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# 530 Review Sheet for MA 26600 at Purdue

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Date Created: 02/06/15
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 f equot an I u39 du including 1 1 Ez P dz partial fractions 1 1391c 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 1 Substitute 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 EQUATIgNS M3y New 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 equation then satis es d Eggmm new Marni Mxy Maui 0 so the solution is of the form Ma y c 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 w ms 3 m2 wj ze 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 dy b z3y22yE0 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 in do terms of 311 and E 5 Find an implicit form of the general solution of the differential equation dy 2 u da my x 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 A 316W 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 2 bis3 j iij y1o a v 31123 110 0 hm v 11115 I Ir 1 III 11 IVll 1111 Il1quotIquot39IquotIVquot Iquot39I f39 Iquot39I we I II I 11 II I IIIquot IIIV Ill 1 1 I ll I II I 11 I 111 1 1quotquot u II 114 Ill39ll II I 111 11quot u 1411 II I 7 I I I IIIIIIvII 11397quot Mtglylr 111 IIIVIII Iir u4 tfl t 411 11711111 azzz u lilil 21 I rrryIIE 1 44ana A441 eaaaa s r a a oo I l I B I I P I B C y 112035 M0 1 d 2 10y 118 110 1 furn 1011Ir1yr1v l 4 I1I11r 1391 39 1 111111 11 1 I11I1111 5 I1111111 1111411v1 1391 11111111 1111113911 1 IIIIEII II 111 1111w111 111 IIIIIII39I 1 11111 11 111 IIIIVIII Ill II39IflI 1 1 7 IIIv 11 a If3939 1 2 c 9 Consider the initial value problem y try 312 y3 1 a Is the solution increasing or decreasing near 1 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 I e 75 32 C39 7 doquot 11 WW1 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 w 3 1 implies y my y2 Rs 2 lt 0 so y is decreasing b 311 44 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 u lj f 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 a 2 311 yo o s 3quot L h ltltL44 ss 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 17 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 1 e2 rm 33 C39 34 7 du39 11 4 25339 r 1 y 2 5 5 2 hll 4quot 6 z y a F 12 1 7 ya 175 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 7 a y 21139 10y 3 0 1 a Lilli Iquot 31139 21 Evaluate 14839 432 Lie ti I 1 b y 21139 10y 8 60530 b Ly y 4y 4y Evaluate Lequot Ltequot Lftzez c y 2 my e mat c Ly y 4y 511 Evaluate Lez cos t Le2 sin t Lsin t 8 Find the general solution of the di 39erential equation y y 4t 2 Suppose that go is a solution of tzy ty y t2 and LM tzyquotty39 y Evaluate myquot t2 2t 1139 9 The di erential equation tzyquot ty y 0 has solution 311 t t 339 Fill the largest open interval for WhjCh the initm Value pmblem a Use the method of reduction of order to nd a differential equation 1 Ey t 211 t y1 3 y l 2 has a solution satis ed by v where yt tvt is a solution of t y ty y 0 4 Sh that d t1 l t h m b Solve the differential equation in a to nd a solution of tagIsl typo y in t an m are so up Ions 0f t e d mung equation tauquot ty y 0 that is not a constant multiple of y 0 Evaluate the wronSkian W rlxt c Find the general solution of the differential equation t y ty y 0 c Find the solution of the initial value problem 191 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 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 l 4 b vytt1t 2t1 cy 3t t 5 a 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 as tA cos3t B sin3te 811 6391 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 gt has particular solution mtgt y t9t Y quot Wy1y2t W1yy2t39d 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 APPLIQATIONS 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 N form fe g39l39ln of n a Spring 1 E39zul39li law39sn ere inon ui m3 k L I mass m W i g l r g 32 ftsec2 98 msec2 980 cmsec gravitational force weightg my orce spring constant k m 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 m M Ma 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 c086 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 8y y0 b y y ti 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 7 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 y tAt B Cte 2 a y Cle Czte Csequot b y Atze Bcost Csint 3 a y Cle ageW cos t2 Cae39m sinwitz 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

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