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# Ordinary Differential Equations MA 26600

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Math 266 Spring 2001 7 REVIEW 3 FIRST ORDER DIFFERENTIAL EQUATIONS You should be able to recognize and know how to solve rst order differential equations that are either separable linear exact or homogeneous You should be able to evaluate integrals of the following types polymom39al das f e du fu du including 7quot 71 f 2201172 das partial fractions You should be able to use given values ya0 go to determine unknown constants in a solution You should know the relation of the graph of the solution of an initial value problem to the corresponding direction eld HOMOGENEOUS EQUATIONS F a Letymsoaijvandu Substitute to obtain 06 v Solve the above separable equation for v in terms of ac Substitute v g to obtain a formula for the solution y of the original hornogeneous equation EXACT EQUATIONS May Ncy 0 is exact if WE93y N9C cy Find a function 11193 y such that 111495 y May and liq95 y Ncywa y f Ma y 6156 My solve 8 May dx h y Ncy for hyA solution y of the exact equation then satis eswxy 111495 wyxy May Ncy 0so the general solution VA is of the form 11cy 0 Numerical Methods For Solving y fty yt0 yo Create an Mi le to de ne the function fty The function name and the le name should be the same Note that Mi les are not entered in the rnatlab command window but are external text les that are created with a text editor EXAMPLE If y t y create an Mi le named f11m function Zf11ty Zsqrtty The general syntax for the Euler tangent line method is gtgt t7yeul7d le t0t naly0stepsize Note that stepsize t nal 7 t0n7 where n is the number of steps EXAMPLE To nd the Euler tangent line approximation of the solution of the initial value problem 7 W7 y1 37 where t 2 using stepsize h 05 gtgt t7yeul f11 1727305 gtgt lt7yl ans 10000 30000 15000 40000 20000 51726 The syntax is the same for the improved Euler method use rk2 in place of eul and the rungei kutta method use rk4 in place of eul To obtain the graph of an approximate solution on a direction eld7 enter gtgtplottyC where 070777X77Ur7 Omit C for a connected graph APPROXIMATE SOLUTIONS The matlab commands eul7 rk27 and rk4 can be used to obtain approximate solutions of the initial value problem 7 ay ya0 yo You should be able to use the formula yn ywl facn17yn1h to evaluate Euler tangent line values by hand Approximation methods may not give good approximations of the solution of the initial value problem 7 ay ya0 yo7 if c The initial value problem does not have a unique solution7 because either f or fy is not continuous at the initial point 9 The approximation extends beyond the interval where the solution is valid7 because either 705 or becomes unbounded o The solution is unstable7 because solutions that have slightly different initial values diverge from the desired solution PROPERTIES OF SOLUTIONS If f has continuous rst partial derivatives7 then solutions ofthe differential equation 7 ashy satisfy 9 MW fy0c7y MW fyI7yfI7y If is a solution of the differentiable equation 7 ay If y gt 0 at a point7 then y is increasing near the point If y lt 0 at a point7 then y is decreasion near the point If y gt 0 at a point7 then y is concave upward and the Euler tangent line approximations are less than or equal to the solution near the point If y lt 0 at a point7 then y is concave downward and the Euler tangent line approximations are greater than or equal to the solution near the point The Taylor expansion of f about cc c is x 7 fc f ca 7 c 7 c2 7 c3 JulSC a 7 c4 7 MA 266 Spring 01 REVIEW 3 PRACTICE QUESTIONS 1 Determine whether each ofthe following differential equations is separable7 homogeneous7 linear7 or exact Brie y justify your answers N 9 a Fquot Sa 7 395 8 Use the formula y m to express the differential equation a 250yc3yj0 b x3y2xy0 cc3y1295y10 d 2xy16210 e21y210 Find the explicit solution of the initial value problem 7 y2 7 17 y0 0 Find the general solution of the differential equation 567 2y x2 dy i m in terms ofa v and div dac 9071 7 7 dac Find an implicit form of the general solution of the differential equation 90212 my 39 Find an implicit solution of the initial value problem 2xy1952 29 07741 71 Determine approximate values at cc 05 of the solution of the initial value problem 7 y7 y0 1 by using the Euler tangent line method with h 025 Use the