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Purdue - MATH - 366
2MA 36600 FINAL REVIEWsuch thati. Each gk (t) is continuous on I . (There are m functions to consider.)ii. Each pij (t) is continuous on I . (There are m n functions to consider.)iii. t0 I .Then there exists a unique solution on the interval I . Thi
Purdue - MATH - 366
MA 36600 FINAL REVIEW3i. A is invertible i.e., there exists B such that A B = B A = I.ii. A is nonsingular i.e., the x = 0 is the only solution to A x = 0.iii. det A = 0. Say that we are given an m n matrix where the entries are functions of time t:
Purdue - MATH - 366
4MA 36600 FINAL REVIEW Given an n n matrix A, consider an equation of the formAx = xwhere x is an n-dimensional vector and is a scalar. If this equation holds for some nonzero vector xand arbitrary scalar , we call x an eigenvector of A, and an eigen
Purdue - MATH - 366
MA 36600 FINAL REVIEW57.5: Homogeneous Linear Systems with Constant Coecients. Consider the homogeneous linear system with constant coecientsdx = A x.dtWe may guess a solution in the form x(t) = ert for some constant vector and constant scalar r.(
Purdue - MATH - 366
6MA 36600 FINAL REVIEW7.7: Fundamental Matrices. Consider the initial value problemdx = P(t) x,x(t0 ) = x0 .dtSay that x(1) , x(2) , . . . , x(n) is a fundamental set of solutions. Then the general solution isx(t) = c1 x(1) (t) + c2 x(2) (t) + +
Purdue - MATH - 366
MA 36600 FINAL REVIEW77.8: Repeated Eigenvalues. Consider 2 2 matrix A with eigenvalues r1 and r2 having corresponding eigenvectors11(1)(2)=and = 12 .2122i. Say that r1 = r2 i.e., we have distinct eigenvalues. Then there exist 2 2 matricesr0
Purdue - MATH - 366
8MA 36600 FINAL REVIEWThe general solution isy(t) = (t) c + (t)t( )1 h( ) d=x(t) = T y(t)in terms of the fundamental matrixrte10(t) = exp (D t) = ...00er2 t...0..00...ern t. Consider the nonhomogeneous systemdx = P(t) x +
Purdue - MATH - 366
MA 36600 FINAL REVIEWChapter 77.1: Introduction. A system of rst order equations is a collection of initial value problemsdxk= Gk (t, x1 , x2 , . . . , xn ) ,dtk = 1, 2, . . . , m.xk (t0 ) = x0 ;kWe say that such a system is a system of rst orde
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, JANUARY 12Say that we have an object of mass m, perhaps measured in kilograms. Also, say that x = x(t) is theposition of an object, perhaps measured in meters, at time t, perhaps measured in seconds. Thendxv==xdtis
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, JANUARY 12Course Information Instructor: Edray Goins. Oce: MATH 612. Extension: 4-1936. E-Mail: egoins@math.purdue.edu.Oce hours will be on Thursdays from 10:00 AM through 12:00 PM. Meeting Times: The class will meet Mo
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 14dv= gdtFigure 2. Slope Field of3.22.41.60.8-4.8-4-3.2-2.4-1.6-0.800.81.62.43.244.8-0.8-1.6-2.4-3.2Figure 3. Slope Field of-7.5-5-2.502.5dv= g vdtm57.510-2.5-5-7.5-10
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 14the rate of change of the size of the population is proportional to the size of the populationas well as the distance the population is from its sustainable size.Another way to say this isdPPdPPP 1==
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 14Basic Mathematical Models (contd)Slope Fields. It is not necessary to nd the exact solution of a dierential equation in order to understandhow the solution behaves. We discuss this concept through several ex
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, JANUARY 16The initial value problem ismdv= m g vdtDierentialEquationv (0) = v0InitialConditionThis is a little more dicult to solve, but not impossible. The problem is we cannot simply integrate bothsides beca
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, JANUARY 163If we did not keep track of the initial condition y (0) = y0 , we would call the solution of the dierentialequation alone the general solution. For example, y (t) = (b/a) + C1 eat would be the general solution
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, JANUARY 16Solutions of Some Differential EquationsGravity without Air Resistance. Say that we have a mass m at a position x = x(t). We wish to discussits motion under the inuence of gravity. We assume for simplicity that
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 21Upon integrating and exponentiating, we nd thattln | =p( ) d + C1=(t) = C2 exptp( ) dwhere C2 = eC1 is some constant. This is the general solution to (ii). Since we want = (t) to satisfy(i), we want
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 213Separable Equations. We introduce a trick to solve separable dierential equations.Divide both sides by the function Y = Y (y ) so that all of the y terms are to one side:dy1 dy= Y (y ) T (t)== T (t).d
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 21First Order Linear EquationsDenitions. So far we have considered initial value problems in the formdy= b + a y,y (0) = y0dtwhere a and b are constants. We showed that the solution isbby (t) = + C eat
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, JANUARY 23The rate of change of the quantity of salt in the solution is the dierence of the rate which ows in and therate which ows out. Hence our initial value problem isdQ1Q(t)= r r,Q(0) = Q0 .dt4100One meth
