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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
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 253Repeated RootsReview. We recap what we know so far. Consider the constant coecient dierential equationa y + b y + c y = 0We know that we can nd solutions by considering the characteristic equationa r2 +
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 25In particular, a y + b y + c y = 0 if and only if v = 0. But we know that the general solution to thisdierential equation isv (t) = c1 t + c2for constants c1 and c2 . Hence the general solution to the di
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 25Complex RootsEulers Formula. We showed in the previous lecture thateit = cos t + i sin t.This is known as Eulers Formula. In general, if we writer = + iin terms of real numbers and , we have the expressi
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27Summary. We summarize the general solution to the dierential equationa y + b y + c y = 0.Say that the characteristic equationa r2 + b r + c = 0has roots r1 and r2 . A fundamental set of solutions for this
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 273Then the general solution to the nonhomogeneous equation is the functiony (t) = c1 y1 (t) + c2 y2 (t) + Y (t)for some constants c1 and c2 . We remark that typically nding y1 and y2 is easy: when the coecient
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27must be the function1y (t) = c1 y1 (t) + c2 y2 (t) + Y (t) = c1 et + c2 e4t + e2t2for some constants c1 and c2 .Example. Consider the dierential equationy 3 y 4 y = 2 sin t.Again, we will nd the general
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27Repeated RootsReview. Say that we wish to nd the general solution to the dierential equationa y + b y + c y = 0where b2 4 a c = 0. First note that the roots of the characteristic polynomial a r2 + b r + c = 0
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, MARCH 2It makes sense to choose A and B such that10 A2A+ 2 B+ 10 B= 8=0We can solve for A and B by multiplying the rst equation by 5; or multiplying the second by 5:50 A2A52 A+10 B+ 10 B= 40=0= 4010 A +
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, MARCH 23for some constants c1 and c2 .Variation of ParametersRecap. Recall that we wish to nd the general solution to the linear second order dierential equationa(t) y + b(t) y + c(t) y = f (t);where a(t), b(t), c(t),
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: MONDAY, MARCH 2This method is called Variation of Parameters. We remark that we can always writey (t) = u1 (t) y1 (t) + u2 (t) y2 (t) = c1 y1 (t) + c2 y2 (t) + Y (t)in terms of the functionf ( ) y1 ( ) y2 (t) y1 (t) y2 ( )d
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, MARCH 2Method of Undetermined CoefficientsUndetermined Coecients. Say that we wish to solve a constant coecient linear second order dierential equation in the forma y + b y + c y = f (t)We know how to nd the general sol
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 4#1. Find a fundamental set of solutions cfw_y1 , y2 to the homogeneous dierential equationa(t) y + b(t) y + c(t) y = 0.#2. Compute the integralstf ( ) y2 ( )u(t) = d + c1a( ) W ( )andu2 (t) =in terms
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 43Newtons Second Law of Motion states that F = m a. That is,m u = sum of forces = Fg + Fs + Fd + Fe = m g k (L + u) u + F (t)Since m g k L = 0, we can rewrite this asm u k u u + F (t)=m u + u + k u = F (t).
