tma4110 - calculus 3 - lecture 6 · consider the second-order homogeneous linear differential...
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TMA4110 - Calculus 3Lecture 6
Toke Meier CarlsenNorwegian University of Science and Technology
Fall 2012
www.ntnu.no TMA4110 - Calculus 3, Lecture 6
Reference group
We need a reference group (referansegruppe) for TMA4110. Sendan email to [email protected] if you are interested inparticipating in the reference group.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 2
Onsager Lecture
The Lars Onsager Lecture for 2012 will be presented by ProfessorArnold J. Levine, Institute for Advance Study, Princeton. Monday,September 10 at 11:15–12:00 in S2 he will present the LarsOnsager Lecture 2012 with title Searching for the Origins ofCancers so as to Inform Treatments.See http://www.ntnu.edu/onsager for more information.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 3
Second-order homogeneous lineardifferential equations with constantcoefficients
Consider the second-order homogeneous linear differentialequation
y ′′ + py ′ + qy = 0
with constant coefficients.
The characteristic polynomial of the equation is the polynomialλ2 + pλ+ q.The roots
λ =−p ±
√p2 − 4q
2of λ2 + pλ+ q are called thecharacteristic roots of the equation.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 4
Second-order homogeneous lineardifferential equations with constantcoefficients
Consider the second-order homogeneous linear differentialequation
y ′′ + py ′ + qy = 0
with constant coefficients.
The characteristic polynomial of the equation is the polynomialλ2 + pλ+ q.The roots
λ =−p ±
√p2 − 4q
2of λ2 + pλ+ q are called thecharacteristic roots of the equation.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 4
Second-order homogeneous lineardifferential equations with constantcoefficients
Consider the second-order homogeneous linear differentialequation
y ′′ + py ′ + qy = 0
with constant coefficients.The characteristic polynomial of the equation is the polynomialλ2 + pλ+ q.
The roots
λ =−p ±
√p2 − 4q
2of λ2 + pλ+ q are called thecharacteristic roots of the equation.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 4
Second-order homogeneous lineardifferential equations with constantcoefficients
Consider the second-order homogeneous linear differentialequation
y ′′ + py ′ + qy = 0
with constant coefficients.The characteristic polynomial of the equation is the polynomialλ2 + pλ+ q.The roots
λ =−p ±
√p2 − 4q
2of λ2 + pλ+ q are called thecharacteristic roots of the equation.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 4
General solutions to second-orderhomogeneous linear differentialequations with constant coefficients
If p2 − 4q > 0, then the characteristic polynomial λ2 + pλ+ qhas two distinct real roots λ1 and λ2, and the general solutionof y ′′ + py ′ + qy = 0 is c1eλ1t + c2eλ2t .If p2 − 4q < 0, then the characteristic polynomial λ2 + pλ+ qhas two distinct complex roots λ1 = a + ib and λ2 = a− ib,and the general solution of y ′′ + py ′ + qy = 0 isc1eat cos(bt) + c2eat sin(bt).If p2 − 4q = 0, then the characteristic polynomial λ2 + pλ+ qjust have one root λ, and the general solution ofy ′′ + py ′ + qy = 0 is c1eλt + c2teλt .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 5
General solutions to second-orderhomogeneous linear differentialequations with constant coefficients
If p2 − 4q > 0, then the characteristic polynomial λ2 + pλ+ qhas two distinct real roots λ1 and λ2, and the general solutionof y ′′ + py ′ + qy = 0 is c1eλ1t + c2eλ2t .
