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Unformatted text preview: MAT 22B-001: Differential Equations Final Exam Solutions Note: There is a table of the Laplace transform in the last page. Name: SSN: Score of this page: Total Score: throughout this exam). © §¥ § ¤ ¨¦ ¤ $ " ©  %#! ¤ § , we have $ &  § © & ¥ ¤ ¤ 0©  & ¤ )(© & & Ans: Using the integration factor ¢ £¡ ¡ Problem 1 (5 pts) Solve the following initial value problem ( represents ¤ '¥ § & ¤ & So, we must have $ & § © & That is, $§ § 1 ©§ 2 & 2 & , so let’s take 8 IG¢ E! §  HF D © 1 & B C¥  A9§ @8 © 6 § & 753 64 From this we have   D 8 H F ¢ D § © P¤ 8 H F ¢ ) Integrating this, we get $ S¥ 8 D © Q¤ 8 D R H F¢ H F¢ "T VU© . So, finally we have $8 YD T  © ¦'XW! § ¤  H F¢ R Using the initial condition, we get 1 . Now, (1) becomes: (1) Score of this page: 8§ § Problem 2 (15 pts) Consider two functions and .  ¦"  Vb "T § .  p© §  ¦"  `9 "T f f f f f "f T 8 § © f §  f© ge 8 d§  c § § 8§ © 8§ except . $8§ 8 9§  h© pi 9d§  8§ on a" V`T c . So, these are linearly independent on . ¥ ¤ Thus " a § (a) (8 pts) Show that they are linearly independent on Ans: Consider the Wronskian of and over  ¦"  `9 "T (b) (7 pts) Can they be a fundamental set of solutions of some differential equations on ? Answer yes or no and state your reasoning clearly. [Hint: Check the Wronskian in this interval.] Ans: They cannot be the solution of this differential equation. The reason is the following. Let , be a fundamental set of solutions of the differential equation of the form , where . Then, from Abel’s theorem, " Wa § § 8 ¤ v¤ u ‰‡ ˆ†"  Vb "T  E© § 2 . Therefore, and 8§ W©xb ¤ ¤ §  C¥ 9 § wq s 8 ¤ v¤ u Rys €7£ q at ¦"  Vb "T ©¤ W9 §  t¥ b § rq ¤ s a" `VT ¥ ¤ 8 u¤ y § ‚  h© …„ƒ WX ¤  v © 0ˆ 8 d§  § c c However, from (a), tal set of solutions. are the fundamental set of solutions. cannot be the fundamen- $ $ A¦" $ ¦" ¥’ i§  ’™ " T ¢ eY fQD 8 S¥ 8 9u B D B ¢ 3 © ! §  g 88 u¤ ¥  §  ¤ S¥  § Af9u B B © ! § ¤  Finally the general solution is: ¥’ j§  ’™ " T © " U!