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Course: MATH 366, Spring 2009
School: Purdue
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36600 MA LECTURE NOTES: FRIDAY, APRIL 3 3 Then there exists a unique solution on the interval I . This is known as the Existence and Uniqueness Theorem for Linear Systems. The statement for linear systems follows from that for nonlinear systems because if we denote n Gi Gi (t, x1 , x2 , . . . , xn ) = pij (t) xj + gi (t) = = pij (t). xj j =1 Note the similarity here with the Existence and Uniqueness Theorem...

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36600 MA LECTURE NOTES: FRIDAY, APRIL 3 3 Then there exists a unique solution on the interval I . This is known as the Existence and Uniqueness Theorem for Linear Systems. The statement for linear systems follows from that for nonlinear systems because if we denote n Gi Gi (t, x1 , x2 , . . . , xn ) = pij (t) xj + gi (t) = = pij (t). xj j =1 Note the similarity here with the Existence and Uniqueness Theorem for second order linear equations presented in Lecture #14 on Wednesday, October 1. Applications Spring-Mass System: One Mass, One Spring. We discuss several examples of oscillations with springs, each increasing in complexity. For simplicity, we assume that (a) the original length of the spring is negligible to the motion of the mass, so that Hookes Law states that the restoring force is Fs = k u in terms a of displacement u; (b) all motion takes place on a frictionless track so that we can ignore any frictional forces; and (c) that the motion is not aected of the force of gravity. First, say that we have a mass m attached to a spring with constant k . Assume that the mass is under the inuence of an external force F (t). Denote u = u(t) as the position of the mass at time t. Equilibrium is seen to be at u = 0, so this also denotes the displacement from equilibrium of the mass at time t. Hookes Law states that the restoring force of the spring on the mass is Fs = k u. Newtons Second Law of Motion states that F = m a is the sum of the forces on the mass: d2 u m 2 = k u + F (t). dt This can be expressed as a system of rst order dierential equations. Indeed, if we denote x1 = u x = x2 1 = x2 = u m x = k x1 + F (t) 2
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Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, APRIL 3Systems of First Order Linear EquationsRecap. Recall that a rst order equation is an ordinary dierential equation in the formy = G(t, y ),y (t0 ) = y0 .In general, an nth order equation is an ordinary dierential
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, APRIL 6Circuits. Consider a circuit that contains three devices:i. an inductor (which behaves like a mass),ii. a resistor (which behaves like friction), andiii. a capacitor (which behaves like a spring).Such a circui
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, APRIL 6ApplicationsSpring-Mass System: One Mass, Two Springs. Now say that we have a mass m attached to twosprings with constants k1 and k2 , respectively. We assume that these springs are attached to the oppositeends o
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, APRIL 10Systems of Linear EquationsLinear Independence. Consider a collection of m-dimensional vectors:a1k a2k x(k) = . ...amkWe say that the set x(1) , x(2) , . . . , x(n) is a linearly independent set if the on
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, APRIL 103Characteristic Polynomials. We give a simple way to compute eigenvalues of an n n matrix A. Considerthe equationAx = x=( I A) x = 0.Since this holds for a nonzero vector x, we see that the matrix I A must be
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, APRIL 10Systems of Linear EquationsSystems of Linear Dierential Equations. We return to the linear systemx1x2xm==...p11 (t) x1p21 (t) x1++= pm1 (t) x1p12 (t) x2p22 (t) x2+ + ...+ pm2 (t) x2+ Usin
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, APRIL 13Then the general solution to the nonhomogeneous equation is in the formx(t) = c1 x1 (t) + c2 x2 (t) + X (t)where c1 and c2 are constants. In terms of matrices, we see that the general solution is in the formx(
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, APRIL 133It suces then to show that x(c) = c for some constant c. Consider an initial value problem in the formd (c)x = P(t) x(c) ,x(c) (t0 ) = x0 .dtAccording to the Existence and Uniqueness Theorem, we know that a
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: MONDAY, APRIL 13When yk = xk (k) we use the fact that x =ipndWd=( )xk (k)dtdtjpij (t) xjp to ndk=1=ni=1=nni=1 j =1nndxi (i) ( )xk (k) =pij (t) ( ) xj (i) xk (k) dti=1 j =1k=ipij (t) W x(1) , . . .
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, APRIL 13Basic Theory of Systems of First Order Linear EquationsSolutions to Systems of Linear Dierential Equations. We continue to focus on a system of rstorder dierential equations in the formx1x2xm==...p11
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 15Hence the denitions are consistent. As for Abels Theorem, we have the trace tP1 (t)tr P(t) = =W (t) = C exp trP( ) d = C exp P0 (t)tP1 ( )d .P0 ( )Homogeneous Systems. Again, consider the homogeneo
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 153The characteristic polynomial isbca r2 + b r + cr1pA (r) = det=r r+ (1) =.c/a r + b/aaaaHence an eigenvalue r of A is a root of the characteristic equation a r2 + b r + c = 0.More generally, con
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 15Basic Theory of Systems of First Order Linear EquationsExample. We explain how the Wronskian dened in the previous lecture is related to the Wronskian wedened during Lecture #24 on Friday, March 13. Consider t
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, APRIL 20Fundamental MatricesRecap. Consider the initial value problemdx = P(t) x,x(t0 ) = x0 .dtAssuming that P(t) = pij (t) is an n n matrix, say that we can nd n functions x(1) , x(2) , . . . , x(n) suchthati.