given direction elds and the graph of an Euler tangent line approximation of a solution of an initial value problem to explain why the approximation is not a good approximation of the solution 374237 35 91 0 a y c y y20857 y0 1 d y 10y 711847 y0 1 Q3 Consider the initial value problem 7 56y 7f y3 71 a Is the solution increasing or decreasing near cc 3 b Is the solution concave upward or downward near cc 3 c Are the Euler tangent line approximations of the solution near cc 3 greater than or less than the solution 10 Find the rst four nonzero terms of the Taylor series about 0 1 of the solution of the initial value problem 7 5527 y1 2 Math 266 Spring 2001 REVIEW 3 PRACTICE QUESTION ANSWERS 1 a hornogeneous7 exact b hornogeneous c none of these typesd linear7 exact e separable7 exact 7 1621 239 y 7 16275 3y g 4 x2 a R l e H 1 e U tun t A Rl d m H E E Q 6 x2yxy21 7 8 a The initial value problem does not have a unique solution near 00 The functions yt 0 and yt t3 are both solutions b The approximation extends beyond where the solution is valid The solution approaches a point where y a becomes unbounded c The approximation extends beyond where the solution is valid The solution approaches a vertical asymptote d The solution is unstable Solutions that have slightly different initial values diverge from the desired solution Pictures c i 7 hskipltruein d 9 a 5077 x 37 71 implies y any y2 m 72 lt 07 so y is decreasing b 5077 x 37 71 implies y 507 7 ny 73 lt 07 so y is concave downward c The Euler approximations of the solution near cc 3 are greater than the solution 10y22x71395712395713Hgt MA 266 Fall 2000 REVIEW 3 FIRST ORDER DIFFERENTIAL EQUATIONS You should be able to recognize and know how to solve rst order differential equations that are either separable linear exact or homogeneous You should be able to evaluate integrals of the following types fpolymonial dz f e du f uquot du including 1 1 am b dz partial fractions X quot1r2 You should be able to use given values yzo ya to determine unknown constants in a solution You should know the relation of the graph of the solution of an initial value problem to the corresponding direction field HOMOGENEOUS EQUATIONS dz 2 25 FE a y 1 3 Lety mvsodx a 1de 2 Substitute to obtain v Fv Solve the above separa 1 equation for v in terms of X Substitute v 4 to obtain a formula for the solution y of the original homogeneous equation EXACT EQUATIONS Mzy Maw o is exact if Mya y N31 y Find a functional 2 y such that 11 2 y M 2 Y and 21x y N z y Wm y f M x 3 dz My solve 5y May dx h y N any for hy A solution y yxof the exact equation then satisfies d d d gown whim wowg May No 2021 0 so the general solution is of the form 1M3 yc Numerical Methods For Solviny f t y yto yo Create an M le to define the function ft y The function name and the le name should be the same Note that Mfamps are not entered in the matlab commandwindow but are external text files that are created with a text editor EXAMPLE If y 2 55 create an Mfile named f11m function zf11ty zsqrtt39 The general syntax for the Euler tangent line method is gtgt t eul d le t0t naly0stepsize Note that stepsize t nal t0 n Where n is the number of steps EXAMPLE To find the Euler tangent line approximation of the solution of the initial value problem y t 111 8 where t 2 using stepsize h 05 quot gtgt tyeul f11 12305 gtgt ty ans 10000 30000 15000 40000 20000 51726 The syntax is the same for the improved Euler method use rk2 in place of eul and the runge kutta method use rk4 in place of eul To obtain the graph of an approximate solution on a direction field enter gtgtplotiquotyC where C o x Omit C for a connected graph APPROXIMATE SOLUTIONS The matlab commands eul rk2 and rk4 can be used to obtain approximate solutions of theqinitial value problem 31 fcy yao yo You should be able to use the formula ya Un l f Sin 1 yn 1h to evaluate Euler tangent line values by hand Approximation methods may not give good approximations of the solution of the initial value problem 3 fay yxo go if c The initial value problem does not have a unique solution because either f or fv is not continuous at the initial point 0 The approximation extends beyond the interval where the solution is valid because either y t or ya becomes unbounded o The solution is unstable because solutions that have slightly different initial values divergafrom the desired solution PROPampTIES OF