Introductor soution manual elemenrary differential equations with boundary value probolems

INSTRUCTOR’S SOLUTIONS MANUAL LEE JOHNSON Virginia Tech JEREMY BOURDON Virginia Tech to accompany ELEMENTARY DIFFERENTIAL EQUATIONS WITH BOUNDARY VALUE PROBLEMS Werner Kohler Virginia Tech Lee Johnson Virginia Tech This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors Copyright © 2004 Pearson Education, Inc Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116 All rights reserved No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher Printed in the United States of America ISBN 0-321-17323-6 SC 08 07 06 05 CONTENTS Chapter 1: Introduction to Differential Equations Chapter 2: First Order Linear Differential Equations Chapter 3: First Order Nonlinear Differential Equations 26 Chapter 4: Second Order Linear Differential Equations 51 Chapter 5: Higher Order Linear Differential Equations 101 Chapter 6: First Order Linear Systems 115 Chapter 7: Laplace Transforms 178 Chapter 8: Nonlinear Systems 214 Chapter 9: Numerical Methods 250 Chapter 10: Series Solutions of Linear Differential Equations 268 Chapter 11: Second Order Partial Differential Equations and Fourier Series ?? Chapter 12: First Order Partial Differential Equations and the Method of Characteristics Chapter 13: Linear Two-Point Boundary Value Problems Chapter Introduction to Differential Equations Section 1.1 This D.E is of order two because the highest derivative in the equation is y ¢¢ Order is This D.E is of order one because the highest derivative in the equation is y ¢ (Note: ( y ¢ ) π y ¢¢¢ ) Order is (a) y = Ce t Differentiating gives us y ¢ = Ce t ◊ t = ty Therefore, y ¢ - ty = for 2 any value of C (b) Substituting into the differential equation yields y (1) = Ce1 = Ce Using the initial condition, y (1) = = Ce Solving for C , we find C = 2e -1 y ¢¢¢ = t3 t2 y ¢¢ = t + c1, y = t + c1t + c , y = + c1 + c t + c Order = (a) arbitrary constants y = C1 sin t + C2 cos t Differentiating gives us y ¢ = 2C1 cos t - 2C2 sin t and y ¢¢ = -4 C1 sin t - C2 cos t = -4 (C1 sin t + C2 cos t) = -4 y Therefore, y ¢¢ + y = -4 y + y = and thus y ( t) = C1 sin t + C2 cos t is a solution of the D.E y ¢¢ + y = (b) y ( p4 ) = C1 (1) + C2 (0) = C1 = and y ¢ ( p4 ) = 2C1 (0) - 2C2 (1) = -2C2 = -2 fi C2 = y = 2e -4 t y ¢ + ky = -8e -4 t + ke -4 t = 2( k - )e -4 t = \ k = y (0) = = y \ k = 4, y = y = ct -1 Differentiating gives us y ¢ = -ct -2 Thus y ¢ + y = -ct -2 + c t -2 = (c - c ) t -2 = Solving this for c , we find that c - c = c (c - 1) = Therefore, c = 0,1 10 y = -e - t + sin t y ¢ + y = g( t), y (0) = y y ¢ = e - t + cos t y ¢ + y = e - t + cos t - e - t + sin t = g \ g( t) = cos t + sin t, y (0) = -1 = y • Chapter Introduction to Differential Equations 11 y = t r Differentiating gives us y ¢ = rt r -1 and y ¢¢ = r( r - 1) t r - Thus t y ¢¢ - ty ¢ + y = r( r - 1) t r - rt r + t r = [ r( r - 1) - r + 2]t r = Solving this for r , we find that r( r - 1) - r + = r - 3r + = ( r - 2)( r - 1) = Therefore, r = 1, 12 y = c1e t + c 2e -2 t y ¢ = 2c1e t - 2c 2e -2 t , y ¢¢ = c1e t + c 2e -2 t = y \ y ¢¢ - y = 13 From (12), y = C1e t + C2e -2 t , which we differentiate to get y ¢ = 2C1e t - 2C2e -2 t Using the initial conditions, y(0) = and y ¢ (0) = , we have two equations containing C1 and C2 : C1 + C2 = and 2C1 - 2C2 = Solving these simultaneous equations gives us C1 = C2 = Thus, the solution to the initial value problem is y = e