MATH 4B–Differential Equations, Fall 2016Final Exam Study Guide

GENERAL INFORMATION AND FINAL EXAM RULES

• The exam will have a duration of 3 hours. No extra time will be given. Failing tosubmit your solutions within 3 hours will result in your exam not being graded.

• The Final Exam is comprehensive. The sections are 1.1–1.3, 2.1–2.6, 3.1–3.7, 6.1–6.4,7.1–7.9.

• 35% of the questions will be from Chapters 1 and 2, 35% from Chapters 3 and 6, and30% from Chapter 7.

• You can bring ONE index card of dimensions up to 5′′× 6′′. This index card should behandwritten and can be filled on both sides. However, note cards of higher dimensionsthan the ones mentioned above or typewritten WILL NOT be allowed.

• Calculators WILL NOT be needed, nor allowed for this exam.

• Last but not least, CHEATING WILL NOT BE TOLERATED.

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SKILL’S LIST

• Verify that a given function is a solution for an Initial Value Problem (IVP).

• Sketch the direction (slope) fields of an ODE. Find equilibrium solutions for Au-tonomous ODEs, and determine whether equilibrium solutions are semistable or not.

• Solve a separable ODE and corresponding IVP.

• Find the general solution to a first order linear ODE.

• Determine if an ODE is exact.

• Solve an exact ODE.

• Read information from a word problem, and establish the corresponding ODE modelingthe situation in the following cases:

1. Free falling object.

2. Population growth and decay.

3. Tank model.

4. Newton’s law of cooling.

5. Springs.

• Determine whether two functions y1(t), y2(t) form a fundamental set of solutions fora second order linear ODE.

• Find the general solution for a second order homogeneous ODE with constant coeffi-cients.

• Given a solution y1(t) for a second order linear ODE, use reduction of order to find asecond (independent) solution of the form y2(t) = u(t)y1(t).

• Find a particular solution for a second order ODE with constant coefficients using themethod of undetermined coefficients, and/or variation of parameters.

• Find the general solution to a second order ODE with constant coefficients.

• Solve Initial Value Problems associated to a second order ODE with constant coeffi-cients.

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• Compute the Laplace transform of a given function.

• Compute inverse Laplace transforms to rational functions and piecewise continuousfunctions.

• Use Laplace transforms to solve second order IVP.

• Compute the determinant of a matrix (2× 2 or 3× 3 would suffice).

• Find the eigenvalues and eigenvectors of a matrix.

• Find the inverse of a matrix.

• Find the canonical Jordan form of a 2× 2 matrix.

• Find the general solution of a homogeneous system of linear ODEs with constantcoefficients.

• Find the fundamental matrix of a system of linear ODEs with constant coefficients.

• Find the solution to IVPs of the form:

x′ = Ax, x(t0) = x0.

• Find the general solution to a 2 × 2 system of linear ODEs with constant matrix A.Must be familiar with:

1. Jordan form method,

2. Undetermined coefficients,

3. Variation of parameters, and

4. Laplace transform.

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SUMMARY

• An Ordinary Differential Equation (ODE) is an equation whose variable is a differen-tiable function and that involves the function and its derivatives:

F (x, y, y′, y′′, . . . , y(n), . . . ) = 0.

• The order of an ODE is the order of the highest derivative that appears in the equation.For instance, y′ + y = x is first order while x3y′′ − 10x7y′ = y4 is second order.

• A first order ODE of the formy′ = f(y)

is called autonomous. Notice that autonomous equations are always separable, how-ever the integration involved may be very complicated (if not impossible). One canin any case study solutions by looking at the direction (slope) field for the differentialequation.

• A constant solution to an Autonomous ODE is called an equilibrium solution. Theseare the zeroes of the function f(y). An equilibrium solution is called semistable ifas t approaches infinity solutions on one side of the equilibrium solution approach theequilibrium solution while solutions on the other side become further and further awayfrom the equilibrium solution; it is called asymptotically stable if for large valuesof t solutions always approach the equilibrium solution, in population dynamics thiswould say that no matter the initial size of a population it will (after a long time)stabilize to the equilibrium size; it is called unstable if for large values of t everysolution stays away from the equilibrium.

