MAT 22B-001: Differential Equations Final Exam Solutions Note: There is a table of the Laplace transform in the last page Name: SSN: Total Score: Problem 1 (5 pts) Solve the following initial value problem ( represents throughout this exam) Ans: Using the integration factor, we have (1) So, we must have That is, have!#"%$ $ '&)( +* From this we, so let s take -, /10&, 0& 2, 0& Now, (1) becomes: Integrating this, we get, 0& 3, 0& 54 Using the initial condition, we get 4 768 So, finally we have 9:6,; 0& 1
Problem 2 (15 pts) Consider two functions and & (a) (8 pts) Show that they are linearly on6<8= = independent '& over>68 Ans: Consider the Wronskian of and? @& A CBBBB & BBBB & 6 & &? @'& ED F '6< F '6< Thus on except So, these are linearly independent on (b) (7 pts) Can they be a fundamental set of solutions of some differential equations GH > JI ' 9 on68<= =9? Answer yes or no and state your reasoning clearly [Hint: Check the Wronskian in this interval] Let LK, Ans: They cannot & be the solution of this differential equation The reason is the following be a fundamental set of solutions of the differential equation of the form GM ' +I > N 9, whereg I%OP4 >68? Q LK Then, from Abel s theorem, & :D 9 SRTUO >68 VXWZY LK & are the fundamental set of solutions However, from (a),? @& X at - Therefore, and & cannot be the fundamental set of solutions 2
& Problem 3 (15 pts) Find the general solution to the following nonhomogeneous differential equation by the method of variation of parameters [ 6\] ^6-\] 5 _ Ans: Let us first solve the homogeneous equation: The characteristic equation is` &a ` 6 \8 ` 6 b` +c 9 have LKA d, & Hence we, and & d, ;fe Now, consider the nonhomogeneous equation, and let the particular solution be of the following form:g A hikb ' fkb h & ' & Then,g -h K LK hjk1 K h & & h & & We choosehik h &, so thath K LK h & & 9 Thus,g hik' K h & &, andg dh K K hik' K h & & h & & Inserting gk [ these g 6l\ to g the nonhomogeneous h K K hik' K equation, h & & h we & & get: hik' K h & & \mqhik' LK h & & h K K h & & hikbq K K 6 \] fk> h & Q & & 6 \] & h K K h & [, & h K 6 c,;fe h & the nonhomogeneous part Therefore,hiK andh & must satisfy the, & h following K,3;fe h 1st & - order differential equations: [, & h K 6 c,;fe h & Representing these in a matrix-vector n, & notation,, ;fe nwe h K have: 3, & 6 c, ;fe Qo n That is: h h K & o n, &, ;fe [, & 6 c, ;fe Qo ; K n o p6, q h & o n o n 6 c, ;fe 6r, ;fe 6s[, &, & 5o n o q Therefore, by integration hikb X 6 by parts, we get: [ t i V>, ; & h & 76 uq c[ 6vV>, e n, ; & 6, e o Hence, g A -hikb ' fkb h & > & A p6 ] j w6 u q c[ 6v 76 c \ Q\ j ) Finally the general solution is: * K' fk) 5* & & g * K', & 5* &,3;fe 6 c \ Q\ i 3
x Problem 4 (15 pts) Solve the following initial value problem using the Laplace transform 6 c 3, e iq Q Ans: Letg x3 y{z L}~tx Taking the Laplace transform of both sides yields: x & g x3w6xv a6 w6 c tx g xw6l iq@ g txa xr6 c Putting the initial conditions, we have: tx & 6 c x g txa x6 c x 6v Sincetx & 6 c x xr6vbtx 6+, we have g x tx 6 ƒxr6+btx 6 c 2tx 6vƒxr6 6 xr6vbtx 6+ Let s represent the right-hand side by the partial fractions: x 6 xr6+ 4 xr6 c bytx 6 ƒxr6+btx 6 c Multiplying both sides and rearranging terms, we have 54 @x & 6q u +c4 @x \ +c 4 x & 6 q x u Comparing the coefficients of both sides, we have p 768 4 In other words, we have: g xx xr6v 6 x 6+ Taking the inverse Laplace transform simply yields: i ˆ -, 6, &,ƒe tx 6 c 4
n2 K & o n 68 c 6s o n K & o Problem 5 (15 pts) Solve the following system of differential equations Ans: The characteristic polynomial of this matrix is:,)š`[ 6 A ŒBBBB ` 6 c 6: ` BBBB ` b` w6\< ` & +c ` 6 u ` 6vb` u A -2 Thus,` 7 6 u Now let s the corresponding eigenvectors For` 7, we need to computež K n by 6 c K 6: c o K K & n o From this, we have K K 6 c K & - Taking K K c, we getž K n c o Similarly for` 76 u, we have: n 6 c 6 c From this, we have & K & & 6: 6s o K & & 9 Taking & Therefore, we have the following general solution: & n o K 7, we getž & n 68 o F * KŽ K, +* & Ž &,3;L * K n c o, 5* & n 68 o,3;l n c]* K>, +* &, ;L * K', 6 * &, ;L o 5
