Ordinary Di erential Equations - Texas A&M Universityroquesol/Math_308_Fall_2016_Test_2_Print.pdfAlong with this di erential equation we will have the following initial conditions

Second Order Differential EquationsHigher Order Linear Differential Equations

Laplace TransformLaplace Transform II

Ordinary Differential Equations

Dr. Marco A Roque Sol

12/01/2015

Dr. Marco A Roque Sol Ordinary Differential Equations



Mechanical and Electrical Vibrations



Second order linear equations with constant coefficients areimportant in two physical processes, namely, Mechanical andElectrical oscillations.

Actually from the Math point of view, both problems are the same.However, from the Physics point of view they are quite different.

For example, the motion of a mass on a vibrating spring, theangular motion of a simple pendulum, the flow of electric currentin a simple series circuit and the electrical charge in an electriccircuit, are just examples of that difference.






Let’s get the situation setup. We are going to start with a springof length l , called the natural length, and we’re going to hook anobject with mass m up to it. When the object is attached to thespring the spring will stretch a length of L. Below is sketch of thespring with and without the object attached to it.






Convention

As denoted in the sketch we are going to assume that all forces,velocities, and displacements in the downward direction will bepositive. All forces, velocities, and displacements in the upwarddirection will be negative.

Also, as shown in the sketch above, we will measure alldisplacement of the mass from its equilibrium position. Therefore,the u = 0 position will correspond to the center of gravity for themass as it hangs on the spring and is at rest.






Now, we need to develop a differential equation that will give thedisplacement of the object at any time t. First, recall Newton’sSecond Law of Motion.

F = ma

In this case we will use the second derivative of the displacement,u, for the acceleration and so Newton’s Second Law becomes,

F (t, u, u′) = mu′′

Here is a list of the forces that will act upon the object.






Gravity, Fg

The force due to gravity will always act upon the object of course.This force is

Fg = mg

Spring, Fs

We are going to assume that Hookes Law will govern the forcethat the spring exerts on the object. This force will always bepresent as well and is

Fs = −k(L + u)Hookes Law tells us that the force exerted by a spring will be thespring constant, k > 0, times the displacement of the spring fromits natural length.






Damping, Fd

The next force that we need to consider is damping. This forcemay or may not be present for any given problem. This force worksto counteract any movement. This damping force is

Fd = −γu′

where, γ > 0 is the damping coefficient.

External Forces, F (t)

If there are any other forces acting on our object we collect themin this term. We typically call F (t) the forcing function.






Putting all of these together gives us the following for NewtonsSecond Law.

mu′′ = mg − k(L + u)− γu′ + F (t)Or, upon rewriting, we get,

mu′′ + γu′ + ku = mg − kL + F (t)Now, when the object is at rest in its equilibrium position,

mg − kL = 0Using this in Newtons Second Law gives us the final version of thedifferential equation that well work with.

mu′′ + γu′ + ku = F (t)Dr. Marco A Roque Sol Ordinary Differential Equations





Along with this differential equation we will have the followinginitial conditions.

u(0) = u0 Initial displacement

u′(0) = u′0 Initial velocity

OBS

If we have a mass m attached to a spring with constant k in asurface with friction c ( or γ) and subject to an external force F (t),






then it satisfy the differential equation

mu′′ + cu′ + ku = F (t)

Free, Undamped Vibrations

This is the simplest case that we can consider. Free or unforcedvibrations means that F (t) = 0 and undamped vibrations meansthat γ = 0. In this case the differential equation becomes,

mu′′ + ku = 0

The characteristic equation has the roots,

r = ±√

k

m= ±ω0i ; ω0 =

√k

m






Where ω0 is called the natural frequency. Recall as well thatm > 0 and k > 0 and so we can guarantee that this quantity willnot be complex. The solution in this case is then

u(t) = c1cos(ω0t) + c2sin(ω0t)

We can write the above equation in the following form

u(t) = Rcos(ω0t − δ)(If c1 = Rcos(δ) c2 = Rsin(δ) =⇒ u(t) = Rcos(δ)cos(ω0t)+Rsin(δ)sin(ω0t) =⇒ u(t) = Rcos(ω0t − δ); R2 = c21 + c22 ;tan(δ) = c2/c1

where R is the amplitude of the displacement and δ is the phaseshift or phase angle of the displacement. T = 2πω0 = 2π

√mk is

called the natural period.Dr. Marco A Roque Sol Ordinary Differential Equations





Example 55A 16 lb object stretches a spring 8/9 ft by itself. There is nodamping and no external forces acting on the system. The springis initially displaced 6 inches upwards from its equilibrium positionand given an initial velocity of 1 ft/sec downward. Find thedisplacement at any time t, u(t).

Solution

We first need to set up the IVP for the problem. We need to findm and k . This is the British system so we’ll need to compute themass.

m =W

g=

16

32=

1

2






Now, lets find k . We can use the fact that mg = kL to find k .We’ll use feet for the unit of measurement for this problem.

k =mg

L=

16

8/9= 18

We can now set up the IVP.

1

2u′′ + 18u = 0; u(0) = −1

2(6inches), u′(0) = 1

Now, the natural frequency, is

ω0 =

√18

1/2= 6






The general solution, along with its derivative, is then,

u(t) = c1cos(6t) + c2sin(6t)

u′(t) = −6c1sin(6t) + 6c2cos(6t)Applying the initial conditions gives

−12

= u(0) = c1 =⇒ c1 = −1

2

1 = u′(0) = 6c2 =⇒ c2 =1

6The displacement at any time t is then

u(t) = −12cos(6t) +

1

6sin(6t)






Now, lets convert this to a single cosine function. First let’s getthe amplitude, R.

R =

√(−1

2

)2+

(1

6

)2=

√10

6= 0.52705

Now let’s get the phase shift.

δ = tan−1(

1/6

−1/2

)= −0.32175

From the above equations, we have two angles

δ1 = −0.32175; δ2 = δ1 + π = 2.81984






We need to decide which of these phase shifts is correct, becauseonly one will be correct. To do this recall that

c1 = Rcos(δ) = −1/2 < 0; c2 = Rsin(δ) = 1/6 > 0

This means that the phase shift must be in Quadrant II and so thesecond angle is the one that we need. Thus, the displacement atany time t is.

u(t) =

√10

6cos(6t − δ); δ = tan−1

(−13

)+ π






Here is a sketch of the displacement for the first 5 seconds.






