Math 211 Math 211 Lecture #33 Harmonic Motion Inhomogeneous - - PowerPoint PPT Presentation

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Math 211 Math 211

Lecture #33 Harmonic Motion Inhomogeneous Equations November 13, 2002

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Harmonic Motion Harmonic Motion

• Spring: y′′ + µ

my′ + k my = 1 mF(t).

• Circuit: I′′ + R

LI′ + 1 LC I = 1 LE′(t).

• Essentially the same equation. Use

x′′ + 2cx′ + ω2

0x = f(t).

We call this the equation for harmonic motion. It includes both the vibrating spring and the RLC

circuit.

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The Equation for Harmonic Motion The Equation for Harmonic Motion

x′′ + 2cx′ + ω2

0x = f(t).

• ω0 is the natural frequency.

Spring: ω0 =

• k/m.

Circuit: ω0 =

• 1/LC.
• c is the damping constant.

Spring: 2c = µ/m. Circuit: 2c = R/L.

• f(t) is the forcing term.

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Simple Harmonic Motion Simple Harmonic Motion

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Simple Harmonic Motion Simple Harmonic Motion

No forcing

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Simple Harmonic Motion Simple Harmonic Motion

No forcing , and no damping.

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Simple Harmonic Motion Simple Harmonic Motion

No forcing , and no damping. x′′ + ω2

0x = 0

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Simple Harmonic Motion Simple Harmonic Motion

No forcing , and no damping. x′′ + ω2

0x = 0

• p(λ) = λ2 + ω2

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Simple Harmonic Motion Simple Harmonic Motion

No forcing , and no damping. x′′ + ω2

0x = 0

• p(λ) = λ2 + ω2

0, λ = ±iω0.

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Simple Harmonic Motion Simple Harmonic Motion

No forcing , and no damping. x′′ + ω2

0x = 0

• p(λ) = λ2 + ω2

0, λ = ±iω0.

• Fundamental set of solutions:

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Simple Harmonic Motion Simple Harmonic Motion

No forcing , and no damping. x′′ + ω2

0x = 0

• p(λ) = λ2 + ω2

0, λ = ±iω0.

• Fundamental set of solutions: x1(t) = cos ω0t &

x2(t) = sin ω0t.

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Simple Harmonic Motion Simple Harmonic Motion

No forcing , and no damping. x′′ + ω2

0x = 0

• p(λ) = λ2 + ω2

0, λ = ±iω0.

• Fundamental set of solutions: x1(t) = cos ω0t &

x2(t) = sin ω0t.

• General solution: x(t) = C1 cos ω0t + C2 sin ω0t.

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Simple Harmonic Motion Simple Harmonic Motion

No forcing , and no damping. x′′ + ω2

0x = 0

• p(λ) = λ2 + ω2

0, λ = ±iω0.

• Fundamental set of solutions: x1(t) = cos ω0t &

x2(t) = sin ω0t.

• General solution: x(t) = C1 cos ω0t + C2 sin ω0t.
• Every solution is periodic at the natural frequency ω0.

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Simple Harmonic Motion Simple Harmonic Motion

No forcing , and no damping. x′′ + ω2

0x = 0

• p(λ) = λ2 + ω2

0, λ = ±iω0.

• Fundamental set of solutions: x1(t) = cos ω0t &

x2(t) = sin ω0t.

• General solution: x(t) = C1 cos ω0t + C2 sin ω0t.
• Every solution is periodic at the natural frequency ω0.

The period is T = 2π/ω0.

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Amplitude and Phase Amplitude and Phase

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Amplitude and Phase Amplitude and Phase

• Put C1 and C2 in polar coordinates:

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Amplitude and Phase Amplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

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Amplitude and Phase Amplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

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Amplitude and Phase Amplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

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Amplitude and Phase Amplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

• A is the amplitude

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Amplitude and Phase Amplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

• A is the amplitude; A =
• C2

1 + C2 2.

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Amplitude and Phase Amplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

• A is the amplitude; A =
• C2

1 + C2 2.