• There are special kinds of ODEs that we know how to solve, these are:

1. Separable: First order ODEs of the form

y′ = F (t)G(y).

These can be solved by dividing both sides by G(y) and then integrating withrespect to t: ∫

y′(t)

G(y(t))dt =

∫F (t) dt,

the left hand side is actually the integral with respect to y since if y = y(t) thendy = y′(t)dt, and so the equation becomes∫

1

G(y)dy =

∫F (t) dt.

4

2. Linear: First order ODEs of the form

y′ + a(t)y = b(t). (1)

These can be solved by multiplying by the integrating factor

µ(t) = e∫a(t) dt

Notice that µ′(t) = a(t)µ(t) and so multiplying (1) by µ(t) we obtain

µ(t)(y′ + a(t)y) = µ(t)b(t)

µ(t)y′ + µ′(t)y = µ(t)b(t)

d

dt(µ(t)y) = µ(t)b(t).

Thus,

y =1

µ(t)

(∫µ(t)b(t) dt

)3. Exact: First order ODEs

M(x, y) +N(x, y)y′ = 0 (2)

that can be written asd

dx(Ψ(x, y(x))) = 0 (3)

for some differentiable function Ψ(x, y). If it happens that M,N, ∂M∂y, ∂N∂x

are allcontinuous in some rectangle then such function Ψ exists if and only if

∂M

∂y=∂N

∂x.

If an equation is exact, then the solutions are given by

Ψ(x, y) = c

where c is a constant. To find Ψ(x, y) we notice that for (2) to be equal to (3) werequire

∂Ψ

∂x= M, and

∂Ψ

∂y= N.

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Therefore

Ψ(x, y) =

∫M(x, y) dx+ C(y)

for some function C(y), depending only on y, and that we determine from the

condition∂Ψ

∂y= N .

• Some ODEs may not fall directly into any of the groups above, but after a modificationor substitution they can be solved by the same methods.

1. Bernoulli equations: ODEs of the from

y′ + a(t)y = b(t)yn.

These equations can be transform into linear equations by using the substitutionu = y1−n.

2. Integrating factors: Assume that

M(x, y) +N(x, y)y′ = 0 (4)

satisfies one of the following conditions:

– The quotient∂M∂y− ∂N

∂x

N= P (x)

is a function depending only on the variable x.

– The quotient∂N∂x− ∂M

∂y

M= Q(y)

is a function depending only on the variable y.

Then (4) can be transformed into an exact equation by multiplying by µ(x) =exp(

∫P (x) dx) in the first case, or by µ(y) = exp(

∫Q(y) dy) in the second case.

• ODEs are used to model real-life situations, some common applications are:

1. Free falling objects: Let v(t) represent the velocity of an object at the time t.If the object falls freely (assuming there is no air resistance) then

mdv

dt= m(9.8) or

dv

dt= 9.8,

6

where 9.8m/s2 is the gravity constant, t is measured in seconds (s), and v(t) inmeters per second (m/s). If there is air resistance proportional to the velocity,say γv(t) for some constant γ, then we obtain the ODE

dv

dt= 9.8− γ

mv,

which is separable. If instead, the object is thrown upwards then the gravity willact as a force of resistance, and under the assumption of an air resistance of theform γv(t), the associated ODE is

dv

dt= −9.8− γ

mv.

2. Exponential growth and decay: P (t) represents the size of some populationat the time t. If we assume that the rate of growth of P (t) is proportional to P (t)then we obtain the separable ODE

dP

dt= kP.

3. Predator/ Prey systems: Again P (t) represents the size of some populationat the time t and the rate of growth of P (t) is proportional to P (t). We alsoassume that due to the presence of some predator there are d deaths per unit oftime (here d is a constant). Then we obtain the ODE

dP

dt= kP − d,

which is again separable.