Free, Damped Vibrations

We are still going to assume that there will be no external forcesacting on the system, with the exception of damping of course. Inthis case the differential equation will be

mu′′ + γu′ + ku = 0

where m, γ, and k are all positive constants. Upon solving for theroots of the characteristic equation we get the following.

r1,2 =−γ ±

√γ2 − 4mk

2m






We will have three cases here.

1.− γ2 − 4mk = 0

In this case we will get a double root out of the characteristicequation and the displacement at any time t will be.

u(t) = c1e− γt

2m + c2te− γt

2m

Notice that as t →∞ the displacement will approach zero.

This case is called critical damping and will happen when thedamping coefficient is,

γ2 − 4mk = 0 =⇒ γ =√

4mk = γCR






The above value of the damping coefficient is called the criticaldamping coefficient and denoted by γCR .

2.− γ2 − 4mk > 0.

In this case let’s rewrite the roots

r1,2 =−γ ±

√γ2 − 4mk

2m= − γ

2m

(1±

√1− 4mk

γ2

)Also notice that from our initial assumption that we have,

4mk

γ2< 1 =⇒ 1− 4mk

γ2< 1 =⇒

√1− 4mk

γ2< 1






This means that the quantity in the parenthesis is guaranteed tobe positive and so the two roots in this case are guaranteed to benegative. Therefore the displacement at any time t is

u(t) = c1e− γ

2m

(1+√

1− 4mkγ2

)t

+ c2e− γ

2m

(1−√

1− 4mkγ2

)t

and will approach zero as t →∞.

This case will occur when

γ2 − 4mk > 0 =⇒ γ >√

4mk = γCR

and is called over damping.






3.− γ2 − 4mk < 0.

In this case we will get complex roots out of the characteristicequation.

r1,2 =−γ2m±√γ2 − 4mk

2m= λ± i µ

where the real part is guaranteed to be negative and so thedisplacement is

u(t) = e−γ2m

t

(c1cos(

√4mk − γ2

2mt) + c2sin(

√4mk − γ2

2mt)

)

u(t) = = Rcos(µt − δ)

Since λ < 0 the displacement will approach zero as t →∞ .Dr. Marco A Roque Sol Ordinary Differential Equations





We will get this case will occur when

γ2 − 4mk < 0 =⇒ γ <√

4mk = γCR

and is called under damping.

Example 56

Take the spring and mass system from the example 55 and considerthere is a damping force to it that will exert a force of 17 lbs whenthe velocity is 2ft/s. Find the displacement at any time t, u(t).

Solution

So, the only difference between this example and the previousexample is damping force. So let’s find the damping coefficient

17 = γ(2) =⇒ γ = 2/17 = 8.5 > γCR =√

4km = 6






So it looks like we have got over damping this time around so weshould expect to get two real distinct roots from the characteristicequation and they should both be negative. The IVP for thisexample is

1

2u′′ +

17

2u′ + 18u = 0; u(0) = −1

2, u′(0) = 1

The roots of the characteristic equation are

r1,2 =−17±

√145

2

The general solution for this example is

u(t) = c1e−17+

√145

2t + c2e

−17−√

1452

t






and after applying initial conditions, the particular solution is

u(t) = −0.52e(−17+

√145

2

)t

+ 0.020e

(−17−

√145

2

)t

Here is a sketch of the displacement for the first 5 seconds.






Example 57

Take the spring and mass system from the example 55 and considerthere is a damping force to it that will exert a force of 12lbs whenthe velocity is 2ft/s. Find the displacement at any time t, u(t).

Solution

The damping coefficient is given by

12 = γ(2) =⇒ γ = 12/2 = 6 = γCRSo it looks like we have got critical damping this time. The IVPfor this problem is

1

2u′′ + 6u′ + 18u = 0; u(0) = −1

2, u′(0) = 1







r1,2 = −6

(2)(1/2)= 6


u(t) = c1e−6t + c2te

−6t

and after applying initial conditions, the particular solution is

u(t) = −12e−6t − 2te−6t






Here is a sketch of the displacement for this example






Example 58

Take the spring and mass system from the example 55 and considerthere is a damping force to it that will exert a force of 5lbs whenthe velocity is 2ft/s. Find the displacement at any time t, u(t).

Solution

The damping coefficient is given by

5 = γ(2) =⇒ γ = 5/2 = 2.5 < γCRSo it looks like we have got under damping this time. The IVP forthis problem is

1

2u′′ +

5

2u′ + 18u = 0; u(0) = −1

2, u′(0) = 1







r1,2 =−5±

√119 i

2


u(t) = e−52t

(c1cos

(√119

2t

)+ c2sin

(√119

2t

))and after applying initial conditions, the particular solution is

u(t) = e−5t2

(−0.5cos

(√119

2t

)− 0.046sin

(√119

2t

))






Lets convert this to a single cosine as we did in the undamped case.

R =√

(−0.5)2 + (.046)2 = 0.502 δ = tan−1 =(−0.046−0.5

)= 0.09

or δ = 0.09 + π = 3.23

This means δ must be in the Quadrant II ( why ?) and so thesecond angle is the one that we want. The displacement is then

u(t) = 0.502e−5t2 cos

(√119

2t − 3.23

)






Undamped, Forced VibrationsWe will first take a look at the undamped case. The differentialequation in this case is

mu′′ + ku = F (t)

This is just a nonhomogeneous differential equation and we knowhow to solve these. The general solution will be

u(t) = uc + UP

There is a particular type of forcing function that we should take alook at since it leads to some interesting results. Lets suppose thatthe forcing function is a simple periodic function of the form

F (t) = F0cos(ωt) or F (t) = F0sin(ωt)






For the purposes of this discussion we will use the first one. Usingthis, the ODE becomes,

mu′′ + ku = F0cos(ωt)

The solution of the associate homogeneus , as pointed out above,is just

uc(t) = c1cos(ω0t) + c2sin(ω0t)

where ω0 is the natural frequency.

We will need to be careful in finding a particular solution. Thereason for this will be clear if we use undetermined coefficients.With undetermined coefficients our guess for the form of theparticular solution would be,






UP(t) = Acos(ωt) + Bsin(ωt)

Now, this guess will have problems if ω = ω0. So, we will need tolook at this in two cases.