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, JANUARY 23m4365S (t + 1)=S (t)0.081+m1.082431.083281.083293mHence if the original investment was $1000 you would earn $82.43 from Bank A and $83.28 from Bank B.The latter bank gives a better return on your i
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, JANUARY 23First Order Nonlinear Equations (contd)Example. Say we wish to solve the initial value problemdy3 t2 + 4 t + 2=,dt2 (y 1)y (0) = 1.First we nd the general solution to the dierential equation, then we det
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, JANUARY 26This is the implicit solution for the initial value problem. (The left-hand side is the change in kinetic energy,whereas the right-hand side is the change in potential energy.)This expression states how the d
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, JANUARY 263Note that (t0 ) y (t0 ) = 1 y0 , so we see that the unique solution to the initial value problem must be t1y (t) =( ) g ( ) d + y0 .(t) t0Existence and Uniqueness Theorems: Nonlinear Case. There is a much
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, JANUARY 26Modeling with First Order Equations (contd)Escape Velocity. Say that we have a rocket of mass m which blasts o from the surface of the Earth atan initial velocity v0 . Gravity pulls back on the rocket no matter
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 28First we nd the general solution to the dierential equation. (Again, we solved this equation in Lecture5.) This dierential equation is separable; we can bring all of the y terms to one side, and all of the
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 28Linear vs. Nonlinear SolutionsExample 1. Consider the initial value problemt y + 2 y = 4 t2 ,y (1) = 2.What is the largest region R such that there exists a unique solution? We solve this problem in three
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, JANUARY 30We will choose these constants so that A = 1 and B A = 0 i.e., we choose A = B = 1. Simply put, theleft-hand side of the dierential equation becomes1dy11dyd=+=ln |y | ln |1 y | .y (1 y ) dty 1 y dt
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, JANUARY 30Example. As an example, consider the functionf (y ) = r y (1 y ) = r y y 2=3f (y ) = r (1 2 y ) .The equilibrium solutions are yL = 0 and yL = 1. We have f (0) = r > 0 and f (1) = r < 0. Hence yL = 0is an u
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, JANUARY 30Population DynamicsLogistic Equations. We discuss an application of autonomous equations by considering population growth.Say that we have a population which has a size P = P (t) at time t. Assume that the rate
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 2 There is a threshold T . That is, if P < T then the population will become extinct: P (t) 0 ast .We assume that the rate of change of the size of the population is proportional to three factors:(1) The size
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 23depends on the initial value P (0) = P0 , we consider the slope eld of the dierential equation. The threecritical points PL = 0, T, K break the graph into four regions, but we just consider three of them: Whe
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 2Total Dierentials. Say that z = f (x, y ) is a function. We will show thatffdz =dx +dy.xyThis is called the total dierential of the function z = f (x, y ). First we make sense of this notation. Saythat
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 2Population Dynamics (contd)Logistic Equation with Threshold. Consider the initial value problemdyy= r y 1 ,y (0) = y0 ;dtTin terms of positive constants r and T . The solution isy0 Ty (t) =.y0 + (T y
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 4Assume rst that statement p is true. Then we haveM f2f2f fN=====.yy xy xx yx yx(We can interchange the partial derivatives when f (x, y ) has partial derivatives which are continuous funct
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 43Integrating Factors. Say that we have a rst order dierential equationdy= G(x, y )dxwhereG(x, y ) = M (x, y ).N (x, y )If = (x, y ) is any nonzero function, we haveG(x, y ) = (x, y ) M (x, y )(x,
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 4We seek a function = (t, y ) such that when we multiply this equation by this function we do have anexact equation. Consider the function t(t, y ) = expp( ) d .Then we have the products t(t, y ) M (t,
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 4Exact Equations (contd)Exact Equations. Say that we have a rst order dierential equation in the formM (x, y ) dx + N (x, y ) dy = 0.Imagine for the moment that we could nd a function z = f (x, y ) such that
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 6Example. Consider the initial value problemdy1= 3 + et y,y (0) = 1.dt2First we nd an exact solution y = y (t), then we nd an approximate solution via Eulers Method.To nd the exact solution, note that th
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 63Unfortunately, this is not entirely useful because both sides of the equation involve y = y (t). We use thisidea to dene a sequence of functions yn = yn (t) which will approximate the exact solution y = y (t).