• φ is the phase

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Amplitude and Phase Amplitude and Phase

• Put C1 and C2 in polar coordinates:

C1 = A cos φ, & C2 = A sin φ.

• Then x(t) = C1 cos ω0t + C2 sin ω0t

= A cos(ω0t − φ).

• A is the amplitude; A =
• C2

1 + C2 2.

• φ is the phase; tan φ = C2/C1.

Return Amplitude & phase

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Examples Examples

Return Amplitude & phase

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Examples Examples

• C1 = 3, C2 = 4

Return Amplitude & phase

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Examples Examples

• C1 = 3, C2 = 4 ⇒ A = 5

Return Amplitude & phase

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Examples Examples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.

Return Amplitude & phase

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Examples Examples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.
• C1 = −3, C2 = 4

Return Amplitude & phase

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Examples Examples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.
• C1 = −3, C2 = 4 ⇒ A = 5

Return Amplitude & phase

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Examples Examples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.
• C1 = −3, C2 = 4 ⇒ A = 5, φ = 2.2143.

Return Amplitude & phase

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Examples Examples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.
• C1 = −3, C2 = 4 ⇒ A = 5, φ = 2.2143.
• C1 = −3, C2 = −4

Return Amplitude & phase

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Examples Examples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.
• C1 = −3, C2 = 4 ⇒ A = 5, φ = 2.2143.
• C1 = −3, C2 = −4 ⇒ A = 5

Return Amplitude & phase

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Examples Examples

• C1 = 3, C2 = 4 ⇒ A = 5, φ = 0.9273.
• C1 = −3, C2 = 4 ⇒ A = 5, φ = 2.2143.
• C1 = −3, C2 = −4 ⇒ A = 5, φ = −2.2143.

Return Amplitude & phase

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Example Example

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

Return Amplitude & phase

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Example Example

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω2

0 = 16

Return Amplitude & phase

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Example Example

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω2

0 = 16 ⇒ ω0 = 4.

Return Amplitude & phase

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Example Example

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω2

0 = 16 ⇒ ω0 = 4.

• General solution:

Return Amplitude & phase

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Example Example

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω2

0 = 16 ⇒ ω0 = 4.

• General solution: x(t) = C1 cos 4t + C2 sin 4t.

Return Amplitude & phase

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Example Example

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω2

0 = 16 ⇒ ω0 = 4.

• General solution: x(t) = C1 cos 4t + C2 sin 4t.
• IC: −2 = x(0) = C1, and 4 = x′(0) = 4C2.

Return Amplitude & phase

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Example Example

x′′ + 16x = 0, x(0) = −2 & x′(0) = 4

• Natural frequency: ω2

0 = 16 ⇒ ω0 = 4.

• General solution: x(t) = C1 cos 4t + C2 sin 4t.
• IC: −2 = x(0) = C1, and 4 = x′(0) = 4C2.
• Solution

x(t) = −2 cos 2t + sin 2t = √ 5 cos(2t − 2.6779).

Return Harmonic motion

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Damped Harmonic Motion Damped Harmonic Motion

x′′ + 2cx′ + ω2

0x = 0

Return Harmonic motion

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Damped Harmonic Motion Damped Harmonic Motion

x′′ + 2cx′ + ω2

0x = 0

• p(λ) = λ2 + 2cλ + ω2

Return Harmonic motion

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Damped Harmonic Motion Damped Harmonic Motion

x′′ + 2cx′ + ω2

0x = 0

• p(λ) = λ2 + 2cλ + ω2

0; roots −c ±

• c2 − ω2

0.

Return Harmonic motion

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Damped Harmonic Motion Damped Harmonic Motion

x′′ + 2cx′ + ω2

0x = 0

• p(λ) = λ2 + 2cλ + ω2

0; roots −c ±

• c2 − ω2

0.

• Three cases

Return Harmonic motion

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Damped Harmonic Motion Damped Harmonic Motion

x′′ + 2cx′ + ω2

0x = 0

• p(λ) = λ2 + 2cλ + ω2

0; roots −c ±

• c2 − ω2

0.