1. ω 6= ω0In this case our initial guess is okay since it wont be thecomplementary solution. Upon differentiating the guess andplugging it into the differential equation and simplifying we get,

mu′′ + ku = F0cos(ωt)

m (Acos(ωt) + Bsin(ωt))′′ + k (Acos(ωt) + Bsin(ωt)) = F0cos(ωt)






m(−ω2Acos(ωt)− ω2Bsin(ωt)

)+ k (Acos(ωt) + Bsin(ωt)) = F0cos(ωt)

(−mω2A + kA

)cos(ωt) +

(−mω2B + kB

)sin(ωt) = F0cos(ωt)

Setting coefficients equal gives us,

cos(ωt)(−mω2 + k

)A = F0 =⇒ A =

F0k −mω2

sin(ωt)(−mω2 + k

)B = 0 =⇒ B = 0






The particular solution is then

F0k −mω2

cos(ωt) =F0

m (k/m − ω2)cos(ωt) =

F0

m(ω20 − ω2

)cos(ωt)Note that we rearranged things a little. Depending on the formthat you’d like the displacement to be in we can have either of thefollowing.

u(t) = c1cos(ω0t) + c2sin(ω0t) +F0

m(ω20 − ω2

)cos(ωt)u(t) = Rcos(ω0t − δ) +

F0

m(ω20 − ω2

)cos(ωt)If we used the sine form of the forcing function we could get asimilar formula.






2. ω = ω0In this case we will need to add in a t to the guess for theparticular solution.

UP(t) = Atcos(ωt) + Btsin(ωt)

Differentiating our guess, plugging it into the differential equationand simplifying gives us the following.(

−mω20 + k)Atcos(ωt) +

(−mω20 + k

)Btsin(ω)t + ...

...+ 2mω0Bcos(ωt)− 2mω0Asin(ωt) = F0cos(ωt)but (

−mω20 + k)

= m(−ω20 + k/m

)= m

(−ω20 + ω20

)= 0






So, the first two terms actually drop out and this gives us

cos(ωt) 2mω0B = F0 =⇒ B =F0

2mω0

sin(ωt) 2mω0A = 0 =⇒ A = 0

In this case the particular will be,

F02mω0

tsin(ω0t)

The displacement for this case is then

u(t) = c1cos(ω0t) + c2sin(ω0t) +F0

2mω0tsin(ω0t)






depending on the form that you prefer for the displacement.

u(t) = Rcos(ω0t − δ) +F0

2mω0tsin(ω0t)

So, what was the point of the two cases here? Well in the firstcase, our displacement function consists of two cosines and is niceand well behaved for all time.

In contrast, the second case, will have some serious issues at tincreases. The addition of the t in the particular solution will meanthat we are going to see an oscillation that grows in amplitude as tincreases. This case is called resonance and we would generallylike to avoid this at all costs.






In this case resonance arose by assuming that the forcing functionwas,

F (t) = F0cos(ω0t)

We would also have the possibility of resonance if we assumed aforcing function of the form.

F (t) = F0sin(ω0t)

We should also take care to not assume that a forcing function willbe in one of these two forms. Forcing functions can come in a widevariety of forms. If we do run into a forcing function different fromthe one that used here you will have to go through undeterminedcoefficients or variation of parameters to determine the particularsolution.






Example 59

A 3 kg object is attached to spring and will stretch the spring 392mm by itself. There is no damping in the system and a forcingfunction of the form

F (t) = 10cos(ωt)

is attached to the object and the system will experience resonance.If the object is initially displaced 20 cm downward from itsequilibrium position and given a velocity of 10 cm/sec upward findthe displacement at any time t.

Solution

Since we are in the metric system we wont need to find mass as itsbeen given to us. Also, for all calculations we will be converting alllengths over to meters. The first thing we need to do is find k.






k =mg

L=

(3)(9.8)

0.392= 75

Now, we are told that the system experiences resonance so let’s goahead and get the natural frequency.

ω0 =

√k

m=

√75

3= 5

The IVP for this is then

3u′′ + 75u = 10cos(5t); u(0) = 0.2, u′(0) = −0.1The complementary solution is the free undamped solution whichis easy to get and for the particular solution we can just use theformula that we derived above. The general solution is then,

u(t) = c1cos(5t) + c2sin(5t) +1

3tsin(5t)






Applying the initial conditions gives

u(t) =1

5cos(5t)− 1

50sin(5t) +

1

3tsin(5t)

The last thing that we’ll do is combine the first two terms into asingle cosine.

R =

√(1

5

)2+

(−150

)2= 0.201 δ1 = tan

−1(−1/50

1/5

)= −0.099

δ2 = δ1 + π = 3.042

In this case c1 > 0 is positive and c2 < 0 . This means that thephase shift needs to be in Quadrant IV and so the first one is thecorrect phase shift this time.






The displacement then becomes,

u(t) =1

5

√101

100cos

(5t + tan−1

(−110

))+

1

3tsin(5t)







Example 60

Solve the initial value problem and plot the solution.

u′′ + u = 0.5cos(0.8t), u(0) = 0, u′(0) = 0

Solution

The general solution of is

u = c1cos(ω0t) + c2sin(ω0t) +F0

m(ω20 − ω2)cos(ωt)

Applying initial conditions, we obtain

c1 = −F0

m(ω20 − ω2); c2 = 0






and the particular solution of the IVP is

u =F0

m(ω20 − ω2)(cos(ωt)− cos(ω0t))

This is the sum of two periodic functions of different periods butthe same amplitude. Making use of the trigonometric identities forcos(A± B) with A = (ω0 + ω)t/2 and B = (ω0 − ω)t/2, we canwrite the above equation in the form

u =

[F0

m(ω20 − ω2)sin

((ω0 − ω)t

2

)]sin

((ω0 + ω)t

2

)






If |ω0 − ω| is small, then ω0 + ω is much greater than it.Consequently, sin(ω0 + ω)t/2 is a rapidly oscillating functioncompared to sin(ω0 − ω)t/2. Thus the motion is a rapid oscillationwith frequency (ω0 + ω)/2 but with a slowly varying sinusoidalamplitude

F0m|ω20 − ω2|

∣∣∣∣sin((ω0 − ωt)2)∣∣∣∣

This type of motion, possessing a periodic variation of amplitude,exhibits what is called a beat. In this case ω0 = 1, = 0.8, andF0 = 0.5, so the solution of the given problem is

u(t) = [2.77sin(0.1t)] sin(0.9t)






Here is a sketch of the displacement for this example.






