tma4110 - calculus 3 - lecture 7 · consider the inhomogeneous second-order linear differential...

TMA4110 - Calculus 3Lecture 7

Toke Meier CarlsenNorwegian University of Science and Technology

Fall 2012

www.ntnu.no TMA4110 - Calculus 3, Lecture 7

General solutions to inhomogeneousequations

If yp is a particular solution to the inhomogeneous equationy ′′+ py ′+ qy = f and y1 and y2 form a fundamental set of solutionsto the homogeneous equation y ′′ + py ′ + qy = 0, then the generalsolution to the inhomogeneous equation y ′′ + py ′ + qy = f is

y = yp + c1y1 + c2y2

where c1 and c2 are constants.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 2

The method of undetermined coefficients

Consider the inhomogeneous second-order linear differentialequation

y ′′ + py ′ + qy = f .

If the function f has a form that is replicated under differentiation,then look for a solution with the same general form as f .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 3

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Example

Let us find the general solution to the equation

y ′′ − 2y ′ + y = t3.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 4

Forced harmonic motion

We will now apply the technique of undetermined coefficients toanalyze harmonic motion with an external sinusoidal forcing term.

The equation we need to solve is

y ′′ + 2cy ′ + ω20y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 5

Forced harmonic motion

We will now apply the technique of undetermined coefficients toanalyze harmonic motion with an external sinusoidal forcing term.The equation we need to solve is

y ′′ + 2cy ′ + ω20y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 5

Forced undamped harmonic motion

Let us first assume that c = 0.

Then the equation becomes

y ′′ + ω20y = A cos(ωt).

The general solution to the homogeneous solution y ′′ + ω20y = 0 is

yh = c1 cos(ω0t) + c2 sin(ω0t).Let us find a particular solution by using the method ofundetermined coefficients. We will first look at the case whereω 6= ω0, and then at the case where ω = ω0.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 6

Forced undamped harmonic motion

Let us first assume that c = 0. Then the equation becomes

y ′′ + ω20y = A cos(ωt).

The general solution to the homogeneous solution y ′′ + ω20y = 0 is

yh = c1 cos(ω0t) + c2 sin(ω0t).Let us find a particular solution by using the method ofundetermined coefficients. We will first look at the case whereω 6= ω0, and then at the case where ω = ω0.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 6

Forced undamped harmonic motion

Let us first assume that c = 0. Then the equation becomes

y ′′ + ω20y = A cos(ωt).

The general solution to the homogeneous solution y ′′ + ω20y = 0 is

yh = c1 cos(ω0t) + c2 sin(ω0t).

Let us find a particular solution by using the method ofundetermined coefficients. We will first look at the case whereω 6= ω0, and then at the case where ω = ω0.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 6

Forced undamped harmonic motion

Let us first assume that c = 0. Then the equation becomes

y ′′ + ω20y = A cos(ωt).

The general solution to the homogeneous solution y ′′ + ω20y = 0 is

yh = c1 cos(ω0t) + c2 sin(ω0t).Let us find a particular solution by using the method ofundetermined coefficients.

We will first look at the case whereω 6= ω0, and then at the case where ω = ω0.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 6

Forced undamped harmonic motion

Let us first assume that c = 0. Then the equation becomes

y ′′ + ω20y = A cos(ωt).

The general solution to the homogeneous solution y ′′ + ω20y = 0 is

yh = c1 cos(ω0t) + c2 sin(ω0t).Let us find a particular solution by using the method ofundetermined coefficients. We will first look at the case whereω 6= ω0, and then at the case where ω = ω0.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 6

The case ω 6= ω0

We will look for a solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) of the

form yp = a cos(ωt) + b sin(ωt).

y ′′p (t) + ω20yp(t)

= −aω2 cos(ωt)− bω2 sin(ωt)

+ aω20 cos(ωt) + bω2

0 sin(ωt)

= a(ω20 − ω2) cos(ωt) + b(ω2

0 − ω2) sin(ωt)

so yp is a particular solution if and only if a = Aω2

0−ω2 and b = 0.