t + e -2 t = cosh(2 t) y ( t) = e t 14 y (0) = c1 + c = 1, 2c1 - 2c = \ c1 = 1, c = 15 From (12), y ( t) = C1e t + C2e -2 t Using the initial condition y(0) = 3, we find that C1 + C2 = From the initial condition lim y ( t) = and the equation for y ( t) given to us in (12), we can t ặ conclude that C1 = (if C1 π , then lim = ±•) Therefore, C2 = and y ( t) = 3e -2 t tặ lim y ( t) = c = \ c1 = 10 & y ( t) = 10e t 16 c1 + c = 10 17 From the graph, we can see that y ¢ = -1 and that y(1) = Thus m = y ¢ - = -1 - = -2 and t ặ- y = y (1) = 18 y ¢ = mt fi y = Also c = -1 19 m t + c From graph, y = -1 only at t = \ t0 = From graph y (1) = -0.5 \ - m = - fi m = 2 We know that this is a freefall problem, so we can begin with the generic equation for freefall g situations: y ( t) = - t + v t + y The object is released from rest, so v = The impact time corresponds to the time at which y = , so we are left with the following equation for the g impact time t : = - t + y Solving this for t yields t = of impact: v = y ¢ = - gt + v = - gt = - gy 20 x ¢¢ = a x ¢ = at + v , v = x = fi x = at + Ê 64 ˆ 88 = a(8) fi a = 11 ft / sec At t = 8, x = 11Á ˜ = 352 ft Ë 2¯ y0 For the velocity at the time g Chapter Introduction to Differential Equations • 21 Ê pt ˆ Ê pt ˆ a = y ¢¢ = 32 - e sinÁ ˜ Integrating gives us y ¢ = -32 t - e cosÁ ˜ + C The object is Ë 4¯ Ë 4¯ p 4 dropped from rest, so y ¢ (0) = = - e + C Solving for C yields C = e , and putting this p p value back into the equation for y ¢ and simplifying gives us y ¢ = -32 t + Ê Ê pt ˆ ˆ e Á1 - cosÁ ˜ ˜ Ë ¯¯ p Ë Ê 4ˆ Ê pt ˆ Integrating again gives us y = -16 t + et - Á ˜ e sinÁ ˜ + C ¢ Since the object is dropped Ë ¯ Ë 4¯ p p from a height of 252 ft (at t = ), y (0) = C ¢ = 252 and thus Ê 4ˆ Ê pt ˆ y = -16 t + et - Á ˜ e sinÁ ˜ + 252 Finally, since y( ) = , Ëp ¯ Ë 4¯ p 2 4e p Ê 4ˆ y( ) = = -16 ◊ + ◊ - Á ˜ e sin(p ) + 252 Solving for e yields e = Ëp ¯ p Section 1.2 (a) The equation is autonomous because y ¢ depends only on y (b) Setting y ¢ = , we have = - y + Solving this for y yields the equilibrium solution: y = (a) not autonomous (b) no equilibrium solutions, isoclines are t = const ant (a) The equation is autonomous because y ¢ depends only on y (b) Setting y ¢ = , we have = sin y Solving this for y yields the equilibrium solutions: y = ± np (a) autonomous (b) y ( y - 1) = 0, y = 0, (a) The equation is autonomous because y ¢ does not depend explicitly on t (b) There are no equilibrium solutions because there are no points at which y ¢ = (a) not autonomous (b) y = is equilibrium solution, isoclines are hyperbolas (a) c = -1: Setting c = -1 gives us - y + = -1 which, solved for y , reads y = This is the isocline for c = -1 c = : Setting c = gives us - y + = which, solved for y , reads y = This is the isocline for c = c = 1: Setting c = gives us - y + = which, solved for y , reads y = , the isocline for c = • Chapter Introduction to Differential Equations (a) - y + t = -1 fi y = t + -y + t = fi y = t -y + t = fi y = t -1 (a) c = -1: Setting c = -1 gives us y - t = -1 which can be simplified to t - y = (a hyperbola) This is the isocline for c = -1 c = : Setting c = gives us y - t = which can be simplified to y = ± t This is the isocline for c = c = 1: Setting c = gives us y - t = (a hyperbola) This is the isocline for c = 10 f (0) = f (2) = y ¢ = y (2 - y ) y ¢ > for < y < 2, y  < for - < y < and < y < • 11 One example that would fit these