SOLUTIONS If f has continuous first partial derivatives then solutions of the differential equation 31 f 2 y satisfy rth fy1 f1zry f97yfxiy If 93 is a solution of the differentiable equation 3 f x 1 If y gt 0 at a point then y is increasing near the point If y lt 0 at a point then 3 is decreasion near the point If yquot gt 0 at a point then y is concave upward and the Euler tangent line approximations are less than or equal to the solution near the point If y lt30 at a point then y is concave downward and the Euler tangent line approximations are greater than or equal to the solution near the point The Taylor expansion of f aboutxCis m fcf cm c 2x c2 xc3gampglx c4 MA 266 Fall 2000 REVIEW 3 PRACTICE QUESTIONS 1 Determine whether each of the following differential equations is separable homogeneous linear or exact Brie y justify your answers d a 23yz3yamp0 bx3y2my0 c z3y121y 1g0 2 lg 01 2xy1 a 1amp0 dy 2 elm 1 g21I a 2 Find the explicit solution of the initial value problem 3 32 1 y0 0 3 Findthe general solution of the differential equation xy 2y 22 zym 33 31 d 4 Use the formula y am to express the d1fferent1al equation dz 2 terms of 31 and 5 Find an implicit form of the general solution of the differential equation zzyz dx my 6 Find an implicit solution of the initial value problem d 2321 1 92 2035 0 111 1 7 Determine approximate values at x 05 of the solution of the initial value problem y 3x y y0 1 by using the Euler tangent line method with I 025 8 Use the given direction fields and the graph of an Euler tangent line ap proximation of a solution of an initial value problem to explain why the ap proximation is not a good approximation of the solution by39 3 2 y1o 3y 4 a y 32123 210 0 hm nitgt114 I rI 1I I I14IEII39gI I I 4 LIIIgllgtlj IIIIIrllll i A 0 quotI r flf wt LIlirl39IlIIIII4IIIIIII I M 11 139 Iquot39LIquotII IquotIquotI 1quot39Ifquotl39 r31 r I 39r IIIIIIIIIIIIIgIrZI 39II TIlglilllilll Ill quotbelilligll llllil llili IIfIIIIIIIIIII Iquotiquot IIII39IIIIIIII Ifi milllJlIlIIII 11391quot u 1 1 1311 I Iquot 14311 I If 11 IIIVIIIIIIII IIIIJIII NIH01 ll3IfAll 42t Il Ill uyIlrglrf lilI u lllll llII I III IgI z Allirill I 42 Al 114 er aafaa aa ra ara 1 l l shunnepuuuus c y y2085y0 1 d y 103 11e y0 1 cm I IgI I rIII I II I I 1114 I Iv r z 7l ll0ll0 ll lliEllll v lali vIlleiIt thIIl llllEIlJOJlEllEllZlElit h IlilJljllIl39fllllll lt1t quotquot39x C litlilllthflllllklott 5 g i 1 I I ll 1 II 1 L I J I I I t t 39 quot quot quot39339 quot39 V llllllflI39llgll39llllglll1 quot 5 I l I I II l I I I I I I i I quotquotT39Vquot quotVquot39lquot39Cquot I 5 1 I3 II I I I I I 5M 3 IIIIEIIEII Ilgl I ll 5151i quotquot tw 39 te I IICIIIIIIIII III u HIMIIHIglzljIlgizIfnilI tr s I I f 1 I leigr 39IT 139quot quotI quotquot I39I u 11 1L E Izzy12311511 z4xlg H1 39 f IAazzlr l l l i 4 s 39 a J HJL t 3 u 1 li b 4 hut 1 VI 4 onuuuuunuuu OMMMICISM 07M 9 Consider the initial value problem y my yz y8 1 a Is the solution increasing or decreasing near a 3 b Is the solution concave upward or downward near X 3 c Are the Euler tangent line approximations of the solution near x 3 greater than or less than the solution 10 Find the rst four nonzero terms of the Taylor series about 6 1 of the solution of the initial value problem 11 223 y1 2 MA 266 Fall 2000 REVIEW 3 PRACTICE QUESTION ANSWERS 1 a homogeneous exact b homogeneous c none of these types 01 linear exact e separable exact 1422 2 y 162x 22 C 3 y 4 3 dv 11 4 Efvlv 1 y 2 6 zzymy21 7 m 175 8 a The initial value problem does not have a unique solution near 30 The functions yt O and yt t3 are both solutions b The approximation extends beyond where the solution is valid The solution approaches a point where y x becomes unbounded c The approximation extends beyond where the solution is valid The solution approaches a vertical asymptote d The solution is unstable Solutions that have slightly different initial values diverge from the desired solution 9 a 523mm 3 1 implies y my yzw 2 lt 0 so y is decreasing b 330 m 3 1 implies y xy y 2311 m 3 lt 0 so 2 is concave downward c The Euler approximations of the solution near x 3 are greater than the solution 10y22x 1 3x12 8x 13 MA 266 SPR 00 REVIEW 3 FIRST ORDER DIFFERENTIAL EQUATIONS You should be able to recognize and know how to solve rst order differential equations that are either separable linear exact or homogeneous You should be able to evaluate integrals of the