Thus yp(t) = Aω2

0−ω2 cos(ωt) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 7

The case ω 6= ω0

We will look for a solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) of the

form yp = a cos(ωt) + b sin(ωt).

y ′′p (t) + ω20yp(t)

= −aω2 cos(ωt)− bω2 sin(ωt)

+ aω20 cos(ωt) + bω2

0 sin(ωt)

= a(ω20 − ω2) cos(ωt) + b(ω2

0 − ω2) sin(ωt)

so yp is a particular solution if and only if a = Aω2

0−ω2 and b = 0.

Thus yp(t) = Aω2

0−ω2 cos(ωt) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 7

The case ω 6= ω0

We will look for a solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) of the

form yp = a cos(ωt) + b sin(ωt).

y ′′p (t) + ω20yp(t)

= −aω2 cos(ωt)− bω2 sin(ωt)

+ aω20 cos(ωt) + bω2

0 sin(ωt)

= a(ω20 − ω2) cos(ωt) + b(ω2

0 − ω2) sin(ωt)

so yp is a particular solution if and only if a = Aω2

0−ω2 and b = 0.

Thus yp(t) = Aω2

0−ω2 cos(ωt) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 7

The case ω 6= ω0

We will look for a solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) of the

form yp = a cos(ωt) + b sin(ωt).

y ′′p (t) + ω20yp(t) = −aω2 cos(ωt)− bω2 sin(ωt)

+ aω20 cos(ωt) + bω2

0 sin(ωt)

= a(ω20 − ω2) cos(ωt) + b(ω2

0 − ω2) sin(ωt)

so yp is a particular solution if and only if a = Aω2

0−ω2 and b = 0.

Thus yp(t) = Aω2

0−ω2 cos(ωt) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 7

The case ω 6= ω0

We will look for a solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) of the

form yp = a cos(ωt) + b sin(ωt).

y ′′p (t) + ω20yp(t) = −aω2 cos(ωt)− bω2 sin(ωt)

+ aω20 cos(ωt) + bω2

0 sin(ωt)

= a(ω20 − ω2) cos(ωt) + b(ω2

0 − ω2) sin(ωt)

so yp is a particular solution if and only if a = Aω2

0−ω2 and b = 0.

Thus yp(t) = Aω2

0−ω2 cos(ωt) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 7

The case ω 6= ω0

We will look for a solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) of the

form yp = a cos(ωt) + b sin(ωt).

y ′′p (t) + ω20yp(t) = −aω2 cos(ωt)− bω2 sin(ωt)

+ aω20 cos(ωt) + bω2

0 sin(ωt)

= a(ω20 − ω2) cos(ωt) + b(ω2

0 − ω2) sin(ωt)

so yp is a particular solution if and only if a = Aω2

0−ω2 and b = 0.

Thus yp(t) = Aω2

0−ω2 cos(ωt) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 7

The case ω 6= ω0

We will look for a solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) of the

form yp = a cos(ωt) + b sin(ωt).

y ′′p (t) + ω20yp(t) = −aω2 cos(ωt)− bω2 sin(ωt)

+ aω20 cos(ωt) + bω2

0 sin(ωt)

= a(ω20 − ω2) cos(ωt) + b(ω2

0 − ω2) sin(ωt)

so yp is a particular solution if and only if a = Aω2

0−ω2 and b = 0.

Thus yp(t) = Aω2

0−ω2 cos(ωt) is a particular solution,

and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 7

The case ω 6= ω0

We will look for a solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) of the

form yp = a cos(ωt) + b sin(ωt).

y ′′p (t) + ω20yp(t) = −aω2 cos(ωt)− bω2 sin(ωt)

+ aω20 cos(ωt) + bω2

0 sin(ωt)

= a(ω20 − ω2) cos(ωt) + b(ω2

0 − ω2) sin(ωt)

so yp is a particular solution if and only if a = Aω2

0−ω2 and b = 0.

