# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

1 Page

### lecture_9 (dragged) 3

Course: MATH 366, Spring 2009
School: Purdue
Rating:

Word Count: 301

#### Document Preview

36600 4 MA LECTURE NOTES: MONDAY, FEBRUARY 2 Total Dierentials. Say that z = f (x, y ) is a function. We will show that f f dz = dx + dy. x y This is called the total dierential of the function z = f (x, y ). First we make sense of this notation. Say that we have a path = (t) in the plane R2 . That is, given any t R, we have a point (x, y ) = (t) R2 . Then the composition z (t) = f (t) is actually a...

Register Now

#### Unformatted Document Excerpt

Coursehero >> Indiana >> Purdue >> MATH 366

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
36600 4 MA LECTURE NOTES: MONDAY, FEBRUARY 2 Total Dierentials. Say that z = f (x, y ) is a function. We will show that f f dz = dx + dy. x y This is called the total dierential of the function z = f (x, y ). First we make sense of this notation. Say that we have a path = (t) in the plane R2 . That is, given any t R, we have a point (x, y ) = (t) R2 . Then the composition z (t) = f (t) is actually a function of t. Choose a small dierence t, and dene the dierences x and y by (x, y ) = (t + t) (t) i.e., (x + x, y + y ) = (t + t). Figure 4. Plot of (t) vs. t 3.2 2.4 1.6 0.8 -4.8 -4 -3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2 4 4.8 -0.8 -1.6 -2.4 -3.2 We have the ratio f + (t t) f (t) z (t + t) z (t) = t t f (x + x, y + y f (x, y ) = t f (x + x, y + y ) f (x, y + y ) f (x, y + y ) f (x, y ) = + t t f (x + x, y + y ) f (x, y + y ) x f (x, y + y ) f (x, y ) y = + . x t y t Upon taking limits, we nd that dz f (x + x, y ) f (x, y ) x f (x, y + y ) f (x, y ) y f dx f dy = lim + = + . t0 dt x t y t x dt y dt This is called the total derivative of the function z = f (x, y ). Note that it involves partial derivatives. This formula is the same as the more familiar result involving the dot product of the gradient: dz f f = (f ) (t) in terms of f (x, y ) = , . dt x y Upon multiplying both sides by the dierential dt we nd the total dierential above.
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

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 &lt; 3 there is one equilibrium solution, but when &gt; 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 &gt; 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,