4. Newton’s law of cooling: Suppose that an object is initially in a room withtemperature T0, and after a while is taken to the outside where the ambienttemperature is Ta. Newton’s Law of Cooling says that the rate of change of thetemperature of the object as the time passes is proportional to the differencebetween the ambient temperature and the temperature of the object, as an IVPthis can be written as:

dT

dt= −k(T − Ta), T (0) = T0,

for some constant k > 0. This ODE is separable.

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5. Tank model: Suppose that a tank contains V0 gallons of brine with y0 poundsof salt in it. Suppose that brine with a concentration of sin pounds of salt/gallonis poured into the tank at a rate of νin gallons/minute, and that the well mixedliquid comes out of the tank at a rate of νout gallons/minute. If y(t) representsthe amount of salt (in pounds) in the tank at the time t (in minutes), then

dy

dt= (rate in)− (rate out)

Since rate in (out)=number of pounds of salt poured in (coming out) per minute,then assuming that the mixture is homogenous we obtain

rate in = sin × νin

rate out =y(t)

V (t)× νout.

Notice that the volume of liquid in the tank at the time t is given by

V (t) = V0 + νint− νoutt.

To determine the amount of salt in the tank at the time t we have to solve theIVP

dy

dt= sin × νin −

y

V (t)× νout, y(0) = y0.

This is a linear ODE.

• A second order linear ODE is a differential equation of the form

a(t)y′′ + b(t)y′ + c(t)y = d(t) (*)

• The associated homogeneous equation is

a(t)y′′ + b(t)y′ + c(t)y = 0 (?)

• Two solutions y1(t) and y2(t) for (?) are said to form a fundamental set of solutions ifthey are linearly independent, i.e., if the Wronskian

W (y1, y2)(t) =

∣∣∣∣y1(t) y2(t)y′1(t) y′2(t)

∣∣∣∣ = y1(t)y′2(t)− y′1(t)y2(t) 6= 0.

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• If y1 and y2 form a fundamental set of solutions for (?), then every other solution is ofthe form

YH(t) = c1y1(t) + c2y2(t)

for some constants c1 and c2. We refer to YH(t) as the general solution to the homo-geneous equation (?).

• Suppose that you know a particular solution YP (t) for (*), then if y(t) is any othersolution one can easily show that y(t)− YP (t) is a solution for (?) and so

y(t) = YH(t) + YP (t)

Thus, the general solution for (*) consists of the general solution for the associatedhomogeneous plus a particular solution.

• When the coefficients of (?) are constant then it is easy to find a fundamental set ofsolutions. Consider the equation

ay′′ + by′ + cy = 0. (z)

When looking for solutions of the form y = ert one finds out that r must satisfy thecharacteristic equation

ar2 + br + c = 0.

Of course, the solutions to the characteristic equation vary depending on the values ofthe constants a, b, c and we have the following cases:

1. Two real solutions: This happens when b2 − 4ac > 0, in which case the solutionsare

r1 =−b+

√b2 − 4ac

2a, and r2 =

−b−√b2 − 4ac

2a.

In this case the solutions y1 = er1t and y2 = er2t form a fundamental set ofsolutions and therefore the general solution for (z) is

y(t) = c1er1t + c2e

r2t

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2. Two complex solutions: This happens when b2 − 4ac < 0, in which case thesolutions are

r1 =−b2a

+ i

√4ac− b2

2a, and r2 =

−b2a− i√

4ac− b22a

.

In this case the solutions y1 = er1t and y2 = er2t are complex, which may ormay not be appropriate depending on what the solutions of the equation shouldrepresent. Fortunately, one is able to prove that if y(t) = A(t)+iB(t) is a solutionfor (z) then so are the real part A(t) and the imaginary part B(t). Using themore convenient notation

r1 = λ+ iω, where λ =−b2a

and ω =

√4ac− b2

2a,

one obtains the solution

y(t) = eλt+iωt = eλteiωt = eλt(cosωt+ i sinωt).