Forced, Damped Vibrations

This is the full blown case where we consider every last possibleforce that can act upon the system. The differential equation forthis case is,

mu′′ + γu′ + ku = F (t)

The displacement function this time will be,

u(t) = uc + UP

where the complementary solution will be the solution to the ( free,damped) homogeneous case and the particular solution will befound using undetermined coefficients or variation of parameters.






There are a couple of things to note here about this case. First,from our work back in the free, damped case we know that thecomplementary solution will approach zero as t →∞. Because ofthis, the complementary solution is often called the transientsolution in this case.

Also, because of this behavior the displacement will start to lookmore and more like the particular solution as t increases and so theparticular solution is often called the steady state solution orforced response.






Example 61

Take the system from the example 59 and add in a damper thatwill exert a force of 45 Newtons when then velocity is 50 cm/sec.Solution

So, all we need to do is compute the damping coefficient for thisproblem then pull everything else down from the previous problem.The damping coefficient is

Fd = γu′ =⇒ 45 = γ(0.5) =⇒ γ = 90






The IVP for this problem is.

3u′′ + 90u′ + 75u = 10cos(5t); u(0) = 0.2, u′(0) = −0.1

The complementary solution for this example is

u(t) = c1e(−15+10

√2)t + c2e

(−15−10√2)t

For the particular solution we the form will be,

UP = Acos(5t) + Bsin(5t)

Plugging this into the differential equation and simplifying gives us,

405Bcos(5t)− 450Asin(5t) = 10cos(5t)






Setting coefficient equal gives,

UP =1

45sin(5t)

The general solution is then

u(t) = c1e(−15+10

√2)t + c2e

(−15−10√2)t +

1

45sin(5t)

Applying the initial condition gives

u(t) = 0.1986e(−15+10√2)t + 0.0014e(−15−10

√2)t +

1

45sin(5t)






Here is a sketch of the displacement for this example.






Electric Circuits

A second example of the occurrence of second order lineardifferential equations with constant coefficients is their use as amodel of the flow of electric current in the simple series circuitshown below






The current I , measured in amperes (A), is a function of time t.The resistance R in ohms (Ω), the capacitance C in farads (F ),and the inductance L in henrys (H) are all positive and areassumed to be known constants.

The impressed voltage E in volts (V ) is a given function of time.Another physical quantity that enters the discussion is the totalcharge Q in coulombs (C ) on the capacitor at time t. The relationbetween charge Q and current I is

I =dQ

dt






The flow of current in the circuit is governed by Kirchhoff’s secondlaw:

(Gustav Kirchhoff (1824 1- 887) was a German physicist andprofessor at Breslau, Heidelberg, and Berlin. http://www.britannica.com/biography/Gustav-Robert-Kirchhoff )vspace3mmIn a closed circuit the impressed voltage is equal to the sum of thevoltage drops in the rest of the circuit. According to theelementary laws of electricity, we know that


http://www.britannica.com/biography/Gustav-Robert-Kirchhoff http://www.britannica.com/biography/Gustav-Robert-Kirchhoff





The voltage drop across the resistor is IR.

The voltage drop across the capacitor is Q/C .

The voltage drop across the inductor is LdI/dt.

Hence, by Kirchhoffs law,

−LdIdt− RI − 1

CQ + E (t) = 0 =⇒ LdI

dt+ RI +

1

CQ = E (t)






The units for voltage, resistance, current, charge, capacitance,inductance, and time are all related:

1volt = 1ohm × 1ampere = 1coulomb/1farad = 1henry × 1ampere/1second .

Substituting dQdt for I , we obtain the differential equation

Ld2Q

dt+ R

dQ

dt+

1

CQ = E (t)

for the charge Q. The initial conditions are

Q(t0) = Q0, Q′(t0) = I (t0) = I0






Thus we must know the charge on the capacitor and the current inthe circuit at some initial time t0. Alternatively, we can obtain adifferential equation for the current I by differentiating the aboveequation with respect to t, and then substituting dQ/dt for I . Theresult is

Ld2I

dt2+ R

dI

dt+

1

CI = E ′(t)

with the initial conditions

I (t0) = I0, I′(t0) = I

′0

it follows from the equation for Q(t) that

I ′0 =E (t0)− RI0 − (1/C )Q0

LDr. Marco A Roque Sol Ordinary Differential Equations





The most important conclusion from this discussion is that theflow of current in the circuit is described by an initial valueproblem of precisely the same form as the one that describes themotion of a springmass system.

m






Second Order Lineat Differential Equations. Reduction ofOrder

In the final part of this section, we will consider the general secondorder linear differential equation

y ′′ + p(t)y ′ + q(t)y = g(t)

and introducing new variables, the idea will be to reduce the aboveequation of second order into something of first order.

In that way we propose

x1 = y x2 = y′






these two varaibles satisfy

x ′1 = y′ = x2

x ′2 = y′′ = −p(t)y ′ − q(t)y + g(t) = −p(t)x2 − q(t)x1 + g(t)

therefore, we obtain

x ′1 = x2

x ′2 = −p(t)x2 − q(t)x1 + g(t)

In this way, we have made a reduction of order, but now instead ofa single equation, we have a system linear first oder differentialequations !!!!!






Thus, we have showed that a second order linear differentialequations can always be transformed into a system of two linearfirst order differential equations.

OBS

1. The above proposition can be generalized for the n dimensionalcase.

”... An nth order linear differential equation is equivalent to asystem of n linear first order differential equations ...”




General Theory of nth Order Linear EquationsHomogeneous Equations with Constant CoefficientsThe Method of Undetermined CoefficientsThe Method of Variation of Parameters

General Theory of nth Order Linear Equations

Example 4.2Determine whether the functions f1(t) = 1, f2(t) = 2 + t,f3(t) = 3− t2, and f4(t) = 4t + t2 are linearly independent ordependent on the interval I : −∞ < t





Thus, can show that the system has infinitely solutions andtherefore k1, k2, k3, and k4 are not necesarily all of them equal tozero. Hence, the set of functions is linearly dependent.

OBS

f4(t) = 4t − t2 = (3− t2)− 4(2 + t) + 5(1) = f3(t)− 4f2(t) + 5f1(t)

therefore the set f1(t) = 1, f2(t) = 2 + t, f3(t) = 3− t2,f4(t) = 4t + t

2 cannot be linearly indepent.