If p2 − 4q < 0, then the characteristic polynomial λ2 + pλ+ qhas two distinct complex roots λ1 = a + ib and λ2 = a− ib,and the general solution of y ′′ + py ′ + qy = 0 isc1eat cos(bt) + c2eat sin(bt).If p2 − 4q = 0, then the characteristic polynomial λ2 + pλ+ qjust have one root λ, and the general solution ofy ′′ + py ′ + qy = 0 is c1eλt + c2teλt .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 5
General solutions to second-orderhomogeneous linear differentialequations with constant coefficients
If p2 − 4q > 0, then the characteristic polynomial λ2 + pλ+ qhas two distinct real roots λ1 and λ2, and the general solutionof y ′′ + py ′ + qy = 0 is c1eλ1t + c2eλ2t .If p2 − 4q < 0, then the characteristic polynomial λ2 + pλ+ qhas two distinct complex roots λ1 = a + ib and λ2 = a− ib,and the general solution of y ′′ + py ′ + qy = 0 isc1eat cos(bt) + c2eat sin(bt).
If p2 − 4q = 0, then the characteristic polynomial λ2 + pλ+ qjust have one root λ, and the general solution ofy ′′ + py ′ + qy = 0 is c1eλt + c2teλt .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 5
General solutions to second-orderhomogeneous linear differentialequations with constant coefficients
If p2 − 4q > 0, then the characteristic polynomial λ2 + pλ+ qhas two distinct real roots λ1 and λ2, and the general solutionof y ′′ + py ′ + qy = 0 is c1eλ1t + c2eλ2t .If p2 − 4q < 0, then the characteristic polynomial λ2 + pλ+ qhas two distinct complex roots λ1 = a + ib and λ2 = a− ib,and the general solution of y ′′ + py ′ + qy = 0 isc1eat cos(bt) + c2eat sin(bt).If p2 − 4q = 0, then the characteristic polynomial λ2 + pλ+ qjust have one root λ, and the general solution ofy ′′ + py ′ + qy = 0 is c1eλt + c2teλt .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 5
Harmonic motion
x = 0
x = x0
We consider a spring suspendedfrom a beam.
The position of thebottom of the spring is thereference point from which wemeasure displacement, so itcorresponds to x = 0.We then attach a weight of massm to the spring. This weightstretches the spring until it isonce more in equilibrium atx = x0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 6
Harmonic motion
x = 0
x = x0
We consider a spring suspendedfrom a beam. The position of thebottom of the spring is thereference point from which wemeasure displacement, so itcorresponds to x = 0.
We then attach a weight of massm to the spring. This weightstretches the spring until it isonce more in equilibrium atx = x0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 6
Harmonic motion
x = 0
x = x0
We consider a spring suspendedfrom a beam. The position of thebottom of the spring is thereference point from which wemeasure displacement, so itcorresponds to x = 0.We then attach a weight of massm to the spring. This weightstretches the spring until it isonce more in equilibrium atx = x0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 6
Harmonic motion
x = 0
x = x0
At this point there are two forcesacting on the mass. There is theforce of gravity mg, and there isthe restoring force of the springwhich we denote by R(x) since itdepends on the distance x thatthe spring is stretched.
Since we have equilibrium atx = x0, the total force on theweight is 0, so R(x0) + mg = 0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 7
Harmonic motion
x = 0
x = x0
At this point there are two forcesacting on the mass. There is theforce of gravity mg, and there isthe restoring force of the springwhich we denote by R(x) since itdepends on the distance x thatthe spring is stretched.Since we have equilibrium atx = x0, the total force on theweight is 0, so R(x0) + mg = 0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 7
Harmonic motion
x = 0
x = x0
We now set the mass in motionby stretching the spring further.
In addition to gravity and therestoring force, there is adamping force D which is theresistance to the motion of theweight due to the mediumthrough which the weight ismoving and perhaps tosomething internal to the spring.We assume that D depends onthe velocity x ′ of the mass, andwrite it as D(x ′).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 8
Harmonic motion
x = 0
x = x0
We now set the mass in motionby stretching the spring further.In addition to gravity and therestoring force, there is adamping force D which is theresistance to the motion of theweight due to the mediumthrough which the weight ismoving and perhaps tosomething internal to the spring.