Thus yp(t) = Aω2

0−ω2 cos(ωt) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 7

The case ω 6= ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).

Let us look at the solution where the motion starts at equilibrium.This means that y(0) = y ′(0) = 0. We then have that0 = y(0) = c1 +

Aω2

0−ω2 and 0 = y ′(0) = c2ω0, so

y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 8

The case ω 6= ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).Let us look at the solution where the motion starts at equilibrium.

This means that y(0) = y ′(0) = 0. We then have that0 = y(0) = c1 +

Aω2

0−ω2 and 0 = y ′(0) = c2ω0, so

y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 8

The case ω 6= ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).Let us look at the solution where the motion starts at equilibrium.This means that y(0) = y ′(0) = 0.

We then have that0 = y(0) = c1 +

Aω2

0−ω2 and 0 = y ′(0) = c2ω0, so

y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 8

The case ω 6= ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).Let us look at the solution where the motion starts at equilibrium.This means that y(0) = y ′(0) = 0. We then have that0 = y(0) = c1 +

Aω2

0−ω2 and 0 = y ′(0) = c2ω0,

so

y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 8

The case ω 6= ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ωt) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + Aω2

0−ω2 cos(ωt).Let us look at the solution where the motion starts at equilibrium.This means that y(0) = y ′(0) = 0. We then have that0 = y(0) = c1 +

Aω2

0−ω2 and 0 = y ′(0) = c2ω0, so

y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 8

The case ω 6= ω0

t

y

y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t))

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 9

The case ω 6= ω0

Consider the solution y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

Let ω = (ω0 + ω)/2 and δ = (ω0 − ω)/2. ω is called the meanfrequency, and δ is called the half difference. We then have that

y(t) =A

ω20 − ω2

(cos(ωt)− cos(ω0t))

=A

4ωδ(cos((ω − δ)t)− cos((ω + δ)t))

=A sin(δt)

2ωδsin(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 10

The case ω 6= ω0

Consider the solution y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

Let ω = (ω0 + ω)/2 and δ = (ω0 − ω)/2.

ω is called the meanfrequency, and δ is called the half difference. We then have that

y(t) =A

ω20 − ω2

(cos(ωt)− cos(ω0t))

=A

4ωδ(cos((ω − δ)t)− cos((ω + δ)t))

=A sin(δt)

2ωδsin(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 10

The case ω 6= ω0

Consider the solution y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

Let ω = (ω0 + ω)/2 and δ = (ω0 − ω)/2. ω is called the meanfrequency, and δ is called the half difference.

We then have that

y(t) =A

ω20 − ω2

(cos(ωt)− cos(ω0t))

=A

4ωδ(cos((ω − δ)t)− cos((ω + δ)t))

=A sin(δt)

2ωδsin(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 10

The case ω 6= ω0

Consider the solution y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

Let ω = (ω0 + ω)/2 and δ = (ω0 − ω)/2. ω is called the meanfrequency, and δ is called the half difference. We then have that

y(t) =A

ω20 − ω2

(cos(ωt)− cos(ω0t))

=A

4ωδ(cos((ω − δ)t)− cos((ω + δ)t))

=A sin(δt)

2ωδsin(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 10

The case ω 6= ω0

Consider the solution y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

Let ω = (ω0 + ω)/2 and δ = (ω0 − ω)/2. ω is called the meanfrequency, and δ is called the half difference. We then have that

y(t) =A

ω20 − ω2

(cos(ωt)− cos(ω0t))

=A

4ωδ(cos((ω − δ)t)− cos((ω + δ)t))

=A sin(δt)

2ωδsin(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 10

The case ω 6= ω0

Consider the solution y(t) = Aω2

0−ω2 (cos(ωt)− cos(ω0t)).