Theorem 4.3

If y1(t), ..., yn(t) is a fundamental set of solutions of the equation

L[y ](t) =dny

dtn+ p1(t)

dn−1y

dtn−1+ ...+ pn−1

dy

dt+ pn(t)y = 0

on an interval I, then y1(t), ..., yn(t) are linearly independent on I.Conversely, if y1(t), ..., yn(t) are linearly independent solutions ofthe above equation on I, then they form a fundamental set ofsolutions on I.






The Nonhomogeneous Equation

Now consider the nonhomogeneous eq

L[y ](t) =dny

dtn+ p1(t)

dn−1y

dtn−1+ ...+ pn−1

dy

dt+ pn(t)y = g(t)

If Y1 and Y2 are any two solutions of the above equation, then itfollows immediately from the linearity of the operator L that

L[Y1 − Y2](t) = L[Y1](t)− L[Y2](t) = g(t)− g(t) = 0






Hence the difference of any two solutions of the nonhomogeneousequation is a solution of the homogeneous equation. Since anysolution of the homogeneous equation can be expressed as a linearcombination of a fundamental set of solutions y1, ..., yn, it followsthat any solution of nonhomogeneus equation can be written as

y(t) = c1y1(t) + c2y2(t) + ...+ cnyn(t) + Y (t)

where Y is some particular solution of the nonhomogeneousequation. The above linear combination is called the generalsolution of the nonhomogeneous equation.





Homogeneous Equations with Constant Coefficients

Let’s take the nth order linear homogeneous differential equation

L[y ](t) = a0dny

dtn+ a1

dn−1y

dtn−1+ ...+ an−1

dy

dt+ any = 0

where a0, a1, ..., an are real constants and a0 6= 0. Again weproposed that y = ert is a solution of the above equation. As amatter of fact,

L[ert ] = ert(a0r

n + a1rn−1 + ...+ an−1r + an

)= ertZ (r)

for all r, where Z (r) = a0rn + a1r

n−1 + ...+ an−1r + an. For thosevalues of r for which Z (r) = 0, it follows that L[ert ] = 0 andy = ert is a solution of homogeneous equation. The polynomialZ (r) is called the characteristic polynomial , and the equationZ (r) = 0 is the characteristic equation .






Since a0 6= 0, we know that Z (r) is a polynomial of degree n andtherefore as n zeros, say, r1, r2, ..., rn, some of which may be equal.Hence we can write the characteristic polynomial in the form(Fundamental Theorem of Algebra: Every non-constantsingle-variable polynomial with complex coefficients has at leastone complex root. )

Z (r) = a0(r − r1)(r − r2) . . . (r − rn).

In general there are three cases, namely

1) Real and Different Roots.

If the roots of the characteristic equation are different, then wehave n distinct solutions er1t , er2t , ..., ernt . These functions arelinearly independent, then the general solution is






y(t) = c1er1t + c2e

r2t + ...+ cnernt

2) Complex Roots .

If the characteristic equation has complex roots, they must occurin conjugate pairs, λ± i µ, since the coefficients a0, a1, a2, ..., anare real numbers. Provided that none of the roots is repeated.Now, just as for the second order equation, we can replace thecomplex-valued solutions z1 = e

(λ+i µ)t and z2 = e(λ−i µ)t by the

real-valued solutions

eλtcos(µt); eλtsin(µt)

obtained as the linear cobinations 12(z1 + z2) and12i (z1 − z2)






3) Repeated Roots.

Let’s assume that some of the roots are repeated. For an equationof order n, if a root of Z (r) = 0, say r = r1, has multiplicity s (that is, it is repeated s times) (where s ≤ n ), then

er1t , ter1t + ...+ tn−1er1t

are corresponding solutions of the differential equation.

If a complex root λ+ i µ is repeated s times, the complexconjugate λ− i µ is also repeated s times. Corresponding to these2s complex-valued solutions, we can find 2s real-valued linearlyindependent solutions:






eλtcos(µt), eλtsin(µt); teλtcos(µt), teλtsin(µt); ...

...; tn−1eλtcos(µt), tn−1eλtsin(µt)

Hence the general solution of the homogeneous equation canalways be expressed as a linear combination of n real-valuedsolutions.

Example 4.3

Find the general solution of the IVP

y (4) + y ′′′ − 7y ′ + 6y = 0; y(0) = 1, y ′(0) = 0, y ′′(0) = −2,

y ′′′(0) = −1Dr. Marco A Roque Sol Ordinary Differential Equations





Solution

The characteristic equation is

r4 + r3 − 7r + 6 = 0The roots of this equation are r1 = 1, r2 = −1, r3 = 2, andr4 = −3. Therefore, the general solution is

y = c1et + c2e

−t + c3e2t + c4e

−3t

and aplying initial conditions we have

c1 + c2 + c3 + c4 = 1

c1 − c2 + 2c3 − 3c4 = 0

c1 + c2 + 4c3 + 9c4 = −2

c1 − c2 + 8c3 − 27c4 = 1Dr. Marco A Roque Sol Ordinary Differential Equations





By solving this system of four linear algebraic equations, we findthat

c1 =11

8, c2 =

5

12, c3 = −

2

3, c4 = −

1

8

Thus the solution of the initial value problem is

y =11

8et +

5

12e−t − 2

3e2t − 1

8e−3t






Example 4.4Find the general solution of

y (4) + y = 0

Solution


r4 + 1 = 0 =⇒ r4 = −1

In this way, we need to find the four roots of -1. Now −1, thoughtof as a complex number, is −1 + 0i . It has magnitude 1 and polarangle π (r = Reθi ). Thus

−1 = cos(π) + i sin(π) = eπi






Moreover, the angle is determined only up to a multiple of 2. Thus

−1 = cos(π + 2mπ) + i sin(π + 2mπ) = e(π+2mπ)i

where m is an integer. Thus

(−1)1/4 = e(π/4+2mπ/4)i = cos(π

4+

2mπ

4

)+ i sin

(π

4+

2mπ

4

)The four fourth roots of −1 are obtained by setting m = 0, 1, 2,and 3;






m = 0 =⇒ cos(π

4

)+ i sin

(π4

)m = 1 =⇒ cos

(π4

+π

2

)+ i sin

(π4

+π

2

)m = 2 =⇒ cos

(π4

+ π)