• Three cases

c < ω0 — underdamped case

Return Harmonic motion

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Damped Harmonic Motion Damped Harmonic Motion

x′′ + 2cx′ + ω2

0x = 0

• p(λ) = λ2 + 2cλ + ω2

0; roots −c ±

• c2 − ω2

0.

• Three cases

c < ω0 — underdamped case c > ω0 — overdamped case

Return Harmonic motion

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Damped Harmonic Motion Damped Harmonic Motion

x′′ + 2cx′ + ω2

0x = 0

• p(λ) = λ2 + 2cλ + ω2

0; roots −c ±

• c2 − ω2

0.

• Three cases

c < ω0 — underdamped case c > ω0 — overdamped case c = ω0 — critically damped case

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Underdamped Case Underdamped Case

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Underdamped Case Underdamped Case

• c < ω0

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Underdamped Case Underdamped Case

• c < ω0
• Two complex roots λ and λ, where λ = −c + iω and

ω =

• ω2

0 − c2 .

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Underdamped Case Underdamped Case

• c < ω0
• Two complex roots λ and λ, where λ = −c + iω and

ω =

• ω2

0 − c2 .

• General solution

x(t) = e−ct[C1 cos ωt + C2 sin ωt]

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Underdamped Case Underdamped Case

• c < ω0
• Two complex roots λ and λ, where λ = −c + iω and

ω =

• ω2

0 − c2 .

• General solution

x(t) = e−ct[C1 cos ωt + C2 sin ωt] = Ae−ct cos(ωt − φ) .

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Overdamped Case Overdamped Case

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Overdamped Case Overdamped Case

• c > ω0 ,

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Overdamped Case Overdamped Case

• c > ω0 , so two real roots

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Overdamped Case Overdamped Case

• c > ω0 , so two real roots

λ1 = −c −

• c2 − ω2

λ2 = −c +

• c2 − ω2

0.

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Overdamped Case Overdamped Case

• c > ω0 , so two real roots

λ1 = −c −

• c2 − ω2

λ2 = −c +

• c2 − ω2

0.

• λ1 < λ2 < 0.

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Overdamped Case Overdamped Case

• c > ω0 , so two real roots

λ1 = −c −

• c2 − ω2

λ2 = −c +

• c2 − ω2

0.

• λ1 < λ2 < 0.
• General solution

x(t) = C1eλ1t + C2eλ2t.

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Critically Damped Case Critically Damped Case

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Critically Damped Case Critically Damped Case

• c = ω0

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Critically Damped Case Critically Damped Case

• c = ω0
• One negative real root λ = −c with multiplicity 2.

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Critically Damped Case Critically Damped Case

• c = ω0
• One negative real root λ = −c with multiplicity 2.
• General solution

x(t) = e−ct[C1 + C2t].

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Inhomogeneous Equations Inhomogeneous Equations

y′′ + py′ + qy = f(t)

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Inhomogeneous Equations Inhomogeneous Equations

y′′ + py′ + qy = f(t)

• The corresponding homogeneous equation is

y′′ + py′ + qy = 0

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Inhomogeneous Equations Inhomogeneous Equations

y′′ + py′ + qy = f(t)

• The corresponding homogeneous equation is

y′′ + py′ + qy = 0

We know how to find a fundamental set of solutions

y1 and y2.

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Inhomogeneous Equations Inhomogeneous Equations

y′′ + py′ + qy = f(t)

• The corresponding homogeneous equation is

y′′ + py′ + qy = 0

We know how to find a fundamental set of solutions

y1 and y2.

The general solution of the homogeneous equation is

yh(t) = C1y1(t) + C2y2(t).

Return Homogeneous equation

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Theorem: Assume

Return Homogeneous equation

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Theorem: Assume

• yp(t) is a particular solution to the inhomogeneous

equation y′′ + py′ + qy = f(t);

Return Homogeneous equation

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Theorem: Assume

• yp(t) is a particular solution to the inhomogeneous

equation y′′ + py′ + qy = f(t);

• y1(t) & y2(t) is a fundamental set of solutions to the

homogeneous equation y′′ + py′ + qy = 0.