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, APRIL 203To see why this suces, rst recall the Product Rule:ddd0x=x = x0 = [A ] x0 = A x.dtdtdtAs for the initial condition:x(0) = (0) x0 = I x0 = x0 .We now show that (t) satises the initial value problem ab
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, APRIL 20Homogeneous Systems with Constant CoefficientsExample #2. We have already seen that the general solution to the homogeneous system1d21x=x1 1dt2is the functioncos t t/2sin t t/2e+ c2e. sin tcos t
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 22Now we return to the general case. Consider an n n matrix A. Say that there exists a nonsingular n nmatrix T such that T1 A T = D is an n n diagonal matrix. Such a matrix A is said to be diagonalizable.We wi
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 22Fundamental MatricesDiagonalizable Matrices. We explain how to exponentiate an arbitrary matrix. First, we consider aspecial case. We say that an n n matrix is a diagonal matrix if it is in the formr1 0 0 0
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, APRIL 23Example. Again, consider the 2 2 matrix1A=411=tr A = 2det A = 3=disc A = 16pA (r) = r2 2 r 3 = (r 3) (r + 1) .Hence the eigenvalues are r1 = 3 and r2 = 1, with corresponding eigenvectors a121a121
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, APRIL 233It is unclear whether we can nd a matrix T such that T1 A T is a diagonal matrix because we only haveone eigenvector. Consider instead the matrix1 1 010101T==T==.1 1111 11Consider the product10
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, APRIL 23Diagonalizing 2 2 MatricesDistinct Eigenvalues. We explain how we chose thewe discuss a general theory which holds especially wellaA = 11a21matrices D and T in the previous example. Indeed,for 2 2 matrices.
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, APRIL 27Matrix Exponentiation Revisited. We saw in the previous lecture thatrtr1 0e10D==exp (D t) =.0 r20er2 tWe show that1 c t rt=exp (J t) =e.01To see why, recall the denition exp (J t) = k Jk tk /k !
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, APRIL 27Repeated EigenvaluesJordan Canonical Form. We explain the general theory for when we can diagonalize a 2 2 matrix A.Say that we have eigenvalues r1 and r2 .i. If r1 = r2 i.e., we have distinct eigenvalues, then
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 29in terms of the fundamental matrixrte10(t) = exp (D t) = ...00..er2 t...000...ern tThe desired solution is x(t) = T y(t)..Example. Consider the systemdx = A x + g(t)dtin terms of2
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 29 By the Product Rule for Matricesd 2dQ dQQ =Q+ Q = QP + PQ = 2QPdtdtdt=3d kQ = k Qk1 P ;dtso that we have the formulad1 d k 1exp Q(t) =Q=k Qk1 P = exp Q(t) P(t).dtk ! dtk!k=0k=0We attemp
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 29We substitute this into the original equation:d (p)x = P(t) x(p) + g(t)dtddu+ u = P(t) u + g(t)dtdtddd(t) u g(t) = P(t) u = 0=(t) u = g(t).dtdtdtUpon inverting and integrating, we nd tha
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 29Nonhomogeneous SystemsConstant Coecients. Consider now the nonhomogeneous systemdx = A x + g(t)dtwhere A is a constant n n matrix. We will show that the general solution to this system is in the formx(t) =
Purdue - MATH - 366
2MA 36600 MIDTERM #1 REVIEW1.3: Classication of Dierential Equations. There are two types of dierential equations: ODEs or ordinary dierential equations are equations which do not involve partial derivatives. PDEs or partial dierential equations are
Purdue - MATH - 366
MA 36600 MIDTERM #1 REVIEW32.3: Modeling with First Order Equations. Mixing Problems: Say that we have a tank containing a volume V gal of a certain liquid. Aconcentration c lbs/gal of a certain substance ows in at a constant rate r gal/min, and a fau
Purdue - MATH - 366
4MA 36600 MIDTERM #1 REVIEWis an equilibrium solution. In general, an equilibrium solution is a constant which is a solution, so itsatises the equationf (yL ) = 0.Any constant yL such that f (yL ) = 0 is called a critical point for the function f (y
Purdue - MATH - 366
MA 36600 MIDTERM #1 REVIEW5#2. Choose a function h(y ) according to the ordinary dierential equation ydhgg= N (x, y ) =h(y ) =N (x, ) (x, ) d .dyyy#3. Choose the function f (x, y ) = g (x, y ) + h(y ). Recall that the solution is f (x, y )
Purdue - MATH - 366
6MA 36600 MIDTERM #1 REVIEW2.9: First Order Dierence Equations. A recursive relation the formyn+1 = (n, yn )is called a rst order dierence equation. Using Eulers Method, the dierential equation y = G(t, y )has the associated dierence equation yn+1 =
Purdue - MATH - 366
MA 36600 MIDTERM #1 REVIEWChapter 11.1: Some Basic Mathematical Models; Direction Fields. Newtons Second Law of Motion is the statement F = m a; it really meansd2 x= the sum of the forces on the object.dt2 Newtons Law of Gravitational Attraction is
Purdue - MATH - 366
2MA 36600 MIDTERM #2 REVIEW Say that y1 = y1 (t) and y2 = y2 (t) are solutions to the homogeneous equationa(t) y + b(t) y + c(t) y = 0.The Principle of Superposition states that the linear combination y (t) = c1 y1 (t) + c2 y2 (t) is also asolution f
Purdue - MATH - 366
MA 36600 MIDTERM #2 REVIEW3A fundamental set of solutions to the dierential equation is cfw_y1 , y2 in terms of the functionset cosh t if b2 4 a c &gt; 0,et sinh t if b2 4 a c &gt; 0,y1 (t) =y2 (t) =et cos tif b2 4 a c &lt; 0.et sin tif b2 4 a c &lt; 0.3.