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 6Numerical SolutionsEulers Method. Consider an initial value problemdy= G(t, y ),dty (t0 ) = y0 ;in terms of some function G(t, y ). In general it is too dicult to solve such an equation. Indeed, we havesee
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 9Difference EquationsFirst Order Equations. Recall that every rst order dierential equation is in the formdy= G(t, y ),dty (t0 ) = y0 ;for some function G = G(t, y ). As discussed in the previous lecture,
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 93We explain the relationship with solutions to these equations. Recall how to solve the dierential equationwhen the equation is homogeneous i.e., when g (t) is the zero function:dy+ p(t) y = 0dtdy= p(t) y
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 9so that we have the equation0 = (1.01)48 10000+1 (1.01)48b1 (1.01)=0 = 16122.261+61.223 b=b=16122.261= 263.338.61.223Hence we must pay at least $263.34 each month in order to pay o the loan in 4 yea
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 95The value = 3 is a point of bifurcation i.e., when < 3 there is one equilibrium solution, but when > 3there are two. As increases eve more, the number of periods increases until there are innitely many!For in
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 9Existence and Uniqueness (contd)Example. We give an example of the proof of the theorem. Consider the initial value problemdy= 2 t (1 + y ),y (0) = 0.dtFirst we nd an exact solution y = y (t), then we nd a
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 16More Denitions. Consider a linear second order dierential equationa(t) y + b(t) y + c(t) y = f (t).If we have initial conditions in the formdy(t0 ) = y0 ;dtwe call this system an initial value problem. N
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 163If the characteristic equation a r2 + b r + c = 0 has two distinct real roots r1 and r2 , then the general solutionto the dierential equation isy (t) = c1 er1 t + c2 er2 tfor some constants c1 and c2 . Reca
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 16Upon multiplying the rst equation by r2 then by r1 , we have the following systems of equations:r2 c1 er1 t0r1 c1 er1 t0(r1 r2 ) c1 er1 t0+ r2 c2 er2 t0=r2 y0r1 c1 er1 t0+r1 c2 er2 t0=r1 y0+=y0r
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 16Second Order Differential EquationsLinear Equations. We briey recall some facts about dierential equations which we have seen over thepast few weeks. Recall that we use the notationdn yy ( n) = ndtto denot
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 18Example. Consider the initial value problem2t 3 t y + t y (t + 3) y = 0;y (1) = 2,y (1) = 1.We do not have enough techniques yet to solve this dierential equation, but we can discuss whether asolution
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 183Wronskian. We have seen that given the homogeneous equation L[y ] = 0, we can nd many solutions whengiven just a few. However, when do we have all solutions? That is, if y1 = y1 (t) and y2 = y2 (t) are sol
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 18Example. Consider the initial value problem2t 3 t y + t y (t + 3) y = 0;y (1) = 2,y (1) = 1.We do not have enough techniques yet to solve this dierential equation, but we can discuss whether asolution
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 18Second Order Differential EquationsExample. Consider the initial value problemy + 5 y + 6 y = 0;wherey (0) = 2,y (0) = 3.To solve this equation, we proceed in two steps: First, we nd the general solutio
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 20We mention in passing that we have the identityy (t ) y2 (t0 )1W0 = det 1 0= dety1 (t0 ) y2 (t0 )00= 1.1Linear IndependenceAbels Theorem. Consider the dierential equationa(t) y + b(t) y + c(t) y =
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 20We explain how this is related to Abels Theorem. Using thethe characteristic equation arebb2 4 a c r1 = 2a 2a=bb2 4 a c r2 = +2a2aHence the Wronskian is in the formbW (t) = C exp tin terms ofa3
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 20Linear Homogeneous EquationsExample. Consider the dierential equationy + 5 y + 6 y = 0.We discuss the Wronskian associated with this equation. We have the characteristic equationr2 + 5 r + 6 = 0.Since the p
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 23Upon multiplying the rst equation by g (t) and the second by g (t), we nd the following system ofequations:f (t) g (t) c1 +g (t) g (t) c2 = 0f (t) g (t) c1 + g (t) g (t) c2 = 0f (t) g (t) f (t) g (t) c1=
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 23Recall that we have the Taylor Series expansionscos x = 1 sin x1214x+x + 224115= x x3 +x + 6120In particular, we nd the identityThis is known as Eulers Formula.==eit = cos t + i sin t. (1)k
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 233is nonzero then y (t) = c1 y1 (t) + c2 y2 (t) is the solution of the initial value problem in terms of the constantsy0 y2 (t0 ) y0 y2 (t0 )y0 y1 (t0 ) y0 y1 (t0 )andc2 = .W0W0But we saw above that W y1
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 25Indeed, we verify that the functionsy1 (t) = et cos tandy2 (t) = et sin tform a fundamental set of solutions to the dierential equation by showing that the Wronskian is a nonzerofunction. We have the d