D ž Ÿ ž ž Ÿ ž ž ž Problem 6 (30 pts) Consider a system of differential equations: ša 9 where is a real number and (a) (10 pts) Use an inductive argument to show that sž ž Ÿ ž š ; K ž ž &; K ž ; & ž ; K [Hint: Induction means that you need to do the following: 1) Prove the statement is true forÿ 9 andÿ 7 ; 2) Assume the statement is true forÿ and prove the statement holds forÿ ] Ans: Let s proceed the induction ForŸ 9, we have So, this is fine ForŸ 7, we have š K ša So, this is fine too Now, suppose this relationship is correct forÿ Then, consider the caseÿ ž K P ž ž Ÿ ž ša š ; K ž ž &; K ž ; & ž ; K ž K Ÿ š ž K ž K Ÿ š ž K Therefore, this formula is correct for allÿ OJ 6 ž ž ž &; K ž ; K Ÿ Ÿ ž K ž ž ž & K ž ; K Ÿ ž V K ž ; K
ž «Ÿ ž ««««(b) (8 pts) Compute, [Hint: ž and ž ž ŸT ž ŸT Ÿ Ÿ Ÿ 6v ž ; K ž K ž ; & & ] Ans: Recall the definition of a matrix exponential:, Using (a) and the hints, we have ž ž Ÿ ž Ÿ š ªša«ž ž) / š,, ž ž ; K Ÿ 6 ž & ž Ÿ ž ; K ž ž &; K ž ; & ž ; K ž K ž # >±[² ; K ž žb &,,,, ž ; & Ÿ 6+ ž & & ž / >± ; & ž K ž >±]² ; K ž žb ƒ³ 7
(c) (6 pts) Find the fundamental matrix satisfying >V% Note that 'N ; This is different from A Ans: For this initial condition, it is clear that X, ; K Using (b), we clearly have: µ d, ; K š, ; K 6v', ; K ; & K, ; K, V ; K 6 >, V ; K, ; K It is easy to verify that > 8
š (d) (6 pts) Suppose that the initial condition: ' š is given Then, solve this IVP Ans: Using the fundamental matrix, we have the following solution: A >V Therefore, F š, ; K 6 V>, ; K ; & K, ; K, V ; K 6 >, V ; K, ; K ªš{ 6v ; & K, ; K ƒ³ > 6 @2, V ; K, ; K / K š &, ; K, V ; K, ; K 9
BBBB n K & o n 6¹ u 6º u o n K & o Problem 7 (20 pts) Solve the following system of differential equations Ans: The characteristic polynomial of this matrix is: ` ¹ 6 u º ` 6 u BBBB ` & u ` u ` & 9S So, we have repeated eigenvalues` p6s Let s compute the eigenvectorž K n \ u K º 6s\ o K K & n o From this, we have\ K K 6 u K & - Taking K K, we getž K n c o Thus, K Ž K, ; & n c o, ; & To compute &, let us assume that & Ž K, ; & Ž &, ; & The vectorž & 6 `3 Ž & Ž K must satisfy the following equation: n 6\ u &, ie, 6º \ o K & & Ž K n c o From this, we have6\ & K u & &» Taking K & 6<, we getž & n 68 68 o Here, K & tož & n ( o orž & n 6 c 6 u o can take other values For example, the other choices lead Thus, the general solution i ¼ is: * K K 5* & & * K n c o, ; & 5* &<½ n c o j n 68 68 o ¾, ; & n * K * & 6 * c* & K c* & 6 * & o,; & n * Ka6 * & * & c* Ka6 * & 5c]* & o,; & 10
Bonus Problem (20 pts) Show that all solutions of 5À * Á approach zero as ÃÂ Ä if ÀV * are all positive real constants [Hint: Think the graph of ` &H là ` l* (a parabola) when ÀV * are positive There are three cases you need to check] Ans: The characteristic equation is clearly ` &ˆ À ` * p LetÅ ` U ` &ˆ À ` * and consider the roots ofå ` Æ À& 6 u *<Ç *8Ç ÀÈÇ Case Ç I:, ie, two distinct real roots In this case, because,,, both roots are negative (Consider the graph ofå ` which is a convex parabola) Let these two roots be` ÊÉ and` ÌË withép=í andë=í Then, the general solution is: * K',ƒÎ 5* &,bï No matter what values * K and * & take, both, Î and, Ï tend to as %Â Ä because É= andëe= Thus,  as Â Ä Case II: Æ À & 6 u * Ð In this case, we have repeated roots` Ñ6 À)(, which is negative since Ç and À^Ç The graph ofå ` in this case is a parabola whose apex is touching on` -axis at` p6 À)( The general solution in this case is: * K>,;fÒ /0&1Ó 5* &,3;fÒ /0&1Ó But both, ;fò /0&1Ó, ;fò /0&1Ó and tend to <Â Ä 6 À)( =Í as since and, Ò /0&1Ó grows much faster than Thus, i  AÂ Ä Æ À& 6 u * =v as Case III: In this case, we have two complex roots, ` 6 ÀAÔ5ÕQÖ u * 6 À & So, we have the general solution: * K',3;fÒ /0&1Ój bøù n Ö u * 6 À & o 5* &,3;fÒ /0&1ÓjÙ Õ" n Ö u * 6 À & Again, because of the factor, ;fò /0&1Ó,  as Â Ä (The solution oscillates but dies out eventually) Therefore, in all cases, we established i  as AÂ Ä o 11
Û Ó Table of Elementary Laplace Transforms Å -y ; K QÚÁx@ ÚÁx3 y{å @ ÛK x Ç 1ÜS ÝÞOJ 1 ÛQß à Ü ² x Ç, Ó Û; K Ó x Ç ƒøù Û Ó x Ç Ù Õ" Û Ó x Ç 12