The real and imaginary parts of y(t) are the solutions

y1(t) = eλt cosωt and y2(t) = eλt sinωt.

After computing W (y1, y1) one sees that y1 and y2 form a fundamental set ofsolutions for (z) and therefore the general solution is

y(t) = c1eλt cosωt+ c2e

λt sinωt

3. Repeated roots: This happens when b2 = 4ac, in which case we obtain only onesolution to the characteristic equation

r = λ =−b2a.

In this case one finds out (using the technique of reduction of order) that afundamental set of solutions is given by y1(t) = eλt and y2(t) = teλt and so thegeneral solution for (z) is

y(t) = c1eλt + c2te

λt

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• We are now interested in finding particular solutions to second order ODEs whoseassociated homogeneous equation has constant coefficients, i.e., equations of the form

ay′′ + by′ + cy = d(t). (zz)

1. Undetermined Coefficients: The idea is to look for a particular solution YP (t)that looks like d(t). We follow the following table:

d(t) YP (t)Pn(t) tsQn(t)Pn(t)e

αt tsQn(t)eαt

Pn(t)eαt sinβt tseαt [Qn(t) cosβt+Rn(t) sinβt]

Pn(t)eαt cosβt tseαt [Qn(t) cosβt+Rn(t) sinβt]

where

Pn(t) = Antn + · · ·+ A1t+ A0

Qn(t) = Bntn + · · ·+B1t+B0

Rn(t) = Cntn + · · ·+ C1t+ C0.

2. Variation of Parameters: If it happens that the function d(t) is nothing likethe functions in the table above, then we can try to look for a particular solutionof the form

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

After one extra assumption one is able to deduce formulas for u1 and u2 in termsof y1, y2, and d. These formulas are:

u1(t) = −∫

y2(t)d(t)

aW (y1, y2)(t)dt and u2(t) =

∫y1(t)d(t)

aW (y1, y2)(t)dt

NOTE: These integrals may be very difficult to compute.

• Suppose that a spring of length ` in vertical position is stretched to a length ` + L,i.e., L units beyond its natural length, by a mass m attached to its end. Then Hooke’sLaw states that there is a constant k (independent of L and m) such that

mg = kL.

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Suppose that the spring is stretched further a distance u(t), then one obtains the secondorder ODE:

mu′′(t) =∑

forces acting on m

= (weight) + (force due to the spring) + (damping force) + (total external force)

= mg − k(L+ u(t))− γu′(t) + F (t)

= (mg − kL)− ku(t)− γu′(t) + F (t)

= −ku(t)− γu′(t) + F (t).

Here we are using the fact that mg− kL = 0, and we are assuming that for small u(t)the damping force (due to air resistance or viscosity) is proportional to the velocityu′(t). The constant γ is called damping constant. Then we obtain a second orderODE with constant coefficients, which will have unique solution once we specify initialposition and velocity, i.e., we have the following IVP:

mu′′(t) + γu′(t) + ku(t) = F (t), u(0) = u0, u′(0) = u′0

• Recall that if a function f(t) is continuous in every interval of the form [0, b] for b > 0,then one can define the infinite integral∫ ∞

0

f(t) dt := limb→∞

∫ b

0

f(t) dt.

If such limit exists (it is a number) then we say that the integral converges, otherwise(the limit is ±∞ or does not exist) we say that the integral diverges.

• Assume that a function f(t) is continuous in every interval of the form [0, b] for b > 0,and assume further that there are constants K, M,α such that

|f(t)| ≤ Keαt, for all t ≥M.

Then the integral

L{f(t)} := F (s) =

∫ ∞0

e−stf(t) dt

converges for all s > α. We refer to the function L{f(t)} as the Laplace Transformof f(t).

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• It follows from the additive properties of the integral that the Laplace transform islinear, i.e.,

L{af(t) + bg(t)} = aL{f(t)}+ bL{g(t)},

for any constants a, and b.