Return Homogeneous equation

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Theorem: Assume

• yp(t) is a particular solution to the inhomogeneous

equation y′′ + py′ + qy = f(t);

• y1(t) & y2(t) is a fundamental set of solutions to the

homogeneous equation y′′ + py′ + qy = 0. Then the general solution to the inhomogeneous equation is y(t) = yp(t) + C1y1(t) + C2y2(t).

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Method of Undetermined Coefficients Method of Undetermined Coefficients

y′′ + py′ + qy = f(t)

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Method of Undetermined Coefficients Method of Undetermined Coefficients

y′′ + py′ + qy = f(t)

• If the forcing term f(t) has a form which is replicated

under differentiation, then look for a particular solution

• f the same general form as the forcing term.

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Exponential Forcing Term Exponential Forcing Term

y′′ + py′ + qy = Ceat

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Exponential Forcing Term Exponential Forcing Term

y′′ + py′ + qy = Ceat

• Example: y′′ + 3y′ + 2y = 4e−3t

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Exponential Forcing Term Exponential Forcing Term

y′′ + py′ + qy = Ceat

• Example: y′′ + 3y′ + 2y = 4e−3t
• Try yp(t) = ae−3t; a to be determined.

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Exponential Forcing Term Exponential Forcing Term

y′′ + py′ + qy = Ceat

• Example: y′′ + 3y′ + 2y = 4e−3t
• Try yp(t) = ae−3t; a to be determined.

Particular solution: yp(t) = 2e−3t.

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Exponential Forcing Term Exponential Forcing Term

y′′ + py′ + qy = Ceat

• Example: y′′ + 3y′ + 2y = 4e−3t
• Try yp(t) = ae−3t; a to be determined.

Particular solution: yp(t) = 2e−3t.

• Homogeneous equation: y′′ + 3y′ + 2y = 0.

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Exponential Forcing Term Exponential Forcing Term

y′′ + py′ + qy = Ceat

• Example: y′′ + 3y′ + 2y = 4e−3t
• Try yp(t) = ae−3t; a to be determined.

Particular solution: yp(t) = 2e−3t.

• Homogeneous equation: y′′ + 3y′ + 2y = 0.

Fundamental set of solutions: e−2t & e−t.

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Exponential Forcing Term Exponential Forcing Term

y′′ + py′ + qy = Ceat

• Example: y′′ + 3y′ + 2y = 4e−3t
• Try yp(t) = ae−3t; a to be determined.

Particular solution: yp(t) = 2e−3t.

• Homogeneous equation: y′′ + 3y′ + 2y = 0.

Fundamental set of solutions: e−2t & e−t.

• General solution to the inhomogeneous equation:

y(t) = 2e−3t + C1e−t + C2e−2t.

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Trigonometric Forcing Term Trigonometric Forcing Term

y′′ + py′ + qy = A cos ωt + B sin ωt

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Trigonometric Forcing Term Trigonometric Forcing Term

y′′ + py′ + qy = A cos ωt + B sin ωt

• Example: y′′ + 4y′ + 5y = 4 cos 2t − 3 sin 2t

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Trigonometric Forcing Term Trigonometric Forcing Term

y′′ + py′ + qy = A cos ωt + B sin ωt

• Example: y′′ + 4y′ + 5y = 4 cos 2t − 3 sin 2t
• Try yp(t) = a cos 2t + b sin 2t

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Trigonometric Forcing Term Trigonometric Forcing Term

y′′ + py′ + qy = A cos ωt + B sin ωt

• Example: y′′ + 4y′ + 5y = 4 cos 2t − 3 sin 2t
• Try yp(t) = a cos 2t + b sin 2t

Particular solution: yp(t) = [28 cos 2t + 29 sin 2t]/65.