We assume that D depends onthe velocity x ′ of the mass, andwrite it as D(x ′).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 8
Harmonic motion
x = 0
x = x0
We now set the mass in motionby stretching the spring further.In addition to gravity and therestoring force, there is adamping force D which is theresistance to the motion of theweight due to the mediumthrough which the weight ismoving and perhaps tosomething internal to the spring.We assume that D depends onthe velocity x ′ of the mass, andwrite it as D(x ′).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 8
Harmonic motion
x = 0
x = x0
According to Newton’s secondlaw, we have
mx ′′ = R(x) + mg + D(x ′).
We assume that R(x) = −kx forsome positive constant k calledthe spring constant, and thatD(x ′) = −µx ′ for somenonnegative constant µ calledthe damping constant.Thus we havemx ′′ = −kx + mg − µx ′.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 9
Harmonic motion
x = 0
x = x0
According to Newton’s secondlaw, we have
mx ′′ = R(x) + mg + D(x ′).
We assume that R(x) = −kx forsome positive constant k calledthe spring constant, and thatD(x ′) = −µx ′ for somenonnegative constant µ calledthe damping constant.
Thus we havemx ′′ = −kx + mg − µx ′.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 9
Harmonic motion
x = 0
x = x0
According to Newton’s secondlaw, we have
mx ′′ = R(x) + mg + D(x ′).
We assume that R(x) = −kx forsome positive constant k calledthe spring constant, and thatD(x ′) = −µx ′ for somenonnegative constant µ calledthe damping constant.Thus we havemx ′′ = −kx + mg − µx ′.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 9
Harmonic motion
x = 0
x = x0
Recall that R(x0) + mg = 0.
Somg = −R(x0) = kx0. If we lety = x − x0, thenmx ′′ = −kx + mg − µx ′ becomesmy ′′ + µy ′ + ky = 0 becausex ′ = y ′ and x ′′ = y ′′.If we let ω0 =
√k/m and
c = µ/2m, then the aboveequation becomes
y ′′ + 2cy ′ + ω20y = 0
where c ≥ 0 and ω0 > 0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 10
Harmonic motion
x = 0
x = x0
Recall that R(x0) + mg = 0. Somg = −R(x0) = kx0.
If we lety = x − x0, thenmx ′′ = −kx + mg − µx ′ becomesmy ′′ + µy ′ + ky = 0 becausex ′ = y ′ and x ′′ = y ′′.If we let ω0 =
√k/m and
c = µ/2m, then the aboveequation becomes
y ′′ + 2cy ′ + ω20y = 0
where c ≥ 0 and ω0 > 0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 10
Harmonic motion
x = 0
x = x0
Recall that R(x0) + mg = 0. Somg = −R(x0) = kx0. If we lety = x − x0, thenmx ′′ = −kx + mg − µx ′ becomesmy ′′ + µy ′ + ky = 0 becausex ′ = y ′ and x ′′ = y ′′.
If we let ω0 =√
k/m andc = µ/2m, then the aboveequation becomes
y ′′ + 2cy ′ + ω20y = 0
where c ≥ 0 and ω0 > 0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 10
Harmonic motion
x = 0
x = x0
Recall that R(x0) + mg = 0. Somg = −R(x0) = kx0. If we lety = x − x0, thenmx ′′ = −kx + mg − µx ′ becomesmy ′′ + µy ′ + ky = 0 becausex ′ = y ′ and x ′′ = y ′′.If we let ω0 =
√k/m and
c = µ/2m, then the aboveequation becomes
y ′′ + 2cy ′ + ω20y = 0
where c ≥ 0 and ω0 > 0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 10
Harmonic motion
The motion described by a solution to the equation
y ′′ + 2cy ′ + ω20y = 0
where c ≥ 0 and ω0 > 0, is called a harmonic motion.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 11
Simple harmonic motion
If c = 0 we say that the system is undamped.
In that case, theequation becomes
y ′′ + ω20y = 0
where ω0 > 0.The general solution to this equation is
y(t) = c1 cos(ω0t) + c2 sin(ω0t).
The motion described by this solution is called a simple harmonicmotion. The number ω0 =
√k/m is called the natural frequency.
The number T = 2π/ω0 = 2π/√
k/m is called the period.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 12
Simple harmonic motion
If c = 0 we say that the system is undamped. In that case, theequation becomes
y ′′ + ω20y = 0
where ω0 > 0.