Let ω = (ω0 + ω)/2 and δ = (ω0 − ω)/2. ω is called the meanfrequency, and δ is called the half difference. We then have that

y(t) =A

ω20 − ω2

(cos(ωt)− cos(ω0t))

=A

4ωδ(cos((ω − δ)t)− cos((ω + δ)t))

=A sin(δt)

2ωδsin(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 10

The case ω 6= ω0

t

y

y(t) = A sin(δt)2ωδ sin(ωt)

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 11

The case ω = ω0

We will look for a solution to y ′′ + 2cy ′ + ω20y = A cos(ω0t). Since

A cos(ω0t) is a solution to the homogeneous equationy ′′ + 2cy ′ + ω2

0y = 0, we will look for a solution of the formyp = t(a cos(ω0t) + b sin(ω0t)).

y ′′p (t) + ω20yp(t)

= 2ω0(−a sin(ω0t) + b cos(ω0t))

+ tω20(−a cos(ω0t)− b sin(ω0t))

+ ω20t(a cos(ω0t) + b sin(ω0t))

= 2ω0(−a sin(ω0t) + b cos(ω0t)).

So yp is a particular solution if and only if a = 0 and b = A2ω0

. Thusyp(t) = A

2ω0t sin(ω0t) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 12

The case ω = ω0We will look for a solution to y ′′ + 2cy ′ + ω2

0y = A cos(ω0t).

SinceA cos(ω0t) is a solution to the homogeneous equationy ′′ + 2cy ′ + ω2

0y = 0, we will look for a solution of the formyp = t(a cos(ω0t) + b sin(ω0t)).

y ′′p (t) + ω20yp(t)

= 2ω0(−a sin(ω0t) + b cos(ω0t))

+ tω20(−a cos(ω0t)− b sin(ω0t))

+ ω20t(a cos(ω0t) + b sin(ω0t))

= 2ω0(−a sin(ω0t) + b cos(ω0t)).

So yp is a particular solution if and only if a = 0 and b = A2ω0

. Thusyp(t) = A

2ω0t sin(ω0t) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 12

The case ω = ω0We will look for a solution to y ′′ + 2cy ′ + ω2

0y = A cos(ω0t). SinceA cos(ω0t) is a solution to the homogeneous equationy ′′ + 2cy ′ + ω2

0y = 0, we will look for a solution of the formyp = t(a cos(ω0t) + b sin(ω0t)).

y ′′p (t) + ω20yp(t)

= 2ω0(−a sin(ω0t) + b cos(ω0t))

+ tω20(−a cos(ω0t)− b sin(ω0t))

+ ω20t(a cos(ω0t) + b sin(ω0t))

= 2ω0(−a sin(ω0t) + b cos(ω0t)).

So yp is a particular solution if and only if a = 0 and b = A2ω0

. Thusyp(t) = A

2ω0t sin(ω0t) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 12

The case ω = ω0We will look for a solution to y ′′ + 2cy ′ + ω2

0y = A cos(ω0t). SinceA cos(ω0t) is a solution to the homogeneous equationy ′′ + 2cy ′ + ω2

0y = 0, we will look for a solution of the formyp = t(a cos(ω0t) + b sin(ω0t)).

y ′′p (t) + ω20yp(t) = 2ω0(−a sin(ω0t) + b cos(ω0t))

+ tω20(−a cos(ω0t)− b sin(ω0t))

+ ω20t(a cos(ω0t) + b sin(ω0t))

= 2ω0(−a sin(ω0t) + b cos(ω0t)).