¦vT ‘™  § qu " T w¦" ¥ j§   “ T© pe §  8 b §  8 h ¤ " u¤ u h © ¥  § ˜f9 § ˜iEe §  g Hence, qu $ e b†vT ‘™  D" § ¢ T© U! §  8 h " D  ¢ 8 Y b†" ¥ i§ t  ‘ " T© #ˆ § ˜ih u Therefore, by integration by parts, we get: e o¢ D § T Y ¢8 D§ n q © " o§  n oS¢ 8 D eD fY rT ¢ ¢ D q ¢ 8 ‘sT e fY D ™ T n ¢D o§ T p©  o ¢ nuY efY D ™ T e fY D ¢ D ¢ 8 ‘ ¢8D n © 8 o h u h n That is: o§ $ © n 8 o h u h n efY D ™ T e fY D o ¢ D ¢ 8 Q ¢8D n ¢ Representing these in a matrix-vector notation, we have: © 8 h f¦D ™ eY ¢  § T u h 8 ‘ D ¢ 8  eY  E© h ¢ fQD u ¥ h¢8 D must satisfy the following 1st order differential equations: $ 8 eY h ¢ f¦D ™ T 88 ¤ h 88 ¥ ¤ h u ¤u ¥ 9ih 8h u ih © § u h¢8 u ¥ ¤ u ¥ ¤ u ¥ ¤ 8 u u ¤ u  8 “¨T 8 ¤’ ¤8 u ¤ ’ ¤ ¥  h ¥ bf“¨T ¤ ¥ ˜ih 88 88 88 u ¤u u ¤u h ¥ v9im’ ¥ ¤ h ¥ 9ih ¥ ¤ h ¥ ¤ h and 8¤8 h Therefore, the nonhomogeneous part © h uh © uh © D ‘ u g © g ’ lT g ¥ ‘k Ans: Let us first solve the homogeneous equation: . The characteristic equation is . Hence we have , and . Now, consider the nonhomogeneous equation, and let the particular solution be of the following form: . Then, . We choose , so that . Thus, , and . Inserting these to the nonhomogeneous equation, we get: 88 ¤ h u ih 88 ¥ ¤ h ¢ u ¤u ¥ 9ih 88 ¤ h eD fY d© 8 ¤ g u uh ¥ ¤ d© 8¤8 h u ¤u ¥ Ijh ¥ D ©u ¢ 8 dev¤  •S“E”¤ © ¤’ T g 88 u ¤u h  W© 8 ¤ 8 h ¥ v¤ u h u ¤ h ¥ 9i0© g u uh  §  8 9 §  8 h ¥  § ˜f9 § ˜i0e §  ¤ u¤ u h © ¥ v¤ E© ©™ W! C¥ ¥ ¤ $§ © ¤’ “'T ¤ – –  ˜%¨T –  ©’ (V¨T 8h g – ¥ —8 ¥ ‘ ¤ Problem 3 (15 pts) Find the general solution to the following nonhomogeneous differential equation by the method of variation of parameters. Score of this page: $u ¥x  q T 8x "  $ ™ $ ƒD e ¢ "  4 D D©  ¥ ¢ 8 'T ¢ Eˆ § i¤ Taking the inverse Laplace transform simply yields: T x t‡ T Cx ¥ " " T "T vrx " © x ˆ¦ g In other words, we have: $ ©  " © … R "T † VU© "  p© „ Comparing the coefficients of both sides, we have R … ¥ „’ ™ C¥ T rx u  xR „ q 0T 8 d S¥ ¥ T x   T x  " T x €t˜%Crƒ¦‚t CTrx … ¥ R … x R™ ¥ d %C¥ ™ and rearranging terms, we have $™ … ¥ „ Multiplying both sides by "T ‚x ¥ „ Let’s represent the right-hand side by the partial fractions: $   T x " T x %C€t˜¦vr   T x  " T x %¨rƒ¦v€tP T " ™ ¥ x T x   T x  " T x €t˜%Crƒ¦‚t " , we have  $" %vT T x ™ ¥ x g © x e¦t "   T x  " T x ©  %C€t˜¦vr(!%  % ¥ x™ T 8 g © x !¦ ¥ x™ T 8 t x Since x t Putting the initial conditions, we have: $™ T rx g © x e¦t "  g   ¤ T  x ¥ d ilw¦ x t ™ T   ¤ T   ¤ x T  x w% '—%†'wQ . Taking the Laplace transform of both sides yields:  $ ©  ¤ ©  !  