The general solution to this equation is
y(t) = c1 cos(ω0t) + c2 sin(ω0t).
The motion described by this solution is called a simple harmonicmotion. The number ω0 =
√k/m is called the natural frequency.
The number T = 2π/ω0 = 2π/√
k/m is called the period.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 12
Simple harmonic motion
If c = 0 we say that the system is undamped. In that case, theequation becomes
y ′′ + ω20y = 0
where ω0 > 0.The general solution to this equation is
y(t) = c1 cos(ω0t) + c2 sin(ω0t).
The motion described by this solution is called a simple harmonicmotion. The number ω0 =
√k/m is called the natural frequency.
The number T = 2π/ω0 = 2π/√
k/m is called the period.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 12
Simple harmonic motion
If c = 0 we say that the system is undamped. In that case, theequation becomes
y ′′ + ω20y = 0
where ω0 > 0.The general solution to this equation is
y(t) = c1 cos(ω0t) + c2 sin(ω0t).
The motion described by this solution is called a simple harmonicmotion.
The number ω0 =√
k/m is called the natural frequency.The number T = 2π/ω0 = 2π/
√k/m is called the period.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 12
Simple harmonic motion
If c = 0 we say that the system is undamped. In that case, theequation becomes
y ′′ + ω20y = 0
where ω0 > 0.The general solution to this equation is
y(t) = c1 cos(ω0t) + c2 sin(ω0t).
The motion described by this solution is called a simple harmonicmotion. The number ω0 =
√k/m is called the natural frequency.
The number T = 2π/ω0 = 2π/√
k/m is called the period.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 12
Simple harmonic motion
If c = 0 we say that the system is undamped. In that case, theequation becomes
y ′′ + ω20y = 0
where ω0 > 0.The general solution to this equation is
y(t) = c1 cos(ω0t) + c2 sin(ω0t).
The motion described by this solution is called a simple harmonicmotion. The number ω0 =
√k/m is called the natural frequency.
The number T = 2π/ω0 = 2π/√
k/m is called the period.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 12
Amplitude and phase angle
It is frequently convenient to put the solutiony(t) = c1 cos(ω0t) + c2 sin(ω0t) into another form that is moreconvenient and more revealing of the nature of the solution.
Let z = c1 + ic2. If we let A = |z| and φ = Arg(z), thenz = A(cos(φ) + i sin(φ)) from which it follows that c1 = A cos(φ)and c2 = A sin(φ), and that
y(t) = c1 cos(ω0t) + c2 sin(ω0t)= A cos(φ) cos(ω0t) + A sin(φ) sin(ω0t) = A cos(ω0t − φ).
The number A is called the amplitude, and the number φ is calledthe phase.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 13
Amplitude and phase angle
It is frequently convenient to put the solutiony(t) = c1 cos(ω0t) + c2 sin(ω0t) into another form that is moreconvenient and more revealing of the nature of the solution.Let z = c1 + ic2.
If we let A = |z| and φ = Arg(z), thenz = A(cos(φ) + i sin(φ)) from which it follows that c1 = A cos(φ)and c2 = A sin(φ), and that
y(t) = c1 cos(ω0t) + c2 sin(ω0t)= A cos(φ) cos(ω0t) + A sin(φ) sin(ω0t) = A cos(ω0t − φ).
The number A is called the amplitude, and the number φ is calledthe phase.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 13
Amplitude and phase angle
It is frequently convenient to put the solutiony(t) = c1 cos(ω0t) + c2 sin(ω0t) into another form that is moreconvenient and more revealing of the nature of the solution.Let z = c1 + ic2. If we let A = |z| and φ = Arg(z), thenz = A(cos(φ) + i sin(φ)) from which it follows that c1 = A cos(φ)and c2 = A sin(φ),
and that
y(t) = c1 cos(ω0t) + c2 sin(ω0t)= A cos(φ) cos(ω0t) + A sin(φ) sin(ω0t) = A cos(ω0t − φ).