So yp is a particular solution if and only if a = 0 and b = A2ω0

. Thusyp(t) = A

2ω0t sin(ω0t) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 12

The case ω = ω0We will look for a solution to y ′′ + 2cy ′ + ω2

0y = A cos(ω0t). SinceA cos(ω0t) is a solution to the homogeneous equationy ′′ + 2cy ′ + ω2

0y = 0, we will look for a solution of the formyp = t(a cos(ω0t) + b sin(ω0t)).

y ′′p (t) + ω20yp(t) = 2ω0(−a sin(ω0t) + b cos(ω0t))

+ tω20(−a cos(ω0t)− b sin(ω0t))

+ ω20t(a cos(ω0t) + b sin(ω0t))

= 2ω0(−a sin(ω0t) + b cos(ω0t)).

So yp is a particular solution if and only if a = 0 and b = A2ω0

. Thusyp(t) = A

2ω0t sin(ω0t) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 12

The case ω = ω0We will look for a solution to y ′′ + 2cy ′ + ω2

0y = A cos(ω0t). SinceA cos(ω0t) is a solution to the homogeneous equationy ′′ + 2cy ′ + ω2

0y = 0, we will look for a solution of the formyp = t(a cos(ω0t) + b sin(ω0t)).

y ′′p (t) + ω20yp(t) = 2ω0(−a sin(ω0t) + b cos(ω0t))

+ tω20(−a cos(ω0t)− b sin(ω0t))

+ ω20t(a cos(ω0t) + b sin(ω0t))

= 2ω0(−a sin(ω0t) + b cos(ω0t)).

So yp is a particular solution if and only if a = 0 and b = A2ω0

.

Thusyp(t) = A

2ω0t sin(ω0t) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 12

The case ω = ω0We will look for a solution to y ′′ + 2cy ′ + ω2

0y = A cos(ω0t). SinceA cos(ω0t) is a solution to the homogeneous equationy ′′ + 2cy ′ + ω2

0y = 0, we will look for a solution of the formyp = t(a cos(ω0t) + b sin(ω0t)).

y ′′p (t) + ω20yp(t) = 2ω0(−a sin(ω0t) + b cos(ω0t))

+ tω20(−a cos(ω0t)− b sin(ω0t))

+ ω20t(a cos(ω0t) + b sin(ω0t))

= 2ω0(−a sin(ω0t) + b cos(ω0t)).

So yp is a particular solution if and only if a = 0 and b = A2ω0

. Thusyp(t) = A

2ω0t sin(ω0t) is a particular solution,

and the generalsolution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A

2ω0t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 12

The case ω = ω0We will look for a solution to y ′′ + 2cy ′ + ω2

0y = A cos(ω0t). SinceA cos(ω0t) is a solution to the homogeneous equationy ′′ + 2cy ′ + ω2

0y = 0, we will look for a solution of the formyp = t(a cos(ω0t) + b sin(ω0t)).

y ′′p (t) + ω20yp(t) = 2ω0(−a sin(ω0t) + b cos(ω0t))

+ tω20(−a cos(ω0t)− b sin(ω0t))

+ ω20t(a cos(ω0t) + b sin(ω0t))

= 2ω0(−a sin(ω0t) + b cos(ω0t)).

So yp is a particular solution if and only if a = 0 and b = A2ω0

. Thusyp(t) = A

2ω0t sin(ω0t) is a particular solution, and the general

solution is y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 12

The case ω = ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ω0t) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).

Let us look at the solution where the motion starts at equilibrium.This means that y(0) = y ′(0) = 0. We then have that0 = y(0) = c1 and 0 = y ′(0) = c2ω0, so y(t) = A

2ω0t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 13

The case ω = ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ω0t) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).Let us look at the solution where the motion starts at equilibrium.

This means that y(0) = y ′(0) = 0. We then have that0 = y(0) = c1 and 0 = y ′(0) = c2ω0, so y(t) = A

2ω0t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 13

The case ω = ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ω0t) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).Let us look at the solution where the motion starts at equilibrium.This means that y(0) = y ′(0) = 0.