0! i¤ " D©¤  ¢ e 0Q ¥ ¤™ g x 8 g  x  } ¤z y ©  x ¦t~v|{0!Q Ans: Let T ¤ Problem 4 (15 pts) Solve the following initial value problem using the Laplace transform. Score of this page: . " o VT " n ¢ ˜Y vQD ˜ vY D 8 B T 9u B  ¢D $ o¢ ˜ vY D 8 C¥ bu “™ B ¢ ¢D B n  o "VT 8 Do uB " n BS¥ ¢ ™n  Ž uB B C¥ ¢ D u ‘ ¢ © ‘8 Ž $ ˜Y vQD ‘8 © ©  Ž 8 , we get o © ™n Ž ‘ u o  " U© ‘8 u ” ” •– 8 ” ‘ 8  ’ u © n , we get . –  o sT ™T ‘8 © ™ ‘ u u ” ” •– 8 ‘ ”’ u  “o u o $ . Taking © n ‘ u ™ ™T by ¥ © p §  —  W© . Taking ‘8 8 ” ¥ ‘8 u ”  XT ™Tn T u U© , we have:  E© ‘ u 8 ” T ™ ‘ u u ”  –  XT  n ‘ u  Ž – " U© – uT "  U© f f 2  XT f ‹ ‘– f© Š ¦AD – Œe „ T , we need to compute . Now let’s compute the corresponding eigenvectors. $ © PEe u 5 Therefore, we have the following general solution: From this, we have Similarly for From this, we have f – " ˜¦vT  (© u – T – ™ C¥ 8 ©’ T `'w% – ¥  ˜¦" ¥ – f – f ¥ ™T  © f " For Thus, ¥ Ans: The characteristic polynomial of this matrix is: o $ 8  o sT ™ ‰ u ‰ ‡n  " VT 8 o ‰ u © n ‰ Pn Problem 5 (15 pts) Solve the following system of differential equations. Score of this page: Score of this page: Problem 6 (30 pts) Consider a system of differential equations:  " " ›  —™š › —„ „ © © —  ›  Wh © is a real number and œ  . › where › (a) (10 pts) Use an inductive argument to show that œ 8 $ ‘ u u Yž u ž› Y› ž Ÿ 8 Yž ž ž› Yž Ÿ ž © s„ ž  › ™š ›  › [Hint: Induction means that you need to do the following: 1) Prove the statement is true for and ; 2) Assume the statement is true for and prove the statement .] holds for , we have Ans: Let’s proceed the induction. For Ÿ " U© $  © ‹  " " © „  ™š "  " ¥Ÿ  W© Ÿ Ÿ  W© œ  Ÿ  So, this is fine. For , we have " U© $ œ  „ " " © ›  —™š › Ÿ u„ ©  › Ÿ u Y œ 8 ž œ u Yž ž› Y› ž Ÿ ¡ u £ž „ ž €„ „ ‘ u u Ÿ 8 Yž ž ¥› u ž› Yž Ÿ œ  ž › ‘ u › ¢ž u¡  ž †› " Y œ u › ž u ‘ › ¢ž u¡ › ž ¦" ¥ › Ÿ  Yž 8 Yž ž ¡ u ¢ž  ¦" ¥ Ÿ  › ž › › ¥ Ÿ 8 ¡ ¢ž ž ž › ¡ u ¢ž  ¦" ¥ Ÿ  › › ›  —™š ›  £ ty Ÿ © " "  › ™š ©  ¡ u ¢ž  › ™š © ¡ u ¢ž   › ™š ©  6 . ¥Ÿ Therefore, this formula is correct for all " So, this is fine too. Now, suppose this relationship is correct for . Then, consider the case . ¯¬ ˜ž ° ®‘¢  ¯ u Y ž ­ ¬ ‘ ²± “b® Y ‘ ¢  ¬  ¯ ‘ 8 ž ° ³ƒœ F bG‘ ¢ ±® « ¨ ¢ž « ¨ u £ž §¦ ¨ 8 8 ¢ž ¦ F¢ « ¦ 7 ¬   ¢D ¬ ¬  ™š ¢D ¢¬ D ¬ 8§ ¬ D ¢ D§ ¢ D F¢ œ ¢ «   ¯¬ ˜ž ­ ¢ž ¨ « « —š  ™ª ®‘¢  ¯¬ Až ­ £ž ¯ u Y ž ° u ¢ž ¨ ¨ ¬ ‘ ¦ ²‘b5‘ ¢ § › ±® ® G‘ ¢ ¦ › ž ›¦ Y› ž  › ™š u Y œ 8 ž ‘ u Ÿ 8 Yž ž u ž› Yž Ÿ © © © Ÿ © ¨ ¢ž § ž§ ž ¦ Using (a) and the hints, we have $ ¨ £ž © Ÿ ž „§ § © ¤ ¥¢ D ¦ Ans: Recall the definition of a matrix exponential: $  © %CT › Ÿ  8 Yž§ ¨ 8 £ž §  ›  © 8Yž 8§ " ¦vT  © „Ÿ Ÿ Ÿ ž§ ¦  ¨ ¢ž § ¦ ¨ u ¢ž " © ¦‚T › Ÿ  Yž§ u  › © §§ Yž u Ÿ and ¨ ¢ž © „Ÿ § ž§ ¦ ¦ ¤ ¥¢ D (b) (8 pts) Compute [Hint: . ‹ © " 7†b ´ § ´ © e  ‹ ¤ u ‘ Y ¢  D© 0ˆ §  ´ ´ $ Y ¢   †¬ Y D ‘  ¶u ¬ ¢ D" 9¦vT §  ‘ u Y ¢  ¶¬  D ™š © ¤ u ‘ Y  8 ‘ u .  ¶¬ Y D ‘  †u ¬ ¢ Y D" b¦–T §  ‘ u ¢  ¶¬ 8 Y D Y œ u ¢ F u ¢ ‘ ‘ ‹ © " !¦b ´ ¢  D© dµ §  ´ It is easy to verify that ´ Using (b), we clearly have: © " x¦9 . Note that ‹ (c) (6 pts) Find the fundamental matrix satisfying This is different from . Ans: For this initial condition, it is clear that ; ´ © e §  — " " ™š œ " 9  ¶¬ Y D ‘ u ¢ †¬  Y D $ ™š § ‘ u  ¶¬¢ Y D ¡8 F œ u ¢ ‘ u G¢  ¶¬ Y D ‘  †¬ u ¢ Y · {š D " Pd¦–T §  ¥ b " ‘ u ¢ ™ª  ¶¬ 8 Y D¸ " Y  ¥ ¦vT §  ¥ " F u ¢ ‘ ³‘ ƒœ u ¢  ¶¬ Y D   ‘  †u ¬ ¢  †¬ Y Y D" b¦–T D  ™š § ‘ u ¢ ‘  ¶¬  ¶u ¬ ¢  ¶¬ 8 Y Y Y D D" b†‚T §  D Y œ u ¢ F u ¢ ‘ ‘ u ¢ ‘ u ¢ ‘ , we have the following solution: " †b —  §  § ´ © © © p §  — Ans: Using the fundamental matrix Therefore, . is given. Then, solve this IVP. " " ™š © " !¦9 — œ " (d) (6 pts) Suppose that the initial condition: u ou T ™Tn ” © ‘8  Ž or " o VT " VT . Here, . ‘8 n " ¾ VT Y D o ¢8 " VT n ‘8 n Ž © ‘8 $ ‘8 Ž must satisfy Ž ‘ u  Ž 8 8 B™ $ 8 Y¦D o § “S¥ B ¢ 8 8 § B ¥ B 8 8 B o  B T § %™   8 B T § 8 B   Y ¦D ¢8 ¥ j§ "T `© o™  $ © n © n ‘ u . Thus, ‘8 u ” Ž . $ ƒW©   % 8 – ¥ © ¥ u %™ B ¥ uBn © TB —u %™ T —u B  n ©   »© ‘8 ” • 8 ’ ’ ‘ 8  ” “o u u © o™ n YD ¢8 ©  © o $ 8 © ¼ § i¤  ” u YD ¢8 ‘ u ” u ‘ u  Ž © ‘ u u ¥8 u ” ’ T n „ T ‘8  — — ‘ u ” 8 T u ‘ u u ” ’ º ’ n ‘ u – ¥   E© ” • 8 ’ sT ’ ‘ u  ” “o u u © n  (© u ¥ ‘8 º T ’ T ‘8  Ž  Ž Q– ‹ ©  ‘  ‘8  Ž u ¥ ¢ 8 Y D u § © ‘8 — ‘ Ž Y ¢8 D‘8 . The vector o™  o ™ ½`8 o™ u BS¥ ¢ 8 Y D n n B  — 8 S¥  §  —uB B ‘ u ‘8 o@"  © , we get , we get Ž . Taking . Taking . Let’s compute the eigenvector ‘ u 10 Thus, the general solution is: can take other values. For example, the other choices lead to From this, we have , i.e., To compute , let us assume that the following equation: From this, we