The number A is called the amplitude, and the number φ is calledthe phase.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 13
Amplitude and phase angle
It is frequently convenient to put the solutiony(t) = c1 cos(ω0t) + c2 sin(ω0t) into another form that is moreconvenient and more revealing of the nature of the solution.Let z = c1 + ic2. If we let A = |z| and φ = Arg(z), thenz = A(cos(φ) + i sin(φ)) from which it follows that c1 = A cos(φ)and c2 = A sin(φ), and that
y(t) = c1 cos(ω0t) + c2 sin(ω0t)= A cos(φ) cos(ω0t) + A sin(φ) sin(ω0t) = A cos(ω0t − φ).
The number A is called the amplitude, and the number φ is calledthe phase.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 13
Amplitude and phase angle
It is frequently convenient to put the solutiony(t) = c1 cos(ω0t) + c2 sin(ω0t) into another form that is moreconvenient and more revealing of the nature of the solution.Let z = c1 + ic2. If we let A = |z| and φ = Arg(z), thenz = A(cos(φ) + i sin(φ)) from which it follows that c1 = A cos(φ)and c2 = A sin(φ), and that
y(t) = c1 cos(ω0t) + c2 sin(ω0t)= A cos(φ) cos(ω0t) + A sin(φ) sin(ω0t) = A cos(ω0t − φ).
The number A is called the amplitude, and the number φ is calledthe phase.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 13
Simple harmonic motion
t
y
y(t) = A cos(ω0t − φ)
φ/ω0
T = 2π/ω0
A
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 14
The underdamped case
If 0 < c < ω0, then the characteristic roots of y ′′ + 2cy ′ + ω20y = 0
are
λ =−2c ±
√(2c)2 − 4ω0
2= −c ±
√c2 − ω2
0 = −c ± i√ω2
0 − c2
so the the general solution to y ′′ + 2cy ′ + ω20y = 0 is
y(t) = e−ct(c1 cos(ωt) + c2 sin(ωt))
where ω =√ω2
0 − c2.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 15
The underdamped case
If 0 < c < ω0, then the characteristic roots of y ′′ + 2cy ′ + ω20y = 0
are
λ =−2c ±
√(2c)2 − 4ω0
2= −c ±
√c2 − ω2
0 = −c ± i√ω2
0 − c2
so the the general solution to y ′′ + 2cy ′ + ω20y = 0 is
y(t) = e−ct(c1 cos(ωt) + c2 sin(ωt))
where ω =√ω2
0 − c2.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 15
The underdamped case
If 0 < c < ω0, then the characteristic roots of y ′′ + 2cy ′ + ω20y = 0
are
λ =−2c ±
√(2c)2 − 4ω0
2= −c ±
√c2 − ω2
0 = −c ± i√ω2
0 − c2
so the the general solution to y ′′ + 2cy ′ + ω20y = 0 is
y(t) = e−ct(c1 cos(ωt) + c2 sin(ωt))
where ω =√ω2
0 − c2.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 15
The underdamped case
t
y
y(t) = e−ct(c1 cos(ωt) + c2 sin(ωt))
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 16
The overdamped case
If c > ω0, then the characteristic roots of y ′′ + 2cy ′ + ω20y = 0 are
λ =−2c ±
√(2c)2 − 4ω0
2= −c ±
√c2 − ω2
0
so the the general solution to y ′′ + 2cy ′ + ω20y = 0 is
y(t) = c1eλ1t + c2eλ2t
where λ1 = −c −√
c2 − ω20 and λ2 = −c +
√c2 − ω2
0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 17
The overdamped case
If c > ω0, then the characteristic roots of y ′′ + 2cy ′ + ω20y = 0 are
λ =−2c ±
√(2c)2 − 4ω0
2= −c ±
√c2 − ω2
0
so the the general solution to y ′′ + 2cy ′ + ω20y = 0 is
