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Course: MATH 366, Spring 2009
School: Purdue
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36600 2 MA LECTURE NOTES: FRIDAY, FEBRUARY 20 We mention in passing that we have the identity y (t ) y2 (t0 ) 1 W0 = det 1 0 = det y1 (t0 ) y2 (t0 ) 0 0 = 1. 1 Linear Independence Abels Theorem. Consider the dierential equation a(t) y + b(t) y + c(t) y = 0. Say that y1 = y1 (t) and y2 = y2 (t) are solutions. (1) The Wronskian of y1 and y2 is the function W (t) = C exp t b( ) d a( ) for some...

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36600 2 MA LECTURE NOTES: FRIDAY, FEBRUARY 20 We mention in passing that we have the identity y (t ) y2 (t0 ) 1 W0 = det 1 0 = det y1 (t0 ) y2 (t0 ) 0 0 = 1. 1 Linear Independence Abels Theorem. Consider the dierential equation a(t) y + b(t) y + c(t) y = 0. Say that y1 = y1 (t) and y2 = y2 (t) are solutions. (1) The Wronskian of y1 and y2 is the function W (t) = C exp t b( ) d a( ) for some constant C . (2) W (t) = 0 for all t if and only if W (t0 ) = 0 for some t0 . We give a proof. Denote the function W = y1 y2 y 1 y 2 . This has the derivative W = y 1 y 2 + y1 y2 y1 y 2 + y1 y2 = y 1 y2 y 1 y 2 . Hence we have the expression a(t) W + b(t) W = a(t) y1 y2 y1 y2 + b(t) y1 y2 y1 y2 = a(t) y1 y2 y1 y2 + b(t) y1 y2 y1 y2 + c(t) y1 y2 y1 y2 = y1 a(t) y2 + b(t) y2 + c(t) y2 a(t) y1 + b(t) y1 c(t) + y1 y2 = 0. Hence W = W (t) satises a rst order dierential equation. Since this equation is separable, we may solve by dividing both sides by W (t): dW = b(t) W dt 1 dW b(t) = W dt a(t) a(t) d b(t) ln |W | = dt a(t) = ln |W | = t b( ) d + (constant). a( ) Upon exponentiating, we can express W = W (t) in the form t b( ) W (t) = W0 exp d t0 a( ) where W0 = W (t0 ). Since the exponential is never zero, we see that W (t) = 0 for any t if and only if W (t0 ) = 0 for some t0 . Example. Consider the constant coecient equation a y + b y + c y = 0. If the characteristic equation a r2 + b r + c = 0 has two real roots r1 and r2 , we know that two solutions to the dierential equation are y1 (t) = er1 t and y2 (t) = er2 t . We have seen that the Wronskian of these two functions is W (t) = y1 (t) y2 (t) y1 (t) y2 (t) = (r2 r1 ) e(r1 +r2 )t .
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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,
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