criteria is y ¢ = -( y - 1) For this autonomous D.E., y ¢ = at y = and y ¢ < for -• < y < and < y < 12 y  = 13 One example that would fit these criteria is y ¢ = sin(2py ) For this autonomous D.E., y ¢ = at y= 14 c 15 f 16 a 17 b 18 d 19 e n Chapter First Order Linear Differential Equations Section 2.1 This equation is linear because it can be written in the form y ¢ + p( t) y = g( t) It is nonhomogeneous because when it is put in this form, g( t) π nonlinear This equation is nonlinear because it cannot be written in the form y ¢ + p( t) y = g( t) nonlinear This equation is nonlinear because it cannot be written in the form y ¢ + p( t) y = g( t) linear, homogeneous This equation is nonlinear because it can be written in the form y ¢ + p( t) y = g( t) nonlinear This equation is linear because it cannot be written in the form y ¢ + p( t) y = g( t) It is nonhomogeneous because when it is put in this form, g( t) π 10 linear, homogeneous 11 (a) Theorem 2.1 guarantees a unique solution for the interval (-•, •) , since t and sin( t) are t +1 both continuous for all t and -2 is on this interval 11 (b) Theorem 2.1 guarantees a unique solution for the interval (-•, •) , since t and sin( t) are t +1 both continuous for all t and is on this interval 11 (c) Theorem 2.1 guarantees a unique solution for the interval (-•, •) , since both continuous for all t and p is on this interval 12 (a) < t < • 12 (b) -2 < t < 12 (c) -2 < t < 12 (d) -• < t < -2 t and sin( t) are t +1 • Chapter First Order Linear Differential Equations 13 (a) For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π Therefore, Theorem 2.1 guarantees a unique solution for ( 3, •) , the largest interval that includes t = 13 (b) For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π Therefore, Theorem 2.1 guarantees a unique solution for (-2, 2) , the largest interval that includes t = - 13 (c) For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π Therefore, Theorem 2.1 guarantees a unique solution for (-2, 2) , the largest interval that includes t = 13 (d) For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π Therefore, Theorem 2.1 guarantees a unique solution for (-•, -2) , the largest interval that includes t = -5 13 (e) For this equation, p( t) is continuous for all t π 2, -2 and g( t) is continuous for all t π Therefore, Theorem 2.1 guarantees a unique solution for (-2, 2) , the largest interval that includes t = 14 ln t + t -1 t-2 ln t t+1 = t-2 undefined at t = 0, 14 (a) < t < • 14 (b) < t < 14 (c) -• < t < 14 (d) -• < t < 15 2 y ( t) = 3e t Differentiating gives us y ¢ = 3e t (2 t) = ty Substituting these values into the given equation yields ty + p( t) y = Solving this for p( t) , we find that p( t) = -2 t Putting t = into the equation for y gives us y = 16(a) y = Ct r y ¢ = Crt r -1 ty ¢ - y = \ 2Crt r - 6Ct r = (2 r - 6)Ct r = fi (2 r - 6) y = fi r - = fi r = y (-2) = C (-2) r = fi C π \ C (-2) = fi C = -1 16 (b) -• < t < since p( t) = 16 (c) y ( t) = - t , - • < t < • -3 t Chapter First Order Linear Differential Equations • 17 y ( t) = satisfies all of these conditions Section 2.2 (a) First, we will integrate p( t) = to find P ( t) = 3t The general solution, then, is y ( t) = Ce - P ( t ) = Ce -3 t (b) y (0) = C = -3 Therefore, the solution to the initial value problem is y = -3e -3 t (a) y¢ - (b) y (-1) = Ce - = 2, C = 2e y=0 t t (e - y )¢ = 0, y = Ce 1 y ( t) = 2e ( t +1 ) (a) We can rewrite