• If L{f(t)} = F (s) then we write L−1{F (s)} = f(t). L−1 is also linear and is calledthe Inverse Laplace Transform.

• For a real constant c we define the step function

uc(t) =

{1 if t ≥ c

0 otherwise.

• The following list of common Laplace transforms will be provided to you in the exam:

13

f(t) = L−1{F (s)} F (s) = L{f(t)}

11

s, s > 0

eat1

s− a, s > a

tn, n positive integern!

sn+1, s > 0

cos ats

s2 + a2, s > 0

sin ata

s2 + a2, s > 0

cosh ats

s2 − a2, s > |a|

sinh ata

s2 − a2, s > |a|

eat cos bts− a

(s− a)2 + b2, s > a

eat sin btb

(s− a)2 + b2, s > a

tneat, n positive integern!

(s− a)n+1, s > a

uc(t)e−cs

s, s > 0

uc(t)f(t− c) e−csF (s)

ectf(t) F (s− c)

14

• Laplace transforms provide a valuable tool when solving IVP with non homogeneouslinear ODEs involving piecewise defined functions. For second order linear ODEs theadvantage becomes evident from the relations

L{y′(t)} = sL{y(t)} − y(0) L{y′′(t)} = s2L{y(t)} − sy(0)− y′(0)

• The algorithm to solve an IVP using Laplace transform is very simple:

1. Apply Laplace transform to both sides of the equation.

2. Use the relations for L{y′′} and L{y′} boxed above.

3. Use the given initial conditions y(0) = y0 and y′(0) = y′0.

4. Solve for L{y}.5. Apply L−1 to obtain y. This step is critical since computing L−1 may be challeng-

ing, the idea is to rewrite the function of s as a linear combination of functionsappearing in the big table of Laplace transforms.

• A system of first order linear ODEs is an equation of the form:

x′(t) = P(t)x(t) + g(t),

where

x(t) =

x1(t)x2(t)

...xn(t)

, P(t) =

p1,1(t) p1,2(t) · · · p1,n(t)p2,1(t) p2,2(t) · · · p2,n(t)

......

. . ....

pn,1(t) pn,2(t) · · · pn,n(t)

, g(t) =

g1(t)g2(t)

...gn(t)

.

• Though the theory we will describe in what follows applies to matrices of any size, wewill focus on the 2 × 2 and 3 × 3 cases. We will also assume that the matrix P(t) isconstant and we will denote it by A.

• A matrix A has an inverse if and only if det(A) 6= 0.

• Assume that the matrix A =

(a bc d

)is invertible. Then its inverse if given by

A−1 =1

ad− bc

(d −b−c a

)

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• Recall that the (column) vectors X(1), X(2), . . . , X(n) are linearly independent if andonly if

det

| | |X(1) X(2) · · · X(n)

| | |

6= 0.

• If X(1)(t), X(2)(t), . . . , X(n)(t) are linearly independent solutions for the homogeneoussystem

x′ = Ax,

then the general solution is of the form

x(t) = c1X(1)(t) + c2X

(2)(t) + · · ·+ cnX(n)(t)

In this case the matrix

Ψ(t) =

| | |X(1) X(2) · · · X(n)

| | |

is called a fundamental matrix for the system. Thus the general solution can bewritten as

x(t) = Ψ(t)c, where c =

c1c2...cn

• To solve the IVP

x′ = Ax, x(t0) = x0,

we look for the the vector c such that

x(t0) = Ψ(t0)c = x0.

Since the matrix Ψ(t) is invertible then

c = (Ψ(t0))−1x0

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• The eigenvalues of a square matrix are the values λ satisfying the equation:

det(A− λI) = 0

• If det(A − λI) = 0 then the system of equations corresponding to the matrix A − λIadmits more than one solution and so it is possible to find a nonzero vector ξ such that

(A− λI)ξ = 0

Such vector is called an eigenvector associated to the eigenvalue λ.