y(t) = c1eλ1t + c2eλ2t
where λ1 = −c −√
c2 − ω20 and λ2 = −c +
√c2 − ω2
0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 17
The overdamped case
If c > ω0, then the characteristic roots of y ′′ + 2cy ′ + ω20y = 0 are
λ =−2c ±
√(2c)2 − 4ω0
2= −c ±
√c2 − ω2
0
so the the general solution to y ′′ + 2cy ′ + ω20y = 0 is
y(t) = c1eλ1t + c2eλ2t
where λ1 = −c −√
c2 − ω20 and λ2 = −c +
√c2 − ω2
0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 17
The overdamped case
t
y
y(t) = c1eλ1t + c2eλ2t
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 18
The critically damped case
If c = ω0, then the equation y ′′ + 2cy ′ + ω20y = 0 only has one
characteristic root
λ =−2c ±
√(2c)2 − 4ω0
2= −c
so the the general solution to y ′′ + 2cy ′ + ω20y = 0 is
y(t) = c1e−ct + c2te−ct .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 19
The critically damped case
If c = ω0, then the equation y ′′ + 2cy ′ + ω20y = 0 only has one
characteristic root
λ =−2c ±
√(2c)2 − 4ω0
2= −c
so the the general solution to y ′′ + 2cy ′ + ω20y = 0 is
y(t) = c1e−ct + c2te−ct .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 19
The critically damped case
If c = ω0, then the equation y ′′ + 2cy ′ + ω20y = 0 only has one
characteristic root
λ =−2c ±
√(2c)2 − 4ω0
2= −c
so the the general solution to y ′′ + 2cy ′ + ω20y = 0 is
y(t) = c1e−ct + c2te−ct .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 19
The critically damped case
t
y
y(t) = c1e−ct + c2te−ct
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 20
Inhomogeneous equations
We now turn to the solution of inhomogeneous second-order lineardifferential equations
y ′′ + py ′ + qy = f
where p = p(t), q = q(t) and f = f (t) are functions of theindependent variable.Suppose we have found a particular solution yp to the equation.If yh is a solution to the homogeneous equation y ′′ + py ′ + qy = 0,then yp + yh is a solution to the inhomogeneous equationy ′′ + py ′ + qy = f because (yp + yh)
′′ + p(yp + yh)′ + q(yp + yh) =
(y ′′p + py ′p + qyp) + (y ′′h + py ′h + qyh) = f + 0 = f .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 21
Inhomogeneous equations
We now turn to the solution of inhomogeneous second-order lineardifferential equations
y ′′ + py ′ + qy = f
where p = p(t), q = q(t) and f = f (t) are functions of theindependent variable.
Suppose we have found a particular solution yp to the equation.If yh is a solution to the homogeneous equation y ′′ + py ′ + qy = 0,then yp + yh is a solution to the inhomogeneous equationy ′′ + py ′ + qy = f because (yp + yh)
′′ + p(yp + yh)′ + q(yp + yh) =
(y ′′p + py ′p + qyp) + (y ′′h + py ′h + qyh) = f + 0 = f .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 21
Inhomogeneous equations
We now turn to the solution of inhomogeneous second-order lineardifferential equations
y ′′ + py ′ + qy = f
where p = p(t), q = q(t) and f = f (t) are functions of theindependent variable.Suppose we have found a particular solution yp to the equation.