We then have that0 = y(0) = c1 and 0 = y ′(0) = c2ω0, so y(t) = A

2ω0t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 13

The case ω = ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ω0t) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).Let us look at the solution where the motion starts at equilibrium.This means that y(0) = y ′(0) = 0. We then have that0 = y(0) = c1 and 0 = y ′(0) = c2ω0,

so y(t) = A2ω0

t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 13

The case ω = ω0

The general solution to y ′′ + 2cy ′ + ω20y = A cos(ω0t) is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) + A2ω0

t sin(ω0t).Let us look at the solution where the motion starts at equilibrium.This means that y(0) = y ′(0) = 0. We then have that0 = y(0) = c1 and 0 = y ′(0) = c2ω0, so y(t) = A

2ω0t sin(ω0t).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 13

The case ω = ω0

t

yy(t) = A

2ω0t sin(ω0t)

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 14

The forced damped harmonic motion

If we add a damping term to the system we get the equation

y ′′ + 2cy ′ + ω20y = A cos(ωt).

Let us assume that c < ω0. Then the general solution to thehomogeneous equation y ′′ + 2cy ′ + ω2

0y = 0 is

yh(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) where η =√ω2

0 − c2.To find a particular solution we will use the technique ofundetermined coefficients and the complex method. This meansthat we will be looking for a solution z(t) = aeiωt to the equationz ′′ + 2cz ′ + ω2

0z = Aeiωt .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 15

The forced damped harmonic motion

If we add a damping term to the system we get the equation

y ′′ + 2cy ′ + ω20y = A cos(ωt).

Let us assume that c < ω0. Then the general solution to thehomogeneous equation y ′′ + 2cy ′ + ω2

0y = 0 is

yh(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) where η =√ω2

0 − c2.To find a particular solution we will use the technique ofundetermined coefficients and the complex method. This meansthat we will be looking for a solution z(t) = aeiωt to the equationz ′′ + 2cz ′ + ω2

0z = Aeiωt .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 15

The forced damped harmonic motion

If we add a damping term to the system we get the equation

y ′′ + 2cy ′ + ω20y = A cos(ωt).

Let us assume that c < ω0.

Then the general solution to thehomogeneous equation y ′′ + 2cy ′ + ω2

0y = 0 is

yh(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) where η =√ω2

0 − c2.To find a particular solution we will use the technique ofundetermined coefficients and the complex method. This meansthat we will be looking for a solution z(t) = aeiωt to the equationz ′′ + 2cz ′ + ω2

0z = Aeiωt .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 15

The forced damped harmonic motion

If we add a damping term to the system we get the equation

y ′′ + 2cy ′ + ω20y = A cos(ωt).

Let us assume that c < ω0. Then the general solution to thehomogeneous equation y ′′ + 2cy ′ + ω2

0y = 0 is

yh(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) where η =√ω2

0 − c2.

To find a particular solution we will use the technique ofundetermined coefficients and the complex method. This meansthat we will be looking for a solution z(t) = aeiωt to the equationz ′′ + 2cz ′ + ω2

0z = Aeiωt .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 15

The forced damped harmonic motion

If we add a damping term to the system we get the equation

y ′′ + 2cy ′ + ω20y = A cos(ωt).

Let us assume that c < ω0. Then the general solution to thehomogeneous equation y ′′ + 2cy ′ + ω2

0y = 0 is

yh(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) where η =√ω2

0 − c2.To find a particular solution we will use the technique ofundetermined coefficients and the complex method.

This meansthat we will be looking for a solution z(t) = aeiωt to the equationz ′′ + 2cz ′ + ω2

0z = Aeiωt .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 15

The forced damped harmonic motion

If we add a damping term to the system we get the equation

y ′′ + 2cy ′ + ω20y = A cos(ωt).