• If A is square matrix whose entries are complex numbers, then A can be written as

A = TJT−1

for some invertible matrix T. The matrix J is upper triangular and carries all therelevant information from A. J is called the canonical Jordan from of A. In the casewhen A is size 2× 2, we have the following possibilities:

1. A has two different eigenvalues λ1 and λ2. In this case

J =

(λ1 00 λ2

), and T =

| |T (1) T (2)

| |

,

where T (1) is an eigenvector for the eigenvalue λ1, and T (2) is an eigenvector forthe eigenvalue λ2.

2. A has only one eigenvalue λ, but it has two linearly independent eigenvectors T (1)

and T (2). In this case

J =

(λ 00 λ

), and T =

| |T (1) T (2)

| |

.

3. A has only one eigenvalue λ and one eigenvector T (1) (up to multiples). In thiscase

J =

(λ 10 λ

), and T =

| |T (1) T (2)

| |

,

where the vector T (2) is solution to

(A− λI)T (2) = T (1).

17

• To solve an homogeneous system of linear ODEs with constant coefficients

x = Ax

we follow the following steps:

1. Compute the canonical Jordan form of the matrix A.

2. Use the substitution x = Ty. In this case we get x′ = Ty′ and therefore weobtain the new system

Ty′ = ATy.

3. Multiplying both sides by T−1 the system becomes

y′ = Jy.

Reading out the two equations we obtain two very simple first order linear ODEs,solving first the equation for y2(t) and then the equation for y1(t) we obtain y.

4. We find x from our substitution x = Ty.

• The output of the algorithm above will depend on the shape of J.

1. If J =

(λ1 00 λ2

)and T =

(T (1) T (2)

)then

x(t) = c1T(1)eλ1t + c2T

(2)eλ2t

The same answer if there is only one eigenvalue (i.e., λ = λ1 = λ2), but twolinearly independent eigenvectors T (1) and T (2).

2. If J =

(λ 10 λ

), T (1) is an eigenvector and T (2) satisfies

(A− λI)T (2) = T (1),

then the solution is

x(t) = c1T(1)eλt + c2

(T (1)teλt + T (2)eλt

)• It could happen that the entries of a matrix A are all real numbers, but A has complex

eigenvalues. In this case, we can take the real and imaginary parts of the generalsolution to obtain a fundamental set of real solutions. We show this with an example:

18

Example 1. Find the general solution to the system

x′ =

(0 1−1 0

)x.

First, we need to find the eigenvalues:

det

(−λ 1−1 −λ

)= λ2 + 1 = 0 ⇒ λ = ±i.

Eigenvector for λ1 = i:((−i 1−1 −i

)(ξ1ξ2

)=

(00

))⇐⇒

(−i 1−1 −i

)−→

(−i 10 0

),

Thus, −iξ1 + ξ2 = 0 and therefore the solutions are of the form(ξ1ξ2

)=

(ξ1iξ1

)= ξ1

(1i

).

We choose the eigenvector T (1) =

(1i

). Similarly for λ2 = −i we obtain the eigenvector

T (2) =

(1−i

).

The Jordan form is J =

(i 00 −i

), and T =

(1 1i −i

).

The general solution is

x(t) = c1

(1i

)eit + c2

(1−i

)e−it

= c1

(1i

)(cos t+ i sin t) + c2

(1−i

)(cos t− i sin t)

=

(c1 cos t+ ic1 sin t+ c2 cos t− ic2 sin t−c1 sin t+ ic1 cos t− c2 sin t− ic2 cos t

)=

(c1 cos t+ c2 cos t−c1 sin t− c2 sin t

)+ i

(c1 sin t− c2 sin tc1 cos t− c2 cos t

)= (c1 + c2)

(cos t− sin t

)+ i(c1 − c2)

(sin tcos t

).

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Thus, the general solution can be written as

x(t) = A

(cos t− sin t

)+B

(sin tcos t

)for constants A and B.