this equation into the conventional form: y ¢ - ty = Then we will integrate p( t) = -2 t to find P ( t) = - t The general solution, then, is y ( t) = Ce - P ( t ) = Ce t (b) y (1) = Ce = Solving for C yields C = 3e -1 Therefore, the solution to the initial value problem is y ( t) = 3e -1e t = 3e( t (a) ty ¢ - y = fi y ¢ - y = t -4 y ¢ - y = ( t y )¢ = t t (b) -1) Ú - t dt = -4 ln t = - ln(t ) \ m= t4 y = Ct y (1) = C = \ y ( t) = t (a) We can rewrite this equation into the conventional form: y ¢ + p( t) = y = Then we will integrate t to find P ( t) = ln t = ln t The general solution, then, is t -4 y ( t) = Ce - P ( t ) = Ce - ln t = Ce ln t = Ct -4 (b) y (1) = C = Therefore, the solution to the initial value problem is y ( t) = t -4 (a) m = exp( t - cos t) \ y ( t) = Ce -( t - cos t ) (b) p Êp ˆ y Á ˜ = Ce - = Ë 2¯ C=e p p y = e 2e -( t - cos t ) = e p - t + cos t (a) First, we will integrate p( t) = -2 cos(2 t) to find P ( t) = - sin(2 t) The general solution, then, is y ( t) = Ce - P ( t ) = Ce sin( t ) (b) y (p ) = C = -2 Therefore, the solution to the initial value problem is y ( t) = -2e sin( t ) (a) (( t + 1) y )¢ = y= C t +1 Chapter 12 First Order Partial Differential Equations and the Method of Characteristics • 343 12 (a) x = t , t = 0, u(t , 0) = e -t t ∂t ∂x = 2, t(0,t ) = fi t = s, x = t + s fi t = x - = 1, x (0,t ) = t ; ∂s ∂s 2 t t 2 ∂u x + ( 2) = su, u(t , 0) = e -t fi u = e -t + s u( x, t) = e ∂s 12 (b) The solution exists on the entire half plane 1 Ê1 ˆ 12 (c) The solution has a maximum at x = ; uÁ ,1˜ = e Ë2 ¯ 13 (a) x = t , t = 0, u(t , 0) = e -t Therefore, t ∂t ∂x = 2, t(0,t ) = fi t = s, x = t + s fi t = x - Then we have = 1, x (0,t ) = t ; ∂s ∂s ∂u s -t -t = u + s fi u = -4 ( s + 1) + Ce ; - + C = e fi C = e + Finally, ∂s Ê t ˆ Ê -( x - t ) ˆ t u( x, t) = -4 Á + 1˜ + e + e ¯ Ë2 ¯ Ë 13 (b) The solution exists on the entire half plane Ê1 ˆ 13 (c) The solution has a maximum at x = ; uÁ ,1˜ ª 2.24 Ë2 ¯ 14 (a) x = t , t = 0, u(t , 0) = e -t s2 t2 ∂t ∂x = -1, t(0,t ) = fi t = - s, x = - + t fi t = x + = t, x (0,t ) = t ; 2 ∂s ∂s 2 2 - ÊË x + t ˆ¯ Ê t t ˆ ∂u s s -t -t = - + t , u s= = e fi u = - + ts + e u( x, t) = - tÁ x + ˜ + e Ë 2¯ ∂s 14 (b) The solution exists on the entire half plane 14 (c) No finite maximum or minimum exists 15 (a) x = t , t = 0, u(t , 0) = e -t Therefore, - 12 dt ∂t ∂x = (2 t - 1) , t(0,t ) = fi x = t + s, = s + C = 1, x (0,t ) = t ; = ds fi 2t - ∂s ∂s (2t - 1) - 12 Substitution gives us = C = Then we have -1 - 2 ˆ 1Ê -t t ∂u = - = - Á1 + s= = 1, u s= = e -t fi u = s + e -t , t = x - s ˜= 2t - 2 Ë t - 1¯ t - 1 - t ∂s t - ( x - t 1- t ) Finally, u( x, t) = +e - 2t 15 (b) The solution exists on -• < x < •, £ t < 15 (c) The solution does not exist at t = 16 (a) x = t , t = 0, u(t , 0) = e -t ∂t ∂x = 1, t(0,t ) = fi t = s, x = t + vs fi t = x - vt = v, x (0,t ) = t ; ∂s ∂s 2 ∂u = -cu, u s= = e -t fi u = e -t - cs u( x, t) = e -( x - vt ) - ct ∂s 344 • Chapter 12 First Order Partial Differential Equations and the Method of Characteristics 16 (b) If v=5 mi/hr, the peak value reaches 20 miles downstream at t=4 hours At t=4 and x=20 mi, u(20, ) = e -5 c = 0.05 fi c = 0.599hr -1 dt dx 17 (b) = v, x (0,t ) = 0, = 1, t(0,t ) = t fi x = vs, t = s + t ds ds du ÔÏ16t (1 - t ) , £ t £ Therefore, = fi u = w (t ) = Ì ds ƠĨ 0, t > Ï Ê x ˆ 2Ê Ê x ˆˆ x ÔÔ16Á t - ˜ Á1 - Á t - ˜ ˜ , £ t - £ v u( x, t) = Ì Ë v ¯ Ë Ë v ¯ ¯ x Ô 0, t - > ƠĨ v x du 17