If yh is a solution to the homogeneous equation y ′′ + py ′ + qy = 0,then yp + yh is a solution to the inhomogeneous equationy ′′ + py ′ + qy = f because (yp + yh)
′′ + p(yp + yh)′ + q(yp + yh) =
(y ′′p + py ′p + qyp) + (y ′′h + py ′h + qyh) = f + 0 = f .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 21
Inhomogeneous equations
We now turn to the solution of inhomogeneous second-order lineardifferential equations
y ′′ + py ′ + qy = f
where p = p(t), q = q(t) and f = f (t) are functions of theindependent variable.Suppose we have found a particular solution yp to the equation.If yh is a solution to the homogeneous equation y ′′ + py ′ + qy = 0,then yp + yh is a solution to the inhomogeneous equationy ′′ + py ′ + qy = f
because (yp + yh)′′ + p(yp + yh)
′ + q(yp + yh) =(y ′′p + py ′p + qyp) + (y ′′h + py ′h + qyh) = f + 0 = f .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 21
Inhomogeneous equations
We now turn to the solution of inhomogeneous second-order lineardifferential equations
y ′′ + py ′ + qy = f
where p = p(t), q = q(t) and f = f (t) are functions of theindependent variable.Suppose we have found a particular solution yp to the equation.If yh is a solution to the homogeneous equation y ′′ + py ′ + qy = 0,then yp + yh is a solution to the inhomogeneous equationy ′′ + py ′ + qy = f because (yp + yh)
′′ + p(yp + yh)′ + q(yp + yh) =
(y ′′p + py ′p + qyp) + (y ′′h + py ′h + qyh) = f + 0 = f .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 21
Inhomogeneous equations
Conversely, if yp1 and yp2 are two different solutions to theinhomogeneous equation y ′′ + py ′ + qy = f , then yh = yp1 − yp2 isa solution to the homogeneous equation y ′′ + py ′ + qy = 0
because (yp1 − yp2)′′ + p(yp1 − yp2)
′ + q(yp1 − yp2) =(y ′′p1
+ py ′p1+ qyp1)− (y ′′p2
+ py ′p2+ qyp2) = f − f = 0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 22
Inhomogeneous equations
Conversely, if yp1 and yp2 are two different solutions to theinhomogeneous equation y ′′ + py ′ + qy = f , then yh = yp1 − yp2 isa solution to the homogeneous equation y ′′ + py ′ + qy = 0because (yp1 − yp2)
′′ + p(yp1 − yp2)′ + q(yp1 − yp2) =
(y ′′p1+ py ′p1
+ qyp1)− (y ′′p2+ py ′p2
+ qyp2) = f − f = 0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 22
Inhomogeneous equations
It follows that if yp is a particular solution to the inhomogeneousequation y ′′ + py ′ + qy = f and y1 and y2 form a fundamental set ofsolutions to the homogeneous equation y ′′ + py ′ + qy = 0, then thegeneral solution to the inhomogeneous equation y ′′ + py ′ + qy = fis
y = yp + c1y1 + c2y2
where c1 and c2 are constants.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 23
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 6, page 24
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 6, page 24
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 6, page 24
Exponential forcing terms
If f (t) = eat , then f ′(t) = aeat , so we will look for a solution of theform beat .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 25
Exponential forcing terms
If f (t) = eat , then f ′(t) = aeat , so we will look for a solution of theform beat .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 25
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 2e−2t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 26
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 2e−2t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 26
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 2e−2t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 26
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 2e−2t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 26
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 2e−2t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 26
Trigonometric forcing terms
If f (t) = A cos(ωt) + B sin(ωt), thenf ′(t) = −ωA sin(ωt) + ωB cos(ωt), so we will look for a solution ofthe form a cos(ωt) + b sin(ωt).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 27
Trigonometric forcing terms
If f (t) = A cos(ωt) + B sin(ωt), thenf ′(t) = −ωA sin(ωt) + ωB cos(ωt), so we will look for a solution ofthe form a cos(ωt) + b sin(ωt).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 27
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 28
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 28
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 28
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 28
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 28
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 28
The complex method
Consider again the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
Since 5 sin(3t) = Im(5e3it), y(t) = Im(z(t)) is a solution toy ′′ + 2y ′ − 3y = 5 sin(3t) if z is a solution to z ′′ + 2z ′ − 3z = 5e3it .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 29
The complex method
Consider again the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
Since 5 sin(3t) = Im(5e3it), y(t) = Im(z(t)) is a solution toy ′′ + 2y ′ − 3y = 5 sin(3t) if z is a solution to z ′′ + 2z ′ − 3z = 5e3it .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 29
The complex method
Consider again the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