following types fpolymonial dz e du f u39 du including 1 1 az dz partial fractions 1 i391c r2 39 You should be able to use given values yzo ya to determine unknown constants in a solution You should know the relation of the graph of the solution of an initial value problem to the corresponding direction eld HOMOGENEOUS EQUATIONS dy y a F 2 d d l 3 E Letyrzvmodx r vand z u Substitilte to obtain 3 v Fv Solve the above e equation for v in terms of 2 Substitute v g to obtain a formula for the solution y of the original homogeneous equation EXACT EQUATIONS M3y Nomi o is exact if Muzy Nzy Find a function 1122 1 such that 4331 M x y and 1pm 1 N z 11 my Mamas My solve 61 Mm 12 my No11 for hy A solution 11 112 of the exact eguation then satis es d d l l Egg0M 2114211 1124111 dz M any N z 1 dz 0 so the general solution is of the form 1101 y c J Numerical Methods For Solving y f t y yto yo Create an M le to de ne the function f t y The function name and the le name should be the same Note that M les are not entered in the matlab comman39dgwindow but are external text les that are created with a text editor EXAMPLE If y m create an M le named f11m function zf11ty zsqrttamp39 The general syntax for the Euler tangent line method is gtgt tyeul39d le t0t naly0stepsize Note that stepsize t nal t0 n where n is the number of steps EXAMPLE To nd the Euler tangent line approximation of the solution of the initial value problem 3 t 111 3 where t 2 using stepsize h 05 39 gtgt t eul f11 12305 gtgt my F ans 10000 30000 15000 40000 20000 51726 The syntax is the same for the improved Euler method use rk2 in place of eul and the rungchkut39ta method use rk4 in place of eul To obtain the graph of an approximate solution on a direction eld enter gtgtplottyC where d o x 39 Omit C for a connected graph APPROXIMATE SOLUTIONS The matlab eul rk2 and rk4 can be used to obtain approximate solutions of tlminitial value problem 31 f 1y yzo go You should be able to use the formula yI 11 fzn 1 Ila 1M to evaluate 1 Euler tangent line values by hand Approximation methods may not give good approximations of the solution of quotthe initial value problem y fzy yzo go if 0 The initial value problem does not have a unique solution because either V f or f is not continuous at the initial point e The approximation extends beyond the interval where the solution is valid because either 1 t or yz becomes unbounded o The solution is unstable because solutions that have slightly di 39erent initial values diverge from the desired solution PROPEETIES OF SOLUTIONS If f has continuous rst partial derivatives then solutions of the differential equation 1 f z y satisfy d w ms 3 m2 103 mm a M2 me u If y 39m a solution of the differentiable equation 1 fz y Ify gt0atapointthenyisincreasingnearthepoint If y lt 0 at a point then y is decreasion near the point If y gt 0 at a point then y is concave upward and the Euler tangent line approximations are less than or equal to the solution near the point If y lt30 at a point then 11 is concave downward and the Euler tangent line apprmdmations are greater than or equal to the solution near the point The39I aylorexpansionoffaboutzci839 m fcf cr c f Iz c2 Q c3I a c MA 266 SPR 00 REVIEW 3 PRACTICE QUESTIONS 1 Determine whether each of the following differential equations is separable homogeneous linear or exact Brie y justify your answers a2iyz3y bz3y22y0 cx3y12zyl0 d22y1z 1g0 e121y21 y0 2 Find the explicit solution of the initial value problem 3 12 1 110 0 3 Findthe general solution of the differential equation zy 2y z 4 Use the formula 31 w to express the differential equation 5 E i Z in terms of a v and E I dz39 5 Find an implicit form of the general solution of the differential equation 9 v e da my 6 Find an implicit solution of the initial value problem 2zy1m 2y 0 y1 1 7 Determine approximate values at z 05 of the solution of the initial value problem y 3 y y0 1 by using the Euler tangent line method with h 025 x 316 f x 8 Use the given direction elds and the graph of an Euler tangent line ap proximation of a solution of an initial value problem to explain why the ap proximation is not a good approximation of the solution 312 a v 31123 110 0 by3y4 y1o 9 Consider the initial value problem y try 312 y3 1 a Is the solution increasing or decreasing near a 3 b Is the solution concave upward or downward near a 3 c Are the Euler tangent line