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# Differential Equations MAT 022B

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This 6 page Study Guide was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Study Guide belongs to MAT 022B at University of California - Davis taught by Staff in Fall. Since its upload, it has received 76 views. For similar materials see /class/187452/mat-022b-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15

Midterm 2 Study Guide MAT 22B Winter 09 The midterm will cover chapter 3 and sections 61 62 A study guide for most of chapter 3 has been posted Make sure you understand both the key concepts linear independence homogeneous vs non homogeneous Wronskian and its use difference between partic ular complementary general solution as well as the different methods solving second order homogeneous non homogeneous how to use the Wronskian how and when to use the methods of variation of parameters or undetermined coef cients reduction of order etc One thing to emphasize that was not emphasized on most study guides for second order homogeneous DE with repeated roots the solution is always of the form yt 016 Cgte For complex roots you solve for r A i w and then rewrite the solution as yt 016 cosm t 026 Sin t Note that there is no 239 in the solution as written It might come in via OZ or even 01 but in most instances that we ll see it won t be there Expect some conceptual questions like you did on the homeworks as this chapter has a lot of deep math ematical concepts For practice problems you can do more problems from the book any problems I would provide would be similar in scope with different numbers Review the concept homework problems Sections 37 38 will not be covered in depth in the exam The main concepts in those questions are already covered in the rest of the material Sections 61 62 These sections cover the Laplace Transform an important mathematical tool There are a few new de nitions covered in this section Know what piecewise continuous function is and exponential growth means You should also know what linearity means for an operator think integral or derivative Wronskian is a great example also and of course the Laplace transform This section deals mostly with integrating functions with against a kernel of 5 This means taking the integral of a function f t multiplied with 5 within the integral sign ie A00 ft e stdt The function f t is being integrated against is known as the kernel Note that this integral is improper If you need to review how to integrate improperly Also review further integration by parts and the method of partial fractions The method of partial fractions is particularly important in doing the inverse Laplace transform As a tip remember to simplify the problem as much as you can and work from there You will Lot be required to memorize different Laplace transforms You will be given the ones you need and possibly a few extra ones as well just to keep you on your toes There are plenty of problems in the book do some extra ones for practice The steps solving a problem using a Laplace transform are as follows 1 Figure out the Laplace transform of each term in the equation including the non homogeneous one 2 Rewrite the equation using the transform Isolate Ys and solve for it 3 Work backwards to express in terms of possible Laplace transforms Try to simplify as much as possible while looking at the chart on p 319 4 Rewrite solution using the inverse Laplace transform Good luck Angelo Carino Mat 22B Chapter 3 Study Guide A 2 d order differential equation DE often has the form PtY QOY Rt Gt 1 Or dividing by Pt y pOY qty gt 2 This is also known as the differential operator denoted by LM y pty qty gt If the 2 d order DE has the form of either equation 1 or 2 it is known as a linear DE otherwise it is a nonlinear DE If gt Gt 0 then the DE is considered to be homogenous otherwise the DE is nonhomogeneous An initial value problem IVP consists of a DB such as equation 1 or 2 together with a pair of initial conditions yto yo and y to y o When solving a homogeneous DE we usually consider that Pt Qt and Rt in equation 1 are constant thus we obtain ay by ey0 3 The characteristic equation using equation 3 is de ned to be ar2 br c 0 Before we can actually solve a 2 d order DE we would like to know if a solution exists this brings us to the Existence and Uniqueness Theorem If we have an IVP with the form of equation 2 and if pt qt and gt are continuous on an open interval I that contains to then there exists a solution to the DE and it exists throughout the interval I Example Find largest interval in which the solution of the IVP exists 930 tY t3y0 Y12Y 11 Step 1 Get IVP into form of equation 2 y tt23ty t3 I39t2 3ty0 Step 2 Find where pt qt and gt are continuous or discontinuous pt1t 3 Qt t3tt 3gt0 Only points of discontinuity are at t 0 and t 3 Step 3 Find largest interval I that contains to to 1 in this case We nd that this largest interval is 0 lt t lt 3 Thus this interval guarantees that the solution exists and is in fact unique Principle of Superposition it y and y are two solutions of the differential operator Ly y e pmy t qtty 0 then any linear combination my t czyz is also a solution for any 0 and cz in R any real number We de ne the Wronskian ot y and 3393 as it i 5 it 3 think of determinant Find the Wronskian for the given pair of functions 2 4 Let yi e l and yg e l 5 t3 pins Hitw a maintain equation 3 31 y I 39 2e and y 3 3e 39l hcrcl ore Wle2 equotll I r be l t given a pair of initial conditions along with lily 0 and that the Wronskian at tois nonzero then there exists a choice ot c and c tor which y cly czyz satis es the DE