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Course: MATH 366, Spring 2009
School: Purdue
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36600 2 MA LECTURE NOTES: WEDNESDAY, FEBRUARY 25 Indeed, we verify that the functions y1 (t) = et cos t and y2 (t) = et sin t form a fundamental set of solutions to the dierential equation by showing that the Wronskian is a nonzero function. We have the derivatives y1 (t) = et cos t y2 (t) = et sin t y1 (t) = et cos t et sin t y2 (t) = et sin t + et cos t This gives the Wronskisn W y1 , y2 (t)...

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36600 2 MA LECTURE NOTES: WEDNESDAY, FEBRUARY 25 Indeed, we verify that the functions y1 (t) = et cos t and y2 (t) = et sin t form a fundamental set of solutions to the dierential equation by showing that the Wronskian is a nonzero function. We have the derivatives y1 (t) = et cos t y2 (t) = et sin t y1 (t) = et cos t et sin t y2 (t) = et sin t + et cos t This gives the Wronskisn W y1 , y2 (t) = y1 (t) y2 (t) y1 (t) y2 (t) = et cos t et sin t + et cos t et cos t et sin t et sin t = e2t cos t sin t + cos2 t cos t sin t + sin2 t = e2t cos2 t + sin2 t = e2t = 0. Hence as long as = 0 that is, we have complex roots then {y1 , y2 } does indeed form a fundamental set of solutions. We remark that when the discriminant b2 4 a c < 0 that there is not just one fundamental set of solutions. Indeed, we can either choose {y1 , y2 } as above, or we may choose the exponentials {er1 t , er2 t }. The dierence is that the exponentials are complex-valued functions, whereas y1 and y2 are real-valued functions. Either choice is ne for most problems, but usually y1 and y2 are more convenient since they do not involve complex numbers. Example. now Consider the initial value problem 16 y 8 y + 145 y = 0, y (0) = 2, y (0) = 1. We will nd the solution by rst nding the general solution to the dierential equation, then nding a particular solution corresponding to the initial conditions. First we nd the general solution to the dierential equation. We do so by assuming that the solution is in the form y (t) = ert for some constant r. The characteristic equation is 16 r2 8 r + 145 = 0. Upon completing the square, we can nd the roots: 2 (4 r 1) + 122 = 0 = Using Eulers Formula, we nd that Hence the general solution is 4 r 1 = 12 i = r= 1 4 3 i. ert = et/4 cos 3t i et/4 sin 3t. y (t) = c1 et/4 cos 3t + c2 et/4 sin 3t for constants c1 and c2 . Now we nd the particular solution corresponding to the initial conditions. We have the derivatives y (0) = c1 y (t) = (c1 cos 3t + c2 sin 3t) et/4 = t/4 t/4 1 y (0) = 1 c1 + 3 c2 y (t) = 4 (c1 cos 3t + c2 sin 3t) e + (3 c1 sin 3t + 3 c2 cos 3t) e 4 We have the system 1 4 c1 c1 + 3 c2 = 2 = 1 = c1 = 2, Hence the solution to the initial value problem is y (t) = 2 et/4 cos 3t + 1 2 et/4 sin 3t. c2 = 1 . 2
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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,
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, MARCH 303This is known as the Method of Undetermined Coecients. (Compare with Lecture #19 from Monday,March 2.)Example. We wish to nd the general solution to the dierential equationy (3) 3 y (2) + 3 y (1) y = 4 et .Fi
Purdue - MATH - 366
MA 36600 LECTURE NOTES: MONDAY, MARCH 30Higher Order Linear EquationssCase #3: Repeated Roots. Say for the moment that we have Z (r) = r rk k for some complex numberrk . We will write the corresponding operator L[y ] in the following form: sk ndj
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 1We will choose the functions ui (t) such thatn duii=1dt(j 1)yi0(t) =G(t)/P (t) for j = n.0Hence we have the expressionndm YPnm (t) m =dtm=0=nnfor j = 1, 2, . . . , (n 1);d m yiui (t)Pnm (
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 13where y1 ( ) (1) y ( )n11n+iK ( , t) = (1)yi (t) ...W y1 , y2 , . . . , yn ( )..i=1 (n2)y( ) 1 y1 ( )y2 ( ) (1)(1) y ( )y2 ( )11....=..W y1 , y2 , . . . , yn ( ) (n2)(n2)
Purdue - MATH - 366
MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 1Higher Order Linear EquationsVariation of Parameters. Recall that when n = 2, a particular solution to the dierential equationtG( )P0 (t) Y + P1 (t) Y + P2 (t) Y = G(t)is given by the integralY (t) =K ( ,
Purdue - MATH - 366
2MA 36600 LECTURE NOTES: FRIDAY, APRIL 3Equivalently, we say it is linear if we have a system in the formx1x2xm==...p11 (t) x1p21 (t) x1= pm1 (t) x1++p12 (t) x2p22 (t) x2+ pm2 (t) x2+ + ...+ ++p1n (t) xnp2n (t) xn+ pmn (t)
Purdue - MATH - 366
MA 36600 LECTURE NOTES: FRIDAY, APRIL 33Then there exists a unique solution on the interval I . This is known as the Existence and Uniqueness Theoremfor Linear Systems. The statement for linear systems follows from that for nonlinear systems because if
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.