approximations of the solution near 2 3 greater than or less than the solution 10 Find the rst four nonzero terms of the Taylor series about c 1 of the solution of the initial value problem 3 may y1 2 MA 266 SPR 00 REVIEW 3 PRACTICE QUESTION ANSWERS 1 a homogeneous exact b homogeneous c none of these types d linear exact e separable exact 22 2y1 c 1e 32 C39 311 74quot doquot 11 437511 17039 l y 3 6 z yzy 1 7y2l75 8 a The initial value problem does not have a unique solution near 10 The functions yt 0 and yt t3 are both solutions b The approximation extends beyond where the solution is valid The solution approachw a point where y z becomes unbounded c The approximation extends beyond where the solution is valid The solution approaches a vertical asymptote d The solution is unstable Solutions that have slightly di erent initial values diverge from the desired solution 9 a 5y is 3 1 implies y my y2 Rs 2 lt 0 so y is decreasing b 311 4 3 1 implies y zy 3 2yy N 3 lt 0 so y is concave downward c The Euler approximations of the solution near a 3 are greater than the solution 10 y22z l3r 123c 13 jl fjl quot 8 Use the given direction elds and the graph of an Euler tangent line ap proximation of a solution of an initial value problem to explain why the ap proximation is not a good approximation of the solution magi 7 y10 a 2 311 yo o I L I s I x Vs 9 Consider the initial value problem 3139 my yz y3 1 a Is the solution increasing or decreasing near a 3 b Is the solution consave upward or downward near a 3 c Are the Euler tangent line approximations of the solution near a 3 greater than or less than the solution 10 Find the rst four nonzero terms of the Taylor series about c 1 of the solution of the initial value problem 3 22y y1 2 MA 266 SPR 00 REVIEW 3 PRACTICE QUESTION ANSWERS 1 a homogeneous exact b homogeneous c none of these types 1 linear exact e separable exact 1e23 W 33 C39 34 7 du39a 11 4 231 1 1 y 2 6z3yaFy21 7y2175 8 a The initial value problem does not have a unique solution near to The fimctions yt 0 and 110 t3 are both solutions b The approximation extends beyond where the solution is valid The solution approaches a point where y z becomes unbounded c The approximation extends beyond where the solution is valid The solution approaches a vertical asymptote d The solution is unstable Solutions that have slightly different initial values diverge from the desired solution 9 a 5y as 3 1 implies y my 312 N 2 lt 0 so y is decreasing b 211 s 3 1 implies y zy y 2yy z 3 lt 0 so 3 is concave downward c The Euler approximations of the solution near 1 3 are greater than the solution 10y22z 13z123z 13 MA 266 SPR 00 REVIEW 4 PRACTICE QUESTIONS 1 a Lip y 31 2y Evaluate Le Le2 Le b Lly y 41 4y Evaluate Lequot Ltequot Lftzez c Ly y 4y 511 Evaluate Lez cos t Le2 sin t Lsin t 2 Suppose that go is a solution of tzy ty y t2 and Ly t y ty y Evaluate Lyo t2 2t 1 3 Fillid the lairzest opeii interval for which the initial value problem I I 1 t1 t 2y t3 y1 3 y 1 2 has a solution 4 a Show that y t and ya tquot1 are solutions of the differential equation t2y tyly0 r b Evaluate the Wronskian Wtt1t c Find the solution of the initial value problem t y y 0 111 2 u 1 4 In Problems 5 7 nd the general solution of the homogeneous differential equa tions in a and use the method of undetermined coef cients to nd the form of a particular solution of the nonhomogeneous equations in b and c 5 ay quot5y396y0 111 5y 6yt2 c y 5y 6y equotcos3t 6ay 6y 9y0 b y 6y 9y t6 0 yquot 6y399ye cos3t 7 a y 2y 10y 0 b y 21139 101 e cos3t c y39 23 10y e cos3t 8 Find the general solution of the di 39erential equation y y 4t 9 The di erential equation tzyquot ty y 0 has solution 311 t t a Use the method of reduction of order to nd a differential equation satis ed by v where yt tvt is a solution of t y ty y 0 b Solve the differential equation in a to nd a solution of tauquot ty y 0 that is not a constant multiple of 31 c Find the general solution of the differential equation t y ty y 0 MA 266 SPR 00 REVIEW 4 PRACTICE QUESTION ANSWERS l a LIB 0 Lequot 0 Le 6e b Le2 0 His o Lt equot 6e2t c Le2 oost 0 Mequot sint 0 Lsint 2sint 4cost 2 Lyot 2t16t 4t1 3 0 lt t lt 2 4 b Wtt1t 2t1 cy 3t t 5 a y Olequot Cae b yAt2BtC c y Ate2t B oos3t C sin3t 6 a y Chequot Cate b y t3At Be339 c y Ae Bcos3t Csin3t 7 a y 01 0033t Ca sin3te b y Ae B so Csin3t c y tA cos3t B sin3te 8 1161 Cze 2t2 4t 9 a tan 3t v39 