+ i sin(π

4+ π

)m = 3 =⇒ cos

(π

4+

3π

2

)+ i sin

(π

4+

3π

2

)

m = 4 =⇒ cos(π

4+

4π

2

)+ i sin

(π

4+

4π

2

)= cos

(π4

)+ i sin

(π4

)!!!






and the solutiones are

1 + i√2,−1 + i√

2,−1− i√

2,

1− i√2

The general solution is

y(t) = et√2

(c1cos

(t√2

)+ c2sin

(t√2

))+

e− t√

2

(c3cos

(t√2

)+ c4sin

(t√2

))





The Method of Undetermined Coefficients

We can find a particular solution YP of the nonhomogeneous nth

order linear equation with constant coefficients

L[y ](t) = a0dny

dtn+ a1

dn−1y

dtn−1+ ...+ an−1

dy

dt+ any = g(t)

using The Method of Undetermined Coefficients, provided thatg(t) is of an appropriate form. We have to be careful when theroots of the characteristic polynomial equation have multiplicity,because now this could be greater than 2.

Example 4.5

Find the general solution of

y ′′′ − 3y ′′ + 3y ′ − y = 4et






Solution

The characteristic polynomial is

r3 − 3r2 + 3r − 1 = (r − 1)3

so the general solution of the homogeneous equation is

yc(t) = c1et + c2te

t + c3t2et

To find a particular solution YP(t) of the nonhomogeneousequation we start by assuming that YP(t) = Ae

t .






However, since et , tet , and t2et are all solutions of thehomogeneous equation, we must multiply this initial choice by t3.Thus our final assumption is

YP(t) = At3et =⇒ 6Aet = 4et =⇒ A = 2

3

Thus, the general solution is

yc(t) = c1et + c2te

t + c3t2et +

2

3t3et






Example 4.6

Find a particular solution of

y ′′′ − 4y ′ = t + 3cos(t) + e−2t

Solution


r3 − 4r = 0the roots are r = 0,±2 and the general solution of thehomogeneous equation is

y(t) = c1 + c2e2t + c3e

−2t

.Dr. Marco A Roque Sol Ordinary Differential Equations





To find a particular solution YP(t) of the nonhomogeneousequation we propose that

YP(t) = (A0t + A1)t + A3cost(t) + A4sin(t) + A5e−2tt

The constants are determined by substituting into the differentialequation. They are

A0 = −1/8,A1 = 0,A3 = 0,A4 = −3/5, and A5 = 1/8

.Hence a particular solution is

YP(t) = −1

8t3 − 3

5sin(t) +

1

8te−2t





The Method of Variation of Parameters

The method of variation of parameters for determining a particularsolution of the nonhomogeneous nth order linear differentialequation

L[y ](t) =dny

dtn+ p1(t)

dn−1y

dtn−1+ ...+ pn−1

dy

dt+ pn(t)y = g(t)

Suppose then that we know a fundamental set of solutionsy1, y2, ..., yn of the homogeneous equation. Then the generalsolution of the homogeneous equation is

y(t) = c1y1(t) + c2y2(t) + ...+ cnyn(t)






The Method of Variation of Parameters for determining aparticular solution of the nonhomogeneous equation, YP(t), restson the possibility of determining n functions u1, u2, ..., un such thatYP(t) is of the form

YP(t) = u1(t)y1(t) + u2(t)y2(t) + ...+ un(t)yn(t)

Since we have n functions to determine, we will have to specify nconditions. One of these is clearly that YP the ODE. The othern − 1 conditions are chosen arbitrarily. as to make the calculationsas simple as possible. Following the same idea used in the secondorder case we have that the first derivative of YP is given by






Y ′P(t) =(u1(t)y

′1(t) + u2(t)y

′2(t) + ...+ un(t)y

′n(t)

)+

(u′1(t)y1(t) + u

′2(t)y2(t) + ...+ u

′n(t)yn(t)

)Thus the first condition that we impose is that

u′1(t)y1(t) + u′2(t)y2(t) + ...+ u

′n(t)yn(t) = 0

We continue this process by calculating the successive derivativesY ′′, ...,Y (n−1). After each differentiation we set equal to zero thesum of terms involving derivatives of u1, ..., un. In this way weobtain n − 2 further conditions similar to the above and put themtogether we have






u′1(t)y(m)1 (t) + u

′2(t)y

(m)2 (t) + ...+ u

′n(t)y

(m)n (t) = 0;

m = 1, 2, ..., n − 1

As a result of these conditions, it follows that the expressions for

Y ′P ,Y′′P , ...,Y

(n−1)P reduce to

Y(m)P = u1(t)y

(m)1 (t) + u2(t)y

(m)2 (t) + ...+ un(t)y

(m)n (t) = 0;

m = 1, 2, 3, ..., n − 1

Finally, we need to impose the condition that YP must be a

solution of the nonhomogeneous equation. By differentiating Y(n1)P

from the above equation, we obtain






Y(n)P = (Y

(n−1)P )

′ =(u1(t)y

(n)1 (t) + u2(t)y

(n)2 (t) + ...+ un(t)y

(n)n (t)

)+

(u′1(t)y

(n−1)1 (t) + u

′2(t)y

(n−1)2 (t) + ...+ u

′n(t)y

(n−1)n (t)

)= g(t)

plug into the equation and considering that L[yi ] = 0; i = 1, 2, ..., n,then the ramaining terms yield the relation

u′1(t)y(n−1)1 (t) + u

′2(t)y

(n−1)2 (t) + ...+ u

′n(t)y

(n−1)n (t) = g(t)

Thus, we have a system of n simultaneos linear nonhomogeneusalgebraic equations for u′1, u

′2, ..., u

′n :






u′1(t)y1(t) + u′2(t)y2(t) + ...+ u

′n(t)yn(t) = 0

u′1(t)y′1(t) + u

′2(t)y

′2(t) + ...+ u

′n(t)y

′n(t) = 0

u′1(t)y′′1 (t) + u

′2(t)y

′′2 (t) + ...+ u

′n(t)y

′′n (t) = 0

...

u′1(t)y(n−1)1 (t) + u

′2(t)y

(n−1)2 (t) + ...+ u

′n(t)y

(n−1)n (t) = g(t)






y1 y2 ... yny ′1 y

′2 ... y

′n

y ′′1 y′′2 ... y

′′n

...

y(n−1)1 y

(n−1)2 ... y

(n−1)n

u′1u′2u′3...u′n

=

000...

g(t)

The above system, is a linear algebraic system for the unknownquantities u′1, u

′2, ..., u

′n. The determinant of coefficients is precisely

W (y1, y2, ..., yn), and it is nowhere zero since y1, ..., yn is afundamental set of solutions of the homogeneous equation. Henceit is possible to determine u′1, u

′2, ..., u

′n using Cramer’s Rule :






u′m(t) =g(t)Wm(t)

W (t); m = 1, 2, ..., n

where W (t) = W (y1, y2, ..., yn)(t) ( The Wronskian ) and Wm(t)is the determinant obtained from W by replacing the mth columnby the column (0, 0, ..., 0, 1).