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 4Variation of ParametersGeneral Method. We explain how to nd the general solution of the nonhomogeneous equationa(t) y + b(t) y + c(t) y = f (t).Say that cfw_y1 , y2 is a fundamental set of solutions to the ho
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, MARCH 6for some constants R, , and . To see why, recall the Angle Dierence Formula for Cosine:We have the equationcos ( ) = cos cos + sin sin .A et cos t + B et sin t = R et cos ( t ) = R et cos t cos + R et sin t sin
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, MARCH 63Resonance. We know that the general solution to the nonhomogeneous equationm u + k u = F (t)is in the form u(t) = uc (t) + U (t) for a transient solution uc (t) and a steady-state solution U (t). We discussthis
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, MARCH 6Mechanical and Electrical VibrationsHookes Law. Say that we have a mass m which is attached to a spring. Consider four forces on the massm: Gravity: Newtons Law of Gravity states that Fg = m g . Restoring Force:
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, MARCH 9Figure 2. Graph of Steady-State Solution U (t) = 9 cos t105-2.502.557.51012.51512.515-5-10Figure 3. Graph of Solution u(t) = uc (t) + U (t)105-2.502.557.510-5-10We have considered the equ
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, MARCH 93Assume for the moment that = 0 . We may pick a particular solutionU (t) = A cos 0 t + B sin 0 t +for any A and B , so in particular chooseA=This gives the functionU (t) = F02m (0 2 )F0cos t2m (0 2 )an
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: MONDAY, MARCH 920Figure 4. Graph of Solution u(t) = uc (t) + U (t) under Resonance16128402.557.51012.51517.520-4-8-12-16-20ii. Ohms Law asserts that the voltage across a resistor is proportional to the curren
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, MARCH 95How does the charge Q(t) change over time t? If there is no outside electromotive force i.e., E (t) is the zerofunction, then the solution depends on the discriminant R2 4 L/C : If R2 4 L/C > 0, then we expect e
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, MARCH 9Resonance. We know that the general solution to the nonhomogeneous equationm u + k u = F (t)is in the form u(t) = uc (t) + U (t) for a transient solution uc (t) and a steady-state solution U (t). We foundin the p
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 11This is a linear dierential equation if we can expressn1G t, y, y (1) , . . . , y (n1) = g (t) p1 (t) y (n1) pn1 (t) y pn (t) y = g (t) That is, if the dierential equation is in the formj =0pnj (t) y (j )
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 11we haveL[y ] = Lmci yi =i=1mci L[yi ] =i=1mi=13ci 0 = 0.Hence y = y (t) is also a solution to the homogeneous problem. As a consequence, we show that the generalsolution to L[y ] = G(t) is y (t) = u
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, MARCH 11Higher Order Linear EquationsReview. Recall that a second order linear equation is in the forma(t) y + b(t) y + c(t) y = f (t).We found that the general solution is in the formy (t) = c1 y1 (t) + c2 y2 (t) +
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, MARCH 13We write this using the product of matrices:y0 Y (t0 )y1 (t0 ) (1)(1) y0 Y (1) (t0 ) y1 (t0 ) y (2) Y (2) (t0 ) y (2) (t0 )0= 1......(n1)y0 Y (n1) (t0 )(n1)y1(t0 )y2 (t0 ) c1 (1)yn (t0 )
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, MARCH 13In general, the determinant of a11 a12...a1n3an n n matrix isa21 an1 a22 an2 ( ) a(1) 1 a(2) 2 a(n) n ,.. =......a a 2nnnwhere the sum is taken over all n! permutations of the set cfw_1, 2,
Purdue - MATH - 366
4MA 36600 LECTURE NOTES: FRIDAY, MARCH 13since the nth row is equal to the (k + 1)st row for k = 0, 1, . . . , (n 2). Finally, to nish o the proof ofAbels Theorem, we nd the general solution to this rst order dierential equation:P1 (t)W (1) = WP0 (
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, MARCH 13Higher Order Linear EquationsFundamental Set of Solutions. Consider the initial value problemnj =0Pnj (t) y (j ) = G(t)where(j 1)y (j 1) (t0 ) = y0i = 1, 2, . . . , n.,Say that we can nd (n + 1) functions
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, MARCH 23Express this as a product of matrices as follows:f1 (t)f2 (t) (1)(1) f (t)f2 (t)1........(n1)(n1)f1(t) f2(t) 0 (1)fn (t) k2 0 = . . ... . ....fn (t)(n1)fn(t)k1kn0n n matr
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, MARCH 233Case #1: Real and Unequal Roots. Say for the moment that we have the characteristic polynomialpZ (r) =r rkk=1for some distinct real numbers rk . We will write the corresponding operator L[y ] in the followin
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, MARCH 23Higher Order Linear EquationsLinear Independence. Consider a collection of functions cfw_f1 , f2 , . . . , fn . We say that this is a linearlyindependent set if the only solution to the equationk1 f1 (t) + k2 f2
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: MONDAY, MARCH 30(3) Y = Y (t) is a solution to the nonhomogeneous equationnj =0Pnj (t) Y (j ) = G(t).Then the general solution to the dierential equation isy (t) = yc (t) + Y (t)whereyc (t) =nCi yi (t).i=1For example,