
• 𝒄𝟐𝒆 −𝟐𝒙

SOLUTIONS OF HOMOGENEOUS LINEAR D.E. WITH CONSTANT COEFFICIENTS CASE 1: Distinct Real Roots Example : Solve the D.E. 𝑦 ′′′ − 𝑦′′ − 2𝑦′ = 0. The D.E. in operator form: 𝐷 3 − 𝐷 2 − 2𝐷 𝑦 = 0 ∅ 𝑚 = 𝑚 3 − 𝑚 2 − 2 𝑚 = 0 ∅ 𝑚 = 𝑚 𝑚 2 − 𝑚 − 2 = 0 ∅ 𝑚 = 𝑚 𝑚 − 2 𝑚 + 1 = 0 The roots of the auxiliary equation are: 𝑚 1 = − 1 , 𝑚 2 = 0 , & 𝑚 3 = 2 Hence, the solution of the given D.E.: 𝒚 = 𝒄𝟏𝒆 −𝒙

• 𝒄𝟐 +𝒄𝟑𝒆 𝟐𝒙

SOLUTIONS OF HOMOGENEOUS LINEAR D.E. WITH CONSTANT COEFFICIENTS CASE 1: Distinct Real Roots Example : Solve the D.E. 𝐷 2

• 2 𝐷 − 1 𝑦 = 0. ∅ 𝑚 = 𝑚 2
• 2𝑚 − 1 = 0 By quadratic formula, 𝑚 =

2 − 4𝐴𝐶 2𝐴

2 − 4 ( 1 )(− 1 ) 2 ( 1 )

The roots of the auxiliary equation are: 𝑚 1 = − 1 + 2 , & 𝑚 2 = − 1 − 2 Hence, the solution of the given D.E.: 𝒚 = 𝒄𝟏𝒆 − 1 + 2 𝒙

• 𝒄𝟐𝒆 − 1 − 2 𝑥

SOLUTIONS OF HOMOGENEOUS LINEAR D.E. WITH CONSTANT COEFFICIENTS CASE 1: Distinct Real Roots Example : Solve the D.E.𝑦" − 9𝑦 = 0 with 𝑥 = 0 , 𝑦 = 0 , and 𝑦 ′ = 3. The D.E. in operator form: 𝐷 2 − 9 𝑦 = 0 ∅ 𝑚 = 𝑚 2 − 9 = 0 ∅ 𝑚 = 𝑚 + 3 𝑚 − 3 = 0 The roots of the auxiliary equation are: 𝑚 1 = − 3 & 𝑚 2 = 3 The general solution of the given D.E.: 𝒚 = 𝒄𝟏𝒆 − 3 𝒙

• 𝒄𝟐𝒆 3 𝑥

SOLUTIONS OF HOMOGENEOUS LINEAR D.E. WITH CONSTANT COEFFICIENTS CASE 1: Distinct Real Roots Example : Solve the D.E.𝑦" − 9𝑦 = 0 with 𝑥 = 0 , 𝑦 = 0 , and 𝑦 ′ = 3. The general solution of the given D.E.: 𝒚 = 𝒄𝟏𝒆 − 3 𝒙

• 𝒄𝟐𝒆 3 𝑥 But when 𝑥 = 0 , 𝑦 ′ = 3 𝒚′ = −𝟑𝒄𝟏𝒆 − 3 𝒙
• 𝟑𝒄𝟐𝒆 3 𝒙 𝟑 = −𝟑𝒄𝟏 + 𝟑𝒄𝟐 𝟏 = −𝒄𝟏 + 𝒄𝟐 (EQ. 2)

SOLUTIONS OF HOMOGENEOUS LINEAR D.E. WITH CONSTANT COEFFICIENTS CASE 1: Distinct Real Roots Example : Solve the D.E.𝑦" − 9𝑦 = 0 with 𝑥 = 0 , 𝑦 = 0 , and 𝑦 ′ = 3. The general solution of the given D.E.: 𝒚 = 𝒄𝟏𝒆 − 3 𝒙

• 𝒄𝟐𝒆 3 𝑥 By solving equations 1 and 2, simultaneously, 𝟎 = 𝒄𝟏 + 𝒄𝟐 𝟏 = −𝒄𝟏 + 𝒄𝟐

- 𝟏 = 𝟐𝒄𝟏 + 𝟎 𝒄𝟏 = −

SOLUTIONS OF HOMOGENEOUS LINEAR D.E. WITH CONSTANT COEFFICIENTS CASE 2: Repeated Real Roots Suppose a homogeneous linear differential equation is given by 𝐷 − 2 2 𝑦 = 0 𝑒 − 2 𝑥 𝐷 − 2 2 𝑦 = 0 𝑒 − 2 𝑥 𝐷 − 2 2 𝑦 = 𝐷 2 (𝑒 − 2 𝑥 𝑦) = 0 𝐷(𝑒 − 2 𝑥 𝑦) = 𝑐 2 𝑒 − 2 𝑥 𝑦 = 𝑐 1 + 𝑐 2 𝑥 Hence, the solution of the differential equation is 𝑦 = 𝑐 1 𝑒 2 𝑥

• 𝑐 2 𝑥𝑒 2 𝑥 𝐷 − 𝑎 𝑛 𝑦𝑒 𝑎𝑥 = 𝑒 𝑎𝑥 𝐷 𝑛 𝑦

SOLUTIONS OF HOMOGENEOUS LINEAR D.E. WITH CONSTANT COEFFICIENTS CASE 2: Repeated Real Roots Thus, if a homogeneous linear differential equation is given by 𝐷 − 𝑎 3 𝑦 = 0 𝑒 −𝑎𝑥 𝐷 − 𝑎 3 𝑦 = 0 𝑒 −𝑎𝑥 𝐷 − 𝑎 3 𝑦 = 𝐷 3 (𝑒 −𝑎𝑥 𝑦) = 0 𝐷 2 𝑒 −𝑎𝑥 𝑦 = 𝑐 3 𝐷 𝑒 −𝑎𝑥 𝑦 = 𝑐 2 + 𝑐 3 𝑥 𝑒 −𝑎𝑥 𝑦 = 𝑐 1 + 𝑐 2 𝑥 + 𝑐 3 𝑥 2 Its solution is 𝑦 = 𝑐 1 𝑒 𝑎𝑥

• 𝑐 2 𝑥𝑒 𝑎𝑥 +𝑐 3 𝑥 2 𝑒 𝑎𝑥