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Trigonometric Forcing Term Trigonometric Forcing Term

y′′ + py′ + qy = A cos ωt + B sin ωt

• Example: y′′ + 4y′ + 5y = 4 cos 2t − 3 sin 2t
• Try yp(t) = a cos 2t + b sin 2t

Particular solution: yp(t) = [28 cos 2t + 29 sin 2t]/65.

• Homogeneous equation: y′′ + 4y′ + 5y = 0

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Trigonometric Forcing Term Trigonometric Forcing Term

y′′ + py′ + qy = A cos ωt + B sin ωt

• Example: y′′ + 4y′ + 5y = 4 cos 2t − 3 sin 2t
• Try yp(t) = a cos 2t + b sin 2t

Particular solution: yp(t) = [28 cos 2t + 29 sin 2t]/65.

• Homogeneous equation: y′′ + 4y′ + 5y = 0
• Fund. set of sol’ns: e−2t cos t & e−2t sin t.

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Trigonometric Forcing Term Trigonometric Forcing Term

y′′ + py′ + qy = A cos ωt + B sin ωt

• Example: y′′ + 4y′ + 5y = 4 cos 2t − 3 sin 2t
• Try yp(t) = a cos 2t + b sin 2t

Particular solution: yp(t) = [28 cos 2t + 29 sin 2t]/65.

• Homogeneous equation: y′′ + 4y′ + 5y = 0
• Fund. set of sol’ns: e−2t cos t & e−2t sin t.
• General solution to the inhomogeneous equation:

Return

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Trigonometric Forcing Term Trigonometric Forcing Term

y′′ + py′ + qy = A cos ωt + B sin ωt

• Example: y′′ + 4y′ + 5y = 4 cos 2t − 3 sin 2t
• Try yp(t) = a cos 2t + b sin 2t

Particular solution: yp(t) = [28 cos 2t + 29 sin 2t]/65.

• Homogeneous equation: y′′ + 4y′ + 5y = 0
• Fund. set of sol’ns: e−2t cos t & e−2t sin t.
• General solution to the inhomogeneous equation:

y(t) = 28 cos 2t + 29 sin 2t 65 + e−2t[C1 cos t + C2 sin t].

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Complex Method Complex Method

x′′ + px′ + qx = A cos ωt

• r

y′′ + py′ + qy = A sin ωt.

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Complex Method Complex Method

x′′ + px′ + qx = A cos ωt

• r

y′′ + py′ + qy = A sin ωt.

• Solve z′′ + pz′ + qz = Aeiωt.

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Complex Method Complex Method

x′′ + px′ + qx = A cos ωt

• r

y′′ + py′ + qy = A sin ωt.

• Solve z′′ + pz′ + qz = Aeiωt.

Try z(t) = aeiωt.

Return

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Complex Method Complex Method

x′′ + px′ + qx = A cos ωt

• r

y′′ + py′ + qy = A sin ωt.

• Solve z′′ + pz′ + qz = Aeiωt.

Try z(t) = aeiωt.

• xp(t) = Re(z(t))

and yp(t) = Im(z(t)).

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Example Example

x′′ + 4x′ + 5x = 4 cos 2t

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Example Example

x′′ + 4x′ + 5x = 4 cos 2t

• Solve z′′ + 4z′ + 5z = 4e2it.

Return

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Example Example

x′′ + 4x′ + 5x = 4 cos 2t

• Solve z′′ + 4z′ + 5z = 4e2it.

Try z(t) = ae2it.

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Example Example

x′′ + 4x′ + 5x = 4 cos 2t

• Solve z′′ + 4z′ + 5z = 4e2it.

Try z(t) = ae2it. Particular solution: z(t) = (4 − 32i)e2it/65.

Return

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Example Example

x′′ + 4x′ + 5x = 4 cos 2t

• Solve z′′ + 4z′ + 5z = 4e2it.

Try z(t) = ae2it. Particular solution: z(t) = (4 − 32i)e2it/65.

• Particular solution to the real equation:

xp(t) = Re(z(t))

Return

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Example Example

x′′ + 4x′ + 5x = 4 cos 2t

• Solve z′′ + 4z′ + 5z = 4e2it.