and the initial conditions lt there is a point tU Where the Wronskian ofyl and yz is nonzero then the family ol solutions y my t cpyw with arbitrary coef cients c and 02 includes every solution of 1M quot 9 mayquot quot qhh 0 H t l A is known as the general solution this along with anonzero Wronskian at any t is known as the fundamental set of solutions l tquot we have a general solution that SilliSllCS My 0 and a nonzero Wronskian at any t this also implies linear independence ll the Wronskian for all t is zero this implies linear dependence Determine whether the pair ot functions are linearly independent tit ost gtt 39 Sint w e m that at K gun Ltttt 1 mincm at anytwill imply that these two r vv iii iiiit iHMiquot gilt HEM iu t witth Wt osttt Sint 39osztt l Sing t l for all t thus these two functions linearly independent hel39s Theorem states that it yl and y3 are solutions ot the DE Ly 0 then the Wronskian of y and y is given by 4 Wtquot MU cxpt J ptttdl t Where is a ctmstnnt that depends on y and y2 thn solvin 11 2M Ol tl39L l ll i ll Utill either ht homo 39CflCOUS Case 1 01 nonhomo CHCOUS g 39ase 2 39ase l w Homogeneous DE thn solving a homogenetms lli we rst nd the characteristic equation 39l hen we find the roots to that equation via quadratic formula or factoring There can be three possible cases for the roots Case A Non complex roots Case B Complex roots Case C Repeated roots Using the roots we form the general solution to the homogeneous DE If we have an IVP problem nd the general solution and the derivative of that solution Plug in the initial conditions and solve for 01 and 02 Case A NonComplex Roots The general solution has the form y cle C263t Where r1 and r2 are the roots to the characteristic equation Example Find general solution of the given DE y 2y 3y0 Step 1 Find characteristic equation r2 2r 3 0 Step 2 Find the roots r3r 10rl3 Step 3 Form the general solution y Clet 026 Case B Complex Roots The general solution has the form y 016 cosut 026 Sinut Where it and 2 are given by the roots of the characteristic equation r l t pi solve for roots via quadratic formula Example Find solution to DB y 4y 0 y0 0 y 0 1 Step 1 Find characteristic equation 940 Step 2 Find roots use quadratic l 414 2 2 i2i Here we have 2 0 and p 2 Step 3 Form general solution yt C Cos2t c2 Sin2t Step 4 Find 01 and C2 Find derivative Plug in initial conditions solve for Cl and C2 We nd that c 0 and c2 12 Thus the solution to this DE is yt 12 Sin2t ase C Repeated Roots The general solution has the form y 016rt c2 Vtert Where r is root of characteristic equation and Vt t Example Find solution to DE 3 4y 4y0 Step 1 Find characteristic equation r2 4r 4 0 Step 2 Find roots r220r2 Step 3 Form general solution 2t 2 y cle c2 te Case 2 Nonhomogeneous DE There are three methods to solving nonhomogeneous DE s 1 Reduction of Order 2 Method of Undetermined Coef cients 3 Variation of Parameters If Y1 and Y2 are two solutions of a nonhomogeneous equation then their difference is a solution of the corresponding homogeneous equation The general solution of a nonhomogeneous equation can be written as y ClY1 02Y2 Yt Where Yt is known as the particular solution and c1y1 c2y2 is known as the complementary solution which is the solution to the corresponding homogeneous DE 1 Reduction of Order This method is only viable when we are already given a solution to the DE Given the rst solution to the DE yl to nd a second solution let yz vt y1t Plug y2 into DE Solve for vt Example Use reduction of order to nd a second solution of given DE tzy 4ty 6y 0 y1t t2 Step 1 Multiply y1t by vt to obtain yz Y2 V Step 2 Plug y2 into DE t2t2v 4tv 2v 4tt v 2tv 6t2v 0 Collect terms and we get t4v 0 Step 3 Solve for vt We know that v must be 0 for that equation to hold Thus v t C and thus vt Ct D put all coe icients into one Therefore yz Ct3 Note In solving for vt we will sometimes have to use separation of variables or the integrating factor method To do this for instance make a substitution such as y v and thus y v 2 Method of Undetermined Coefficients This method involves making a guess about the form of the particular solution Yt based on gt but with the coef cients not speci ed We then substitute Yt into the DE and attempt to determine the coef cients to satisfy the DE If we are successful then Yt is a solution to the DE if we are not we then modify Yt and try again until we are successful If gt is a product we must take into account all possible derivatives For sums of exponential polynomials etc solve individually for each term then add all of them up If gt is a solution of homogeneous DE then Yt tgt or Yt t2gt Example Find solution to DB y 4y 2e3t Step 1 Guess Yat based on gt Yt Ae Step 2 Plu into DE and solve for coef cients 9Ae 39 4Ae3 2e3t 9A 4A 2 A 25 Thus Yt 25 e3t If unsuccessful modify Yt and try the method once more 3 Variation of Parameters More general way to nd nonhomogeneous DE DE must be in form of equation 2 Find the complementary equation to the DE Find the Wronskian using the y and y2 from the complementary equation The particular solution is given by 0le y2gtay2j y1gipd Wy1ay2 W i 2 The general solution is of the form y c1y1 czyz Yt Example Find the general solution to the DE y 2y ye t2 1 Step 1 Find the complementary solution Note DE is already in form of equation 2 We nd that the complementary solution is y clequot cltet so we have y1 et and y tet Step 2 Find the Wronskian Wet tet em Step 3 Find the particular solution Plugging into the equation for Yt and sim lifying we get Yt 12etln 1 t2te tan39 t Step 4 Form the general solution yt clet clte39 12 etln 1 t2 t e tanquott

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