KEY. Math 334 Midterm III Fall 2008 sections 001 and 003 Instructor: Scott Glasgow

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1 KEY Mah 334 Miderm III Fall 28 secions and 3 Insrucor: Sco Glasgow Please do NOT wrie on his exam. No credi will be given for such work. Raher wrie in a blue book, or on your own paper, preferably engineering paper. Wrie your name, course, and secion number on he blue book, or on your own pile of papers. Again, do no wrie his or any oher ype of informaion on his exam. Good Pracice: A differenial equaion has he propery ha one can check wheher a given funcion saisfies i. So check your soluions! If one doesn work, ry again, or a leas noe ha your proposed soluion doesn work. You will ge much more parial credi if you know wheher or no you answer is correc.

2 a. Compue he Laplace ransform L f ( s) of f ( ) e cosdirecly from he definiion. For which values of s does your formula hold? Assume here ha a is a real number. 2 poins By definiion of he Laplace ransform, and by using inegraion by pars, and provided s a, we ge L s a cos sa sa e e d cos s a sa sa e sin d sin de sa sa sa sa sa sine e sa sa dsin sa cose d f ( s). sa sa L sa sa s a s s a f () s e f () d e e cosd e cosd cosde Noe ha s ais also necessary for he inegral o converge. Solving he las equaion algebraically for L f ( s) we ge () s a s a s a L f s. (.2) 2 s a 2 sasa s a s a (.)

3 2 2. Solve he following IVP. Using he Laplace ransform and knowing how o deal wih piecewise defined funcions in his ransform should make hings easier. Wha is he value of y ()? Do no express his value in erms of an absrac formula, bu raher as a concree number. To compue his value concreely, you will need o know ha he funcion sin( ) is 2 periodic.,, y 4 y f( ), y() y(). 2, 2, 2 (.3) 25 poins Taking he Laplace ransform of (.3) we ge 2 2 L L L s 2 y ( s) y 4 y ( s) f ( s) L, (.4) y () s f (). s 2 2 s 2 L To compue L f ( s) we rewrie f in (.3) as: 2 u() 2 2 u2() 2 u2() f() u () u () 2 u () u () u () 2 u () 2 u (). 2 4 (.5) So, according o he relevan heorem, and given 2 per he able provided you), hen L g () s when g () u() (as s s 2s L f () s L g () s 2 e L g () s e L g () s s 2s s 2s Lg( s) 2e e 2 2 e e. s (.6) Thus, explicily, (.4) is L s 2s y() s 2 2 f () s e e. s 2 L s 2 s (.7)

4 3 and we already know hen ha he soluion is of he form y() h() u () 2 h( ) u () h( 2 ) u (), (.8) 2 where hen we need only find h ( ), whose Laplace ransform is L s 2 s s s s 2 s 2 h() s (.9) Thus, according o he able, or memorized formulae, we have ha he soluion is given by (.8) and by h () sin(2). 4 2 (.) Noe ha for any 2, including, we have y () hu () () 2 h ( ) u () h ( 2 ) u (), 2 h () 2 h ( ) h ( 2 ) sin(2 ) 2 sin 2 2 sin sin(2 ) 2 sin 2 2 sin (.) 3. Solve he following iniial value problem in erms of he convoluion inegral: D D y g( ), y() y(). (.2) Here I expec you o wrie he soluion y y ( ) of (.2) as y() y g () y( ) g( ) d, (.3) so ha he problem is effecively only o deermine he funcion y ( ) in (.3). 2 2 Recall DDy D y y y, bu realize he represenaion in (.2) may be a leas as useful as wriing y y on he lef-hand side. (Hin: Use

5 4 9 poins he Laplace ransform and he convoluion heorem. Noe he poin value his is sraighforward if you recall he relevan heory.) By he Laplace Transform we have from (.2) ha L L L s s y () s D D y() s g () s Ly() s Lg() s Lg() s (.4) ss s s L e e () s g() s sinh() s g() s 2 L L L L sinhg so ha evidenly in (.3) we have y () sinh(). (.5) 4. Suppose ha boh x sin ( ) and 2( ) sin x (.6) solve he sysem and ha solves he sysem an sec x sec an x (.7) cos x p () cos (.8) x an sec sec an x. (.9) Then wha is he soluion of he IVP

6 5 x an sec, ()? sec an x x (.2) This problem ess your general undersanding of he srucure of soluions of linear sysems wihou requiring you o regurgiae a heorem (wih all of he inricacies of is ariculaion). Noe he poin value his problem is sraighforward if you recall he heory. 2 poins According o he informaion given, he general soluion of (.2) can be expressed as x sin cos x () sin c cos, (.2) Choosing and using he iniial daa in(.2), (.2) becomes () x c c c. (.22) So (.2) becomes x sin cos sin cos sin cos x (). sin cos cos cos (.23) 5. Find he fundamenal marix of soluions ( ) o he sysem he one ha has he propery ha() I. 2 poins x 2 x, (.24)

7 6 Firs we find a represenaion of he general soluion of he sysem(.24). This can be expressed as xx() cξ e c ξ e, (.25) 2 provided he ξ s are independen eigenvecors associaed wih he eigenvalues and of he marix in (.24). To deermine his we noe 2 ξ ξ unless de 2 2 i i, i :, (.26) So i i. 2 i 2 i i ξ ξ ξ (.27) Thus, explicily, (.25) is x i i x () c e c2 e. i i (.28) As per he usual heory, we can find a real-valued represenaion by finding he real and imaginary pars of eiher of he above complex-valued soluions: x i cos sin (): e cosisin i, i i cossin cossin (.29) whence a real-valued represenaion of he general soluion is x cos sin () c c 2. cossin cossin (.3)

8 7 A fundamenal marix of soluions ( ), one no necessarily having he desired propery, can be found from he above general soluion (.3) as in cos sin () cossin cossin. (.3) The desired fundamenal marix ( ) can be obained from ( ) via cos sin () () () cos sin cos sin cos sin cos sin cos sin cos sin sin. 2sin cossin (.32) 6. Solve he iniial value problem given by he sysem of problem 5 and he iniial daa 6 poins Using he fundamenal marix of problem 5 we have x (). (.33) cossin sin cos sin sin x() () x() 2sin cos sin 2sin cos sin (.34) cos. cos sin 7. Calculae 6 poins 2 e. (.35)

9 8 We have, from problem 5 and he general heory, 2 cos sin sin e ( ). 2sin cos sin (.36) 8. Find a represenaion of he general soluion of he sysem 2 poins x x. (.37) The marix in (.37) has a repeaed eigenvalue wih only one eigenvecor. Hence he general soluion is of he form 2 x x() cξe c ξη e (.38) where is he sole eigenvalue,ξ is one of is eigenvecors, and ηis an associaed pseudo eigenvecor: Firs ξ ξ unless 2 2 de,. (.39) So, ξξ and. ηξ η η (.4) Thus, explicily, (.38) is

10 9 xx() c e c e 2 c c 2. (.4) 9. Find he fundamenal marix of soluions () for he sysem of problem 8 ha saisfies () I. 5 poins From (.4) we have a fundamenal marix of soluions whence he one desired is (), (.42) () () () (). (.43). Solve he iniial value problem obained from combining he differenial equaion of problem 8 wih he iniial daa 2 x (). (.44) 2

11 6 poins From problem 9 we have 2 x() () x() (). x 2 (.45)