Since 5 sin(3t) = Im(5e3it),
y(t) = Im(z(t)) is a solution toy ′′ + 2y ′ − 3y = 5 sin(3t) if z is a solution to z ′′ + 2z ′ − 3z = 5e3it .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 29
The complex method
Consider again the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
Since 5 sin(3t) = Im(5e3it), y(t) = Im(z(t)) is a solution toy ′′ + 2y ′ − 3y = 5 sin(3t) if z is a solution to z ′′ + 2z ′ − 3z = 5e3it .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 29
The complex method
Consider again the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
Since 5 sin(3t) = Im(5e3it), y(t) = Im(z(t)) is a solution toy ′′ + 2y ′ − 3y = 5 sin(3t) if z is a solution to z ′′ + 2z ′ − 3z = 5e3it .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 29
The complex method
Consider again the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
Since 5 sin(3t) = Im(5e3it), y(t) = Im(z(t)) is a solution toy ′′ + 2y ′ − 3y = 5 sin(3t) if z is a solution to z ′′ + 2z ′ − 3z = 5e3it .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 29
The complex method
Consider again the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
Since 5 sin(3t) = Im(5e3it), y(t) = Im(z(t)) is a solution toy ′′ + 2y ′ − 3y = 5 sin(3t) if z is a solution to z ′′ + 2z ′ − 3z = 5e3it .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 29
The complex method
Consider again the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
Since 5 sin(3t) = Im(5e3it), y(t) = Im(z(t)) is a solution toy ′′ + 2y ′ − 3y = 5 sin(3t) if z is a solution to z ′′ + 2z ′ − 3z = 5e3it .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 29
The complex method
Consider again the equation
y ′′ + 2y ′ − 3y = 5 sin(3t).
Since 5 sin(3t) = Im(5e3it), y(t) = Im(z(t)) is a solution toy ′′ + 2y ′ − 3y = 5 sin(3t) if z is a solution to z ′′ + 2z ′ − 3z = 5e3it .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 29
Polynomial forcing terms
If f (t) = antn + an−1tn−1 + . . . a1t + a0, thenf ′(t) = nantn−1 + (n − 1)an−1tn−2 + . . . a1, so we will look for asolution of the form bntn + bn−1tn−1 + . . . b1t + b0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 30
Polynomial forcing terms
If f (t) = antn + an−1tn−1 + . . . a1t + a0, thenf ′(t) = nantn−1 + (n − 1)an−1tn−2 + . . . a1, so we will look for asolution of the form bntn + bn−1tn−1 + . . . b1t + b0.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 30
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 3t + 4.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 31
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 3t + 4.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 31
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 3t + 4.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 31
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 3t + 4.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 31
Example
Let us find a particular solution to the equation
y ′′ + 2y ′ − 3y = 3t + 4.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 31
Exceptional cases
The method of undetermined coefficients looks straightforward.There are, however, some exceptional cases to look out for. If theforcing term f , and hence the proposed solution, is a solution to thehomogeneous equation y ′′ + py ′ + qy = 0, then the proposedsolution wouldn’t work.
Instead we have to multiply the proposed solution by t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 32
Exceptional cases
The method of undetermined coefficients looks straightforward.There are, however, some exceptional cases to look out for. If theforcing term f , and hence the proposed solution, is a solution to thehomogeneous equation y ′′ + py ′ + qy = 0, then the proposedsolution wouldn’t work.Instead we have to multiply the proposed solution by t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 32
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 33
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 33
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 33
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 33
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 33
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 33
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 33
Combination forcing terms
If yf is a solution the differential equation y ′′ + py ′ + qy = f , yg is asolution the differential equation y ′′ + py ′ + qy = g, and c1 and c2are constants, then y(t) = c1yf (t) + c2yg(t) is a solution to thedifferential equation y ′′ + py ′ + qy = c1f + c2g.
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 34
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = e−2t − 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 35
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = e−2t − 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 35
Example
Let us find a particular solution to the equation
y ′′ − y ′ − 2y = e−2t − 3e−t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 35
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36
Example
Let us find the general solution to the equation
y ′′ + 4y ′ + 4y = 4− t .
www.ntnu.no TMA4110 - Calculus 3, Lecture 6, page 36