Let us assume that c < ω0. Then the general solution to thehomogeneous equation y ′′ + 2cy ′ + ω2

0y = 0 is

yh(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) where η =√ω2

0 − c2.To find a particular solution we will use the technique ofundetermined coefficients and the complex method. This meansthat we will be looking for a solution z(t) = aeiωt to the equationz ′′ + 2cz ′ + ω2

0z = Aeiωt .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 15

The forced damped harmonic motion

If z(t) = aeiωt ,

then

z ′′(t) + 2cz(t)′ + ω20z(t) =

((iω)2 + 2c(iω) + ω20)aeiωt = P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) . The function H(iω) is

called the transfer function.Let R = |P(iω)| and φ = Arg(P(iω)). Then P(iω) = Reiφ andH(iω) = 1

R e−iφ. So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),and yp(t) = Re(z(t)) = A

R cos(ωt − φ) is a particular solution toy ′′ + 2cy ′ + ω2

0y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

If z(t) = aeiωt , then

z ′′(t) + 2cz(t)′ + ω20z(t) =

((iω)2 + 2c(iω) + ω20)aeiωt = P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) . The function H(iω) is

called the transfer function.Let R = |P(iω)| and φ = Arg(P(iω)). Then P(iω) = Reiφ andH(iω) = 1

R e−iφ. So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),and yp(t) = Re(z(t)) = A

R cos(ωt − φ) is a particular solution toy ′′ + 2cy ′ + ω2

0y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

If z(t) = aeiωt , then

z ′′(t) + 2cz(t)′ + ω20z(t) = ((iω)2 + 2c(iω) + ω2

0)aeiωt

= P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) . The function H(iω) is

called the transfer function.Let R = |P(iω)| and φ = Arg(P(iω)). Then P(iω) = Reiφ andH(iω) = 1

R e−iφ. So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),and yp(t) = Re(z(t)) = A

R cos(ωt − φ) is a particular solution toy ′′ + 2cy ′ + ω2

0y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

If z(t) = aeiωt , then

z ′′(t) + 2cz(t)′ + ω20z(t) = ((iω)2 + 2c(iω) + ω2

0)aeiωt = P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) . The function H(iω) is

called the transfer function.Let R = |P(iω)| and φ = Arg(P(iω)). Then P(iω) = Reiφ andH(iω) = 1

R e−iφ. So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),and yp(t) = Re(z(t)) = A

R cos(ωt − φ) is a particular solution toy ′′ + 2cy ′ + ω2

0y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

If z(t) = aeiωt , then

z ′′(t) + 2cz(t)′ + ω20z(t) = ((iω)2 + 2c(iω) + ω2

0)aeiωt = P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) .

The function H(iω) iscalled the transfer function.Let R = |P(iω)| and φ = Arg(P(iω)). Then P(iω) = Reiφ andH(iω) = 1

R e−iφ. So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),and yp(t) = Re(z(t)) = A

R cos(ωt − φ) is a particular solution toy ′′ + 2cy ′ + ω2

0y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

If z(t) = aeiωt , then

z ′′(t) + 2cz(t)′ + ω20z(t) = ((iω)2 + 2c(iω) + ω2

0)aeiωt = P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) . The function H(iω) is

called the transfer function.

Let R = |P(iω)| and φ = Arg(P(iω)). Then P(iω) = Reiφ andH(iω) = 1

R e−iφ. So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),and yp(t) = Re(z(t)) = A

R cos(ωt − φ) is a particular solution toy ′′ + 2cy ′ + ω2

0y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

If z(t) = aeiωt , then

z ′′(t) + 2cz(t)′ + ω20z(t) = ((iω)2 + 2c(iω) + ω2

0)aeiωt = P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) . The function H(iω) is

called the transfer function.Let R = |P(iω)| and φ = Arg(P(iω)).