W =

∣∣∣∣∣∣∣∣∣∣∣

y1 y2 ... yny ′1 y

′2 ... y

′n

y ′′1 y′′2 ... y

′′n

...

y(n−1)1 y

(n−1)2 ... y

(n−1)n

∣∣∣∣∣∣∣∣∣∣∣Dr. Marco A Roque Sol Ordinary Differential Equations





Wm =

∣∣∣∣∣∣∣∣∣∣∣∣∣

m − thy1 ... 0 ... yny ′1 ... 0 ... y

′n

y ′′1 .... 0 ... y′′n

...

y(n−1)1 ... 0 ... y

(n−1)n

∣∣∣∣∣∣∣∣∣∣∣∣∣And integrating the above equations we have that the particularsolution is given by

YP(t) =n∑

m=1

ym

∫ tt0

g(s)WmW (s)

ds






where t0 is an arbitrary point. As we can see, things get morecomplicated than in the second order case. In some cases thecalculations may be simplified to some extent by using Abel’sidentity

W (t) = W (y1, y2, ..., yn)(t) = cexp

[−∫

p1(t)dt

]The constant c can be determined by evaluating W at someconvenient point.






Example 4.7Given that y1(t) = e

t , y2(t) = tet , and y3(t) = e

−t are solutionsof the homogeneous equation corresponding to

y ′′′ − y ′′ − y ′ + y = g(t)

determine a particular in terms of an integral.

Solution

Let’s determine first the Wronskian

W = W (et , tet , e−t)(t) =

∣∣∣∣∣∣et tet e−t

et (t + 1)et −e−tet (t + 2)et e−t

∣∣∣∣∣∣Dr. Marco A Roque Sol Ordinary Differential Equations





Factoring et from each of the first two columns pause and e−t

from the third column, we obtain

W = W (et , tet , e−t)(t) = et

∣∣∣∣∣∣1 t 11 (t + 1) −11 (t + 2) 1

∣∣∣∣∣∣Then, by subtracting the first row from the second and third rows,we have

W = W (et , tet , e−t)(t) = et

∣∣∣∣∣∣1 t 10 1 −20 2 0

∣∣∣∣∣∣Dr. Marco A Roque Sol Ordinary Differential Equations





Finally, evaluating this determinant by minors using the firstcolumn, we find that

W (t) = 4et ; W1(t) = −2t − 1; W2(t) = 2; W3(t) = e2t

Thus, the particular solution is given by

YP(t) =3∑

m=1

ym

∫ tt0

g(s)WmW (s)

YP(t) = et

∫ tt0

g(s)(−1− 2s)4es

+ tet∫ tt0

g(s)(2)

4es+ e−t

∫ tt0

g(s)e2s

4es+

YP(t) =1

4

∫ tt0

[et−s (−1 + 2(t − s)) + e−(t−s)

]g(s)ds




Definition of The Laplace TransformSolution of Initial Value ProblemsStep Functions

Definition of The Laplace Transform

Laplace Transform

Among the tools that are very useful for solving linear differentialequations are integral transforms. An integral transform is arelation of the form

F (s) =

∫ βα

K (s, t)f (t)dt

where K (s, t) is a given function, called the kernel of thetransformation, and the limits of integration α and β are alsogiven. It is possible that α = −∞ or β =∞ or both. The relation,introduced above, transforms the function f into another functionF , which is called the transform of f .






There are several integral transforms that are useful in appliedmathematics, but we consider only the Laplace Transform (https://en.wikipedia.org/wiki/Pierre-Simon_Laplace )(... Napoleon asked Laplace where God fit into his mathematicalwork ” Traite de mecanique celeste ”, and Laplace famously replied”Sir, I have no need of that hypothesis ”... ).

Laplace Transform

Let f (t) be given for t ≥ 0. Then the Laplace transform of f ,which we will denote by L {f (t)} = F (s), is defined by theequation

L {f (t)} = F (s) =∫ ∞0

e−st f (t)dt

whenever this improper integral converges.Dr. Marco A Roque Sol Ordinary Differential Equations

https://en.wikipedia.org/wiki/Pierre-Simon_Laplace





The Laplace transform makes use of the kernel K (s, t) = e−st . Inparticular for linear second order differential equations withconstant coeficients is particular useful, since the solutions arebased on the exponential function.

The general idea in using the Laplace transform to solve adifferential equation is as follows:

1. Use the relation L {f (t)} = F (s) to transform an initial valueproblem for an unknown function f in the t-domain (time domain)into an algebraic problem for F in the s-domain (frequencydomain).






2. Solve this algebraic problem to find F .

3. Recover the desired function f from its transform F . This laststep is known as inverting the transform (which in general involvecomplex integration) and denoted by L −1{F (s)}(= limω→∞

12πi

∫ σ+iωσ−iω F (s)e

stds).

OBS The full power of Laplace Transform becomes available onlywhen we regard F (s) as a function of a complex variable. However,for our purposes it will be enough to consider only real values for s.

The Laplace transform F of a function f exists if f satisfies certainconditions:






Theorem 6.1

Suppose that

1. f is piecewise continuous on the interval 0 ≤ t ≤ A for anypositive A.

2. |f (t)| ≤ Keat when t ≥ M. In this inequality, K , a, and M arereal constants, K and M necessarily positive.

Then the Laplace transform L {f (t)} = F (s), defined by

L {f (t)} = F (s) =∫ ∞0

e−st f (t)dt

exists for s > a.






Remember that a function, f (t), is piecewise continuous on theinterval α ≤ t ≤ β if the interval can be partitioned by a finitenumber of points α = t0 < t1 < . . . < tn = β so that

1. f is continuous on each open subinterval ti−1 < t < ti .

