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