0 b y t 1 or y alt391 cat 1 a 0 c yClt Cat MA 266 SPR 00 REVIEW 5 VARIATION OF PARAMETERS If y and y are solutions of y py qy 0 with Wy1y2 a 0 we can nd a particular solution of y py qy g of the form yt u ty1 tu2 tyg t The simplifying assumption 1 Itin uayz 0 gives 3 ulyi uzy Then substitution of yy and y into y py qy g and simplifying gives the di 39erential equation 2 11in nay 9 Equations 1 and 2 can be solved for u 1 and 11 This gives the following If p q and g are continuous on an open interval I y and ya are solutions of the homogeneous differential equation 1 pty qty 0 and Wy1 m g 0 then the nonhomogeneous di erential equation 11quot pty qty gthas particular solutiozi m t g t y t g t Y quot1 quot Wm ant d 0 Wehyzxt and general solution 3 c1y1t Gays Yt HIGHER ORDER LINEAR EQUATIONS The theory is similar to that for second order linear equations This includes the interval in which solutions exist the form of general solutions of homo geneous and nonhomogeneous equations with constant coef cients and the method of undetermined coef cients APPLICATIONS Springmass system muquot 71 Icy Fo coswt The mass ofan39unforoed system mu 39nky 0 does not oscillate if the system is either critically damped 39y 2Vkm or overdamped 39y gt 2Wcm Otherwise the system oscillates A forced undamped system becomes unbounded as t 00 if and only if w k7m A damped system is always bounded Know how to nd steady state solutions Know how to interpret initial conditions and graphs of solutions Know how to use the formulas Rcos6 A Rsin6 B R V145 Bi A003wot Bsinwot Rcoswot 6 SPRINGMASS SYSTEMS Hm I IllIII I A Natural i 4 hagH of n sprin 3 397 Eta lap in VPQSIiIOI m3 k L 39 I mass m M r g 32 ftsec2 98 msec2 980 cmsec gravitational force weight my 3 rin constant k F rce p g 39 Displacement ut displacement from equilibrium position where my kL spring force lcL null orce damping constant 391 Speed damping force 390 applied external force Ft mass acceleration sum of all external forces mum my ML 0 710 F t Fnu39xt Mt hum Fa um uo u39o ua S M M Ems mo English feet slugs seconds pounds mks meters kilograms seconds newtons cgs centimeters grams seconds dynes F4coswot B sinwot Rcoswot 6M where Rcoswot 6 Rcoswot cos6 sinwot sin 6 so Rcos6 A Rsin6 B R VA5 3 tan6 BA MA 266 SPR 00 REVIEW 5 PRACTICE QUESTIONS In Problems 1 3 nd the general solution of the homogeneous differential equa tions in a and use the method of undetermined coefficients to nd the form of a particular solution of the nonhomogeneous equation in b 1 a y y 0 ym y t et 2 a y y y y 0 b y y y ye cost 3 ay y0 b y y te39 2 COS t2 4 Find the general solution of the differential equation y y t2 5 Find the solution of the initial value problem 3 2y y 0 310 2 y390 0 yquot0 1 6 a the general solution of the differential equation 11 5yquot 6y 26 cos2t b Find the steadystate solution of the differential equation y 51 By 26 cos2t 7 Use the formulas Rcos6 A Rsin6 B R A5 BE A wot Bsin wot Rcos wot 6 to nd R and 6 such that 3 cos 2t 4sin 2t Rcos 2t 6 8 For what nonnegative values of m will the the solution of the initial value problem mu 4n 8 cos 4t u0 4 u 0 0 become unbounded as t co 9 For what nonnegative values of 7 will the the solution of the initial value problem u 711 4n 0 110 4 u 0 0 oscillate 10 A mass that weighs 4 pounds stretches a spring 025 feet The mass is acted updn by an external force of 2cost pounds and moves in a medium that imparts a viscous force of 6 pounds when the speed of the mass is 3 feetsec At time t 0 the mass is 05 feet below the equilibrium position of the system and the was is moving upward at 5 feetsec Set up an initial value problem that describes the motion of the mass You do not need to solve the initial value problem 11 The equation t y ty 0 has solutions y1 1 and y t2 Use the method of variation of parameters to nd a solution of tzy tyquot 4t 12 The di erential equation tzy 2ty 39 21 0 has solution y t Find the general solution of t y 2ty 21 2t MA 266 SPR 00 REVIEW 5 PRACTICE QUESTION ANSWERS 1 a 1 01 028quot 038 b 11 tAt B Cte 2 a y Cle Cate Csequot b y Atze Bcost Csint 3 a y Cle Cge39V2 cos t2 Caequot2 sin t2 b y te39 At B cos t2 Ct D sin t2 1 4 1101 C2c08t038lnt t3 2t 5y iquot e te 6 a y Clea2 Cge 3 cos2t gsin2t b ysteady state gamut gsin2t 7 R 5 6 tan1 43 1r m 2214 s m 14 9 0 S 39y lt 4 10 45 215 16u Zoost 110 u 0 5 11 yu1t2u2 t4t t2 44 12 y Clt Czt2 2t2 lnt MA 266 Fall 