Try z(t) = ae2it. Particular solution: z(t) = (4 − 32i)e2it/65.

• Particular solution to the real equation:

xp(t) = Re(z(t)) = [4 cos 2t + 32 sin 2t] /65.

Return

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Polynomial Forcing Term Polynomial Forcing Term

y′′ + py′ + qy = P(t)

Return

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Polynomial Forcing Term Polynomial Forcing Term

y′′ + py′ + qy = P(t)

• Example: y′′ − 3y′ + 2y = 1 − 4t.

Return

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Polynomial Forcing Term Polynomial Forcing Term

y′′ + py′ + qy = P(t)

• Example: y′′ − 3y′ + 2y = 1 − 4t.

Try y(t) = a + bt.

Return

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Polynomial Forcing Term Polynomial Forcing Term

y′′ + py′ + qy = P(t)

• Example: y′′ − 3y′ + 2y = 1 − 4t.

Try y(t) = a + bt. Particular solution: y(t) = −5 − 2t.

Return

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Polynomial Forcing Term Polynomial Forcing Term

y′′ + py′ + qy = P(t)

• Example: y′′ − 3y′ + 2y = 1 − 4t.

Try y(t) = a + bt. Particular solution: y(t) = −5 − 2t.

• General solution

y(t) = −5 − 2t + C1et + C2e2t.

Return

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Exceptional Cases Exceptional Cases

Return

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Exceptional Cases Exceptional Cases

• Example: y′′ − 3y′ + 2y = 3et.

Return

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Exceptional Cases Exceptional Cases

• Example: y′′ − 3y′ + 2y = 3et.

Try y(t) = aet

Return

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Exceptional Cases Exceptional Cases

• Example: y′′ − 3y′ + 2y = 3et.

Try y(t) = aet The method does not work because et is a solution

to the associated homogeneous equation.

Return

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Exceptional Cases Exceptional Cases

• Example: y′′ − 3y′ + 2y = 3et.

Try y(t) = aet The method does not work because et is a solution

to the associated homogeneous equation.

• Try y(t) = atet

Return

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Exceptional Cases Exceptional Cases

• Example: y′′ − 3y′ + 2y = 3et.

Try y(t) = aet The method does not work because et is a solution

to the associated homogeneous equation.

• Try y(t) = atet

Particular solution: yp(t) = −3tet.

Return

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Exceptional Cases Exceptional Cases

• Example: y′′ − 3y′ + 2y = 3et.

Try y(t) = aet The method does not work because et is a solution

to the associated homogeneous equation.

• Try y(t) = atet

Particular solution: yp(t) = −3tet.

• General solution: y(t) = −3tet + C1et + C2e2t.

Return

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Exceptional Cases Exceptional Cases

• Example: y′′ − 3y′ + 2y = 3et.

Try y(t) = aet The method does not work because et is a solution

to the associated homogeneous equation.

• Try y(t) = atet

Particular solution: yp(t) = −3tet.

• General solution: y(t) = −3tet + C1et + C2e2t.
• If the suggested particular solution does not work,

multiply it by t and try again.

Theorem Previous UDC

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Combination Forcing Term Combination Forcing Term

Theorem Previous UDC

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Combination Forcing Term Combination Forcing Term

Example y′′ + 5y′ + 6y = 2e2t − 5 cos t

Theorem Previous UDC

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Combination Forcing Term Combination Forcing Term

Example y′′ + 5y′ + 6y = 2e2t − 5 cos t

• Solve

y′′

1 + 5y′ 1 + 6y1 = 2e2t

y′′

2 + 5y′ 2 + 6y2 = −5 cos t

Theorem Previous UDC

21

Combination Forcing Term Combination Forcing Term

Example y′′ + 5y′ + 6y = 2e2t − 5 cos t

• Solve

y′′

1 + 5y′ 1 + 6y1 = 2e2t

y′′

2 + 5y′ 2 + 6y2 = −5 cos t

• Set y(t) = y1(t) + y2(t).