have – sTp© f f f – ©fu So, we have repeated eigenvalues f – T f f º uT ¹ –f ¥ Ans: The characteristic polynomial of this matrix is: o $ 8 ‰ u‰ n º T ou ¹ T u n 8 o ‰ u ‰n © Problem 7 (20 pts) Solve the following system of differential equations. Score of this page: Score of this page: © #Á¤ –¥ ¤ S¥ ¤ B À are positive. There are three and  – Ç ÈÀ À¥ ‚ˆ8 © … –¿ –  Å ¿  p© ‚¥ B . Let B ‚¥ BÀ ¢†¢ B l¥ À¥ –ˆ8 – – À¥ lr8 –¿ –¿ © 0! Ç VB  ¿ Ans: The characteristic equation is clearly consider the roots of . BÀ ¢†¢ ¿ (a parabola) when  ç [Hint: Think the graph of cases you need to check.] approach zero as Ä Bonus Problem (20 pts) Show that all solutions of if are all positive real constants.  – Å Ç `B  T À 8 u © , i.e., two distinct real roots. In this case, because , , , both roots are negative. (Consider the graph of which is a convex parabola.) Let these two roots be and with and . Then, the general solution is: a Í#Ë ÍpÉ a Å ¢ $  as  AÀ Ñ© @T tend to  7§ ¢ ÏD Ï ˜D 8 S¥ ƒ9u B ÎD B ¢ ¢ and Ä Ë Ì© ÎD É Ê© – Æ  Ç ¿ – © ! § ¤  take, both as . – Ä B 8  !§ and  uB    § ¤  No matter what values and . Thus,  ¿ Case I: because a ‚iË B  Щ T 8À a –'É u © Case II: . In this case, we have repeated roots , which is negative since and . The graph of in this case is a parabola whose apex is touching on -axis at . The general solution in this case is: –  –   Å  AÀ p© @T ¿ $ I8 fQD § 8 S¥ I8 f¦bu B B Ó G¢ Ò Y Ó G¢ Ò Y D H H  AÀ T @ Ä Â `§ Ä Â e§ . ¿ since  Ía and grows TB ¿    § i¤ ÓHÒ I8 G¢ fY D §   ©  ¿ ¿ © ! § ¤  ÖÕ Ô SeÀ T u Æ – . In this case, we have two complex roots, 8À ¿ Ç  – tend to as as Ò ÓH I8 G¢ D Ç ”À ÓH I8 G¢  va B § Ò fY D ¿ u T À 8 © Æ Case III: ¿ But both and much faster than . Thus, – So, we have the general solution: o TB ¿  ¿ u Ön 4 8 Õ jI8 G¢ fQD S¥ B ÙÓ HÒY Ä Â €§ . 8À Ä Â e§ o 8À § TB ¿    § ¤    ¿ u Ön ÓHÒ I8 G¢ fY D    § i¤ 11 as as § ÙØ× Ó H ÒYD %˜jI8 G¢ fQ9u B © ! § ¤  Therefore, in all cases, we established , $ Again, because of the factor dies out eventually.) . (The solution oscillates but  Çx   Çx  ¿ FÓ ¡FÛ Ó § FÓ ¡FÛ Û Ó u Çx Å  ¿ § YÛ àß ² £Û  ¯Ü Çx    d §  12 Û Çx u  y ©  x {0!QÁÚ ¿ 4 ÕÙ ÙØ %ƒ× ¢ÓD £ y Ý Ü tÞ£ƒI§ 1  x Ú d¦Á Y E! §  y© u Å Table of Elementary Laplace Transforms ...
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