00 REVIEW 7 THEOREM 711 If F1 Fm and all of the firstiorder partial d t 1 1 n er1va1ves 8w18wn 8w1 83371 R o lt t lt al lt 331 lt ldn lt 331 lt and ta1v is in that region then there is an interval t tol lt h in Which there is a unique solu tion 331t 1t 01316 nt of the system of firstmrder differential equations are continuous in a region 331 F1ta1vn 33 Fnta1vn that satis es 331t0 331 33t0 339 THEOREM 712 If p11p12 pmg1 g are continuous in an open interval or lt t lt and to is in that interval then there is a unique solution 331t 1t 01316 nt of the system of firstiorder linear differential equations 331 p11tw1 39 39 39 91 that satis es 331t0 331 33t0 339 The solution is valid for all of I UNDETERMINED COEFFICIENTS The method can be used to solve systems of the form X Px gt Where P is a constant matrix and the components of g are polynomial exponential or sinusoidal functions or sums of products of these The procedure is similar to that for linear second order differential equations The coefficients are vectors in the case of systems Also if gt cc Where 56M is a solution of the corresponding homogeneous equation try ateAtereM instead of ate MIXING PROBLEMS Flow through several connected tanks leads to a system of linear differential equations You should be able to set up and solve these problems 2 331 7 a b 31 To solve the system i lt0 d 332 b first7 solve the equation d A 0 for eigenvalues For each eigenvalue 7 solution of the system of linear equations G A 1b 20 051d A 20 gives a corresponding eigenvector lt51 gt The system of differential equations then has solution 61 gt lt51 6 562 52 If there are two distinct real eigenvalues7 the corresponding two solutions form a fundamental solution set If there is a complex eigenvalue7 the real and imaginary parts of the corre sponding complex solution form a fundamental solution set Note that eltg m cat cos t z39sin t and 6017507 cat cos t z39sin t You need to use only one of the two eigenvalues or z39 and or z39 to nd a fundamental solution set If there is only one real eigenvalue7 then a fundamental solution set is formed 331 51 2 by the corresponding solution eAt and the solution 331 1 teAtJr 771 6 where m isasolution ofthe system 562 52 772 772 a771 5772 51 0771 d A772 52 This system has a solution in the case of a double root A solution of the system gt a b 331 is of the form 32 0 d 32 31 a As t varies7 the solution traces a parametric curve in the 132 132 alum plane7 called a trajectory of the system The slope of the trajectory in the wag plane at a point on the trajectory is given by the formula for the slope of a parametric curve7 dilig E cow dial 331 dt Note that the formula 331 a b 331 551 75102 32 C d 132 0331 dw2 can be used to nd 331 and 332 at any point in the wag plane Also7 the signs of 331 and 332 give the direction of the trajectory at the point That is7 31 is increasing as t increasing if 331 gt 0 and 31 is decreasing as t increases if 331 lt 0 EXAMPLE The trajectory of the system lt3 1 1 gt at 17 2 2 32 1 2 satis es 331 i 3 1 1 i 3 2 i 1 w 2T12 2T1 4T 339 332 3 The slope is then w ll T 3 331 1 gt 0 implies 31 is increasing and 32 3 lt 0 implies 32 is decreasing See graph The MATLAB command pplane can be used to nd trajectories of a sys tem of two rstiorder linear differential equations The Graph menu on the PPLANE Display Window can be used to plot the solution variables against the independent variable t 4 MA 266 SPR 00 REVIEW 7 PRACTICE PROBLEMS 1 Find the solution of the initial value problem 331 1 1 31 3310 1 332 1 1 32 3320 3 39 2 Find the general solution of the system Find the general solution of the system gt lt1 0 1 2 3 1 1 132 4 Find the general solution of w 2 0 1 3 6 32 1 1 32 1 331 i 0 1 31 2 5 Find the general solution of 7 lt1 0 2 lt3 it 6 Find the general solution of w 2 0 1 66 32 1 1 32 1 7 Express the differential equation 3 y y y as a system of rstiorder differential equations 8 The equations 31 y 32 y transform the secondiorder equation t2y 2233 2y 0 into the rstmrder system 331 32 16210 2 2165102 2501 07 ltgtClltigt02lt igt39 Use the above information to nd the solution of the initial value problem Which has solution 76221 2762 2y 0 311 1 y 1 339 5 9 Find the trajectories that corresponds to each of the given systems of differential equations we M solution 3 31 lt lt3 ogtltEZgt7 32gt altgteto2ltltgttetltgtetgt 1 7 1 t 37 solution C1 1 e 02 