Purdue - MATH - 366
4MA 36600 MIDTERM #2 REVIEW3.6: Variation of Parameters. To nd the general solution of the nonhomogeneous equation a(t) y + b(t) y + c(t) y = f (t), performthe following steps:#1. Find a fundamental set of solutions cfw_y1 , y2 to a(t) y + b(t) y +
Purdue - MATH - 366
MA 36600 MIDTERM #2 REVIEW5 Let Q = Q(t) denote the charge in an electric circuit at time t; it is measured in coulombs. Similarly,I=dQdtdenotes the current; it is measured in amperes. We think of Q as the displacement u(t), and I asthe velocity u
Purdue - MATH - 366
6MA 36600 MIDTERM #2 REVIEWChapter 44.1: General Theory of nth Order Equations. An nth order linear dierential equation is an equation of the formdn ydn1 ydy+ P1 (t) n1 + + Pn1 (t)+ Pn (t) y = G(t).dtndtdtWe wish to solve the initial value pr
Purdue - MATH - 366
MA 36600 MIDTERM #2 REVIEW Say that we can factorZ (r) = a0 pk=17q sk skr rkr [k + i k ] r [k i k ].k=1real rootscomplex rootsThen the general solution is to the constant coecient homogeneous equation isspkm1y (t) =Ckm terk tk=1+
Purdue - MATH - 366
MA 36600 MIDTERM #2 REVIEW Say that we can factorZ (r) = a0 pk=17q sk skr rkr [k + i k ] r [k i k ].k=1real rootscomplex rootsThen the general solution is to the constant coecient homogeneous equation isspkm1y (t) =Ckm terk tk=1+
Purdue - MATH - 366
8MA 36600 MIDTERM #2 REVIEW4.4: The Method of Variation of Parameters. Consider the nth order nonhomogeneous linear dierential equationdn Ydn1 YdY+ P1 (t) n1 + + Pn1 (t)+ Pn (t) Y = G(t).ndtdtdtSay that cfw_y1 , y2 , . . . , yn is a fundamen
Purdue - MATH - 366
MA 36600 MIDTERM #2 REVIEWChapter 33.1: Homogeneous Equations with Constant Coecients. An ordinary dierential equation in the formd2 ydy= G t, y,dt2dtis called a second order dierential equation. If we have initial conditions in the formy (t0 )
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 283for some function f (y ).We have seen several examples of autonomous equations: If v = v (t) denotes the velocity at time t of anmass m falling under the inuences of gravity and air resistance, thendv= g
Rutgers - MARKETING - 33:630:301
CHAPTER5:UNDERSTANDINGCONSUMERBEHAVIOR Purchasedecisionprocessconsistsoffivestages:ProblemrecognitionInformationsearchAlternativeevaluationPurchasedecisionPostpurchasebehavior CognitivedissonanceLEVELOFINVOLVEMENT:3CHARACTERISTICS Expensiveit
Rutgers - MARKETING - 33:630:301
Chapter 18: IMCPromotional Mix Inform Prospective Buyers Persuade Them To Try Remind Them of the BenefitsIntegrated MarketingCommunications (IMC)FIGURE 18-1 The communication processFIGUREconsists of six key elementsFIGURE 18-2 The five elements
Emory - MATH - 112
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Emory - MATH - 112
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Emory - MATH - 112
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Emory - MATH - 112
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Emory - MATH - 112
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Emory - MATH - 112
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Emory - MATH - 112
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Emory - MATH - 112
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Emory - MATH - 112
daj 3 D~r-.lCL&lt;-[.fC!h o.nc1lY1od~S~H-&quot;II ct:1Lor's - :Theorem .~.-'&quot;-&quot;-'i00! '(x) an-I:Z; al'Jx-c-)~=fc &quot;) (c.Jn11 I-t?(x) =- y'\ : =.!:'iIril- then&quot;Y'lsO-Exo.YYJp I e: F.~r-1 jr)(:Jl Qx..j2(XJ = C DstCx);:C -Ds X