2. f approaches a finite limit as the endpoints of each subintervalare approached from within the subinterval.






Example 6.1

Find the Laplace transform for f (t) = 1, t ≥ 0

Solution

L {f (t)} = F (s) =∫ ∞0

e−st f (t)dt

L {1} = F (s) =∫ ∞0

e−stdt = − limA→∞

e−st

s

∣∣∣A0

=1

s; s > 0






Example 6.2

Find the Laplace transform for f (t) = eat , t ≥ 0

Solution

L {f (t)} = F (s) =∫ ∞0

e−st f (t)dt

L {eat} = F (s) =∫ ∞0

eate−stdt =

∫ ∞0

e−(s−a)tdt =

− limA→∞

e−(s−a)t

s − a

∣∣∣A0

=1

s − a; s > a






Example 6.3

Find the Laplace transform for

f (t) =

1 0 ≤ t < 1k t = 10 t > 1

where k is a constant. In engineering contexts f (t) oftenrepresents a unit pulse, perhaps of force or voltage.

Solution

L {f (t)} = F (s) =∫ ∞0

f (t)e−stdt =

∫ 10

e−stdt =

−e−st

s

∣∣∣10

=1− e−s

sDr. Marco A Roque Sol Ordinary Differential Equations





Example 6.4

Find the Laplace transform for f (t) = sin(at), t ≥ 0

Solution

L {f (t)} = F (s) =∫ ∞0

e−st f (t)dt =

∫ ∞0

sin(at)e−stdt = Int by Parts

F (s) = limA→∞

[−e−stcos(at)

a

∣∣∣A0− s

a

∫ A0

e−stcos(at)dt

]= Int by Parts

F (s) =1

a− s

2

a2

∫ ∞0

sin(at)e−stdt =1

a− s

2

a2F (s)






Hence, solving for F(s), we have

F (s) =a

s2 + a2

Now, the Laplace transform is a linear operator, that is, supposethat f1 and f2 are two functions whose Laplace transforms exist fors > a1 and s > a2, respectively. Then, for s > max{a1, a2}

L {c1f1 + c2f2} =∫ ∞0

e−st{c1f1 + c2f2}dt =

c1

∫ ∞0

e−st f1dt + c2

∫ ∞0

e−st f2dt = c1L {f1}+ c2L {f2}

Thus, we have

L {c1f1 + c2f2} = c1L {f1}+ c2L {f2}Dr. Marco A Roque Sol Ordinary Differential Equations





OBS

L −1{d1F1 + d2F2} = d1L −1{F1}+ d2L −1{F2}Example 6.5

Find the Laplace transform for f (t) = 5e−2t − 3sin(4t), t ≥ 0

Solution

L {5e−2t − 3sin(4t)} = 5L {e−2t} − 3L {sin(4t)} =

L {5e−2t − 3sin(4t)} = 5s + 2

− 12s2 + 16

; s > 0

L −1{ 5s + 2

− 12s2 + 16

} = 5L −1{ 1s + 2

} − 3L −1{ 4s2 + 42

} =

5e−2t − 3sin(4t)Dr. Marco A Roque Sol Ordinary Differential Equations




Solution of Initial Value Problems

To see how we can apply the method of Transform Laplace tosolve linear differential equations with constant coefficients. Weestablish the following results.

Theorem 6.2

Suppose that f is continuous and f ′ is piecewise continuous on anyinterval 0 ≤ t ≤ A. Suppose further that there exist constantsK , a, and M such that |f (t)| ≤ Keat for t ≥ M. Then

L {f ′} = sL {f } − f (0)proof

L {f , (t)} =∫ ∞0

e−st f ′(t)dt = limA→∞

∫ A0

e−st f ′(t)dt =






limA→∞

[∫ t10

e−st f ′(t)dt +

∫ t2t1

e−st f ′(t)dt +

∫ t3t2

e−st f ′(t)dt + ...+

∫tn−1

tn = Ae−st f ′(t)dt

]=

and integrating by parts, we have

limA→∞

{e−st f (t)

∣∣∣t10

+ e−st f (t)∣∣∣t2t1

+ ...+ e−st f (t)∣∣∣tn=Atn−1

+

s

[∫ t10

e−st f ′(t)dt +

∫ t2t1

e−st f ′(t)dt + ...+

∫ tn=Atn−1

e−st f ′(t)dt

]}=






limA→∞

[e−sAf (A)− f (0) + s

∫ A0

e−st f (t)dt

]= s

∫ A0

e−st f (t)dt − f (0)

In his way we obtain

L {f ′} = sL {f } − f (0)

As a corollary, we have the following

Corollary 6.2.1

Suppose that the functions f , f ′, ..., f (n−1) are continuous and thatf (n) is piecewise continuous on any interval 0 ≤ t ≤ A.






Suppose further that there exist constants K , a, and M such that|f (t)| ≤ Keat , |f ′(t)| ≤ Keat , ..., |f (n−1)(t)|Keat for t ≥ M. ThenL {f (n)(t)} exists for s > a and is given by

L {f (n)(t)} = snL {f (t)} − sn−1f (0)

−sn−2f ′(0) . . .− sf (n−2)(0)− f (n−1)(0)

We use the previous results to solve IVP’s using LaplaceTransform. It is most useful for problems involvingnonhomogeneous differential equations. However, just to illustratethe method we will start with a homogeneus case






Example 6.6

Consider the IVP

y ′′ − y ′ − 2y = 0; y(0) = 1, y ′(0) = 0Solution

Using the traditional method we find that the general solution is

y(x) = c1e−t + c2e

2t

and applying initial conditions we get c1 = 2/3 and c2 = 1/3.Hence, the particular solution is

y(x) =2

3e−t +

1

3e2t






Now, let us solve the same problem by using the Laplacetransform. We start off with the differential equation

y ′′ − y ′ − 2y = 0Applying the Laplace Transform

L {y ′′ − y ′ − 2y = 0}because of linearity

L {y ′′} −L {y ′} − 2L {y} = 0and using corollary 6.2.1

s2L {y} − sy(0)− y ′(0)− [sL {y} − y(0)]− 2L {y} = 0




Definition of The Laplace T

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Ordinary Di erential Equations - Texas A&M Universityroquesol/Math_308_Fall_2016_Test_2_Print.pdfAlong with this di erential equation we will have the following initial conditions