Then P(iω) = Reiφ andH(iω) = 1

R e−iφ. So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),and yp(t) = Re(z(t)) = A

R cos(ωt − φ) is a particular solution toy ′′ + 2cy ′ + ω2

0y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

If z(t) = aeiωt , then

z ′′(t) + 2cz(t)′ + ω20z(t) = ((iω)2 + 2c(iω) + ω2

0)aeiωt = P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) . The function H(iω) is

called the transfer function.Let R = |P(iω)| and φ = Arg(P(iω)). Then P(iω) = Reiφ andH(iω) = 1

R e−iφ.

So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),and yp(t) = Re(z(t)) = A

R cos(ωt − φ) is a particular solution toy ′′ + 2cy ′ + ω2

0y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

If z(t) = aeiωt , then

z ′′(t) + 2cz(t)′ + ω20z(t) = ((iω)2 + 2c(iω) + ω2

0)aeiωt = P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) . The function H(iω) is

called the transfer function.Let R = |P(iω)| and φ = Arg(P(iω)). Then P(iω) = Reiφ andH(iω) = 1

R e−iφ. So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),

and yp(t) = Re(z(t)) = AR cos(ωt − φ) is a particular solution to

y ′′ + 2cy ′ + ω20y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

If z(t) = aeiωt , then

z ′′(t) + 2cz(t)′ + ω20z(t) = ((iω)2 + 2c(iω) + ω2

0)aeiωt = P(iω)aeiωt

where P(λ) = λ2 + 2cλ+ ω20 is the characteristic polynomial.

Thus z(t) = AP(iω)e

iωt = H(iω)Aeiωt is a solution toz ′′+ 2cz ′+ω2

0z = Aeiωt where H(iω) = 1P(iω) . The function H(iω) is

called the transfer function.Let R = |P(iω)| and φ = Arg(P(iω)). Then P(iω) = Reiφ andH(iω) = 1

R e−iφ. So z(t) = H(iω)Aeiωt = 1R e−iφAeiωt = A

R ei(ωt−φ),and yp(t) = Re(z(t)) = A

R cos(ωt − φ) is a particular solution toy ′′ + 2cy ′ + ω2

0y = A cos(ωt).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 16

The forced damped harmonic motion

The general solution to

y ′′ + 2cy ′ + ω20y = A cos(ωt).

is y(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) + AR cos(ωt − φ).

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 17

The forced damped harmonic motion

t

yy(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) + A

R cos(ωt − φ)

y(t) = AR cos(ωt − φ)

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 18

The forced damped harmonic motion

t

yy(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) + A

R cos(ωt − φ)y(t) = A

R cos(ωt − φ)

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 18

Steady-state and transient terms

The general solution to

y ′′ + 2cy ′ + ω20y = A cos(ωt).

is y(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) + AR cos(ωt − φ).

The term e−ct(c1 cos(ηt) + c2 sin(ηt)) is called the transition term,and the particular solution yp(t) = Re(z(t)) = A

R cos(ωt − φ) iscalled the steady-state term.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 19

Steady-state and transient terms

The general solution to

y ′′ + 2cy ′ + ω20y = A cos(ωt).

is y(t) = e−ct(c1 cos(ηt) + c2 sin(ηt)) + AR cos(ωt − φ).

The term e−ct(c1 cos(ηt) + c2 sin(ηt)) is called the transition term,and the particular solution yp(t) = Re(z(t)) = A

R cos(ωt − φ) iscalled the steady-state term.

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 19

Example

Let us find the steady-state solution to the equation

y ′′ + 2y ′ + 2y = 3 sin 4t .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 20

Example

Let us find the steady-state solution to the equation

y ′′ + 2y ′ + 2y = 3 sin 4t .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 20

Example

Let us find the steady-state solution to the equation

y ′′ + 2y ′ + 2y = 3 sin 4t .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 20

Example

Let us find the steady-state solution to the equation

y ′′ + 2y ′ + 2y = 3 sin 4t .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 20

Example

Let us find the steady-state solution to the equation

y ′′ + 2y ′ + 2y = 3 sin 4t .

www.ntnu.no TMA4110 - Calculus 3, Lecture 7, page 20