• As in the case of second order ODEs, the general solution to the system of linear ODEs

x′(t) = Ax(t) + g(t)

can be written as x(t) = xH(t) + xP (t), where xH is the general solution to theassociated homogeneous system and xP is a particular solution.

• To solve a non-homogeneous system of linear ODEs of the form

x′(t) = Ax(t) + g(t)

we follow the following steps:

1. Compute the canonical Jordan form of the matrix A (this includes J, T, andT−1).

2. Use the substitution x = Ty. In this case we get x′ = Ty′ and therefore weobtain the new system

Ty′ = ATy + g(t).

3. Multiplying both sides by T−1 the system becomes

y′ = Jy + T−1g(t).

Reading out the two equations we obtain two first order linear ODEs, solving firstthe equation for y2(t) and then the equation for y1(t) we obtain y.

4. We find x from our substitution x = Ty.

• Even though for 2 × 2 systems you can just solve the equation by finding the Jordanform of the matrix A and using the substitution x = Ty as explained above, forsystems of more variables it is sometimes quicker to find a particular solution by usingeither undetermined coefficients or variation of parameters:

20

1. Undetermined Coefficients: The idea is to look for a particular solution xP (t)that looks like g(t). We follow the table:

g(t) xP (t)Pn(t) tsQn(t)Pn(t)e

αt tsQn(t)eαt

Pn(t)eαt sinβt tseαt [Qn(t) cosβt+Rn(t) sinβt]

Pn(t)eαt cosβt tseαt [Qn(t) cosβt+Rn(t) sinβt]

where

Pn(t) = antn + · · ·+ a1t+ a0

Qn(t) = bntn + · · ·+ b1t+ b0

Rn(t) = cntn + · · ·+ c1t+ c0.

NOTE: Here an, . . . , a0,bn, . . . ,b0, cn, . . . , c0 are constant vectors.

2. Variation of Parameters: We know that the solution to the associated homo-geneous system x′ = Ax can be written as

xH(t) = Ψ(t)c.

The idea is to look for a particular solution of the form

xP (t) = Ψ(t)u(t)

for some vector

u(t) =

u1(t)u2(t)

...un(t)

.

Computing derivatives we obtain

x′P (t) = Ψ′(t)u(t) + Ψ(t)u′(t).

Replacing into the equation we get

x′P (t) = Ψ′(t)u(t) + Ψ(t)u′(t) = AΨ(t)u(t) + g(t).

The second equality can be rewritten as

(Ψ′(t)u(t)−AΨ(t)u(t))︸ ︷︷ ︸=0

+Ψ(t)u′(t) = g(t)

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The first term vanishes because Ψ′(t)c = AΨ(t)c for every vector c, includingu(t). Therefore we must have

Ψ(t)u′(t) = g(t), or u′(t) = (Ψ(t))−1g(t).

Thus

u(t) =

∫ t

t0

(Ψ(t))−1g(t) dt

As for the case of second order ODEs with constant coefficients, these integralsmay be very hard to compute.

• As in the case of second order ODEs with constant coefficients, it is possible to useLaplace transform to solve IVP of the form

x′(t) = Ax(t) + g(t), x(0) = x0.

Define

L(x(t)) = L

x1(t)x2(t)

...xn(t)

=

L{x1(t)}L{x2(t)}

...L{xn(t)}

=

X1(s)X2(s)

...Xn(s)

= X(s).

Then it follows that

L{x′(t)} = sL{x(t)} − x(0) = sX(s)− x0

Applying Laplace Transform to the system we obtain

sX(s)− x0 = AX(s) + G(s),

where G(s) = L{g(t)}. Solving for X(s) the equation becomes

X(s) = (sI−A)−1(G(s) + x0).

Therefore, the solution is given by

x(t) = L−1{

(sI−A)−1(G(s) + x0)}

To compute the Laplace inverse one needs to compute (sI − A)−1(G(s) + x0) as amatrix of s and as usual apply partial fractions to write each entry of the matrix ascombinations of the functions appearing in the table for Laplace transforms.

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