6 1 H 6 10 Pure water ows into Tank 1 at a rate of 5 galmin The wellimixed solution from Tank 1 then ows at a rate of 5 galmin into Tank 2 The solution in Tank 2 ows out at a rate of 5 galmin Set up and solve an initial value problem that gives the amount of salt in Tank 17 011L 7 and the amount of salt in Tank 27 332t if Tank 1 initially holds 50 gallons of brine With concentration 1 lb gal and Tank 2 initially holds 25 gallons of brine With concentration 3 lb gal 11 Tank 1 initially holds 50 gallons of brine With concentration 1 lb gal and Tank 2 initially holds 25 gallons of brine With concentration 3 lb gal The solution in Tank 1 ows at a rate of 5 gal min into Tank 27 While the solution in Tank 2 ows back to Tank 1 at a rate of 5 galmin Set up and solve an initial value problem that gives the amount of salt in Tank 17 011L 7 and the amount of salt in Tank 27 32 12 Find the slope of the trajectory in the wlwg plane of the solution of the 33 2 3 31 system lt lt gt lt gt at the pomt 17 1 and sketch a short 32 3 1 32 segment of the trajectory as it passes through the point 17 1 Your sketch should indicate the slope and direction of the trajectory at that point MA 266 SPR 00 REVIEW 7 PRACTICE QUESTION ANSWERS 131 i 1 1 27 1 50921 46 31 i cost t sint t 239 ltw2gti01lts1ntgte C2lt C0stgt 31 i 0 t 0 t 1 t 3 ltw2gt01lt1gte 02 lt1gtte 0 6 31 i 0 7t 1 727 1 t 4 ltw2gt01lt1gte C2lt1gte lt1 6 31 i 1 t 1 it 3 5 ltw2gt01lt1gte Cglt1gte 2 31 i 0 t 1 2t 2 it 0 6 ltw2gt01lt1gte Cglt1gt lt1 e 1 131 2 7 332 333 8 y t2t2 33g wg wnglert 9a G b B c A dC QM 512 35 lt2 2896 ltgt50lt1gt1025ltgt5 1112 11130 355 lt2 Z133 lt32 ltgtltigtlt31gtW 12 Slope 2 Math 266 Review 4a Spring 2001 Complex numbers and the complex exponential function The complex numbers are needed to provide roots for certain polynomials such as 02 71 The roots of this particular polynomial are called it The convention is that the complex number a bi is graphed as the point a7 b in the plane We can also think of the complex numbers in terms of polar coordinates so that a bi rcos6 isln6 where 7quot a2 b2 and 6 arctangentba Example 1 i 0057T4 BMW4 ew4 Arithmetic with complex numbers for addition and multiplication is not so bad Example 1 22 3 7 3i 1 3 2 7 4 7i and 2i3 7 5i 6710i3i75i2677i75716577i1177i For division it s convenient to have the idea of complex conjugation 7 the conjugate of Z abz39 is 2 a 7bi Here a 7 is often called the part of Z a bi and 77b 7 is the imaginary part of a bi NOT the imaginary part of Z is a real number The absolute value or magnitude of Z a 2b is xE V02 b2 Then division 0 dia bi 20 a 7 bi c clz39a2 b2 Example 23i17i 1i23i1i17i 2735i11 Euler7s Formula I eltamt eatcosbt islnbt In particular I em 005wt isinwt Conversely 005wt em e M2 and 5inwt em 7 e M2i Examples Use of Euler7s formula to nd roots and powers Integral powers are easy if Z 5e2quotquot5 then 23 53e6739i5 To solve TN 1 note that 1 e027 egM for any integer M Hence the numbers rk 827mm are all solutions of TN 17 for any integer k lt suf ces to take 0 S k S N 7 1 in order to get N distinct roots of unity Example The third roots of unity 17821137847 13 17 71i 7SD2 Complex numbers and Euler7s formula allow one to write the solutions to linear constant coef cient DE as either complex exponentials or as products of real exponentials with sines and cosines Fundamental Theorem of Algebra The polynomial pa x alxn l a0 0 has n complex roots counting multiplicities If the coef cients are real7 then the roots are either real7 or appear in complex conjugate pairs If n is odd7 at least one of the roots is real Example r3 71 0 Roots are 7quot 17 71 i Theorem If PD D a1Dn 1 a0 is a linear differential operator with constant REAL coef cients7 then if 205 is a solution to PDz 07 then so is the complex conjugate 215 and each ofthe real and imaginary parts ofzt Example y 4y 0 One solution is 82 The conjugate is 8 2 the real part is 005215 the imaginary part is 5m2t Each of these latter three is a solution also Example y y 07y0 1770 0 Method 1 One fundamental solution set is 6 7 e it Set y A8 Be Then 7 iAe 7 Be Then y0 A B 0 and y 0 iA 7 B 0 Hence A B Thus A B 127 and yt 8 e t2 00505 Method 2 Another fundamental solution set is 005t75mt Set yt 0100505 025mt and y 701527105 02005t So y0 1 011020 and y 0 7010 021 0 That is7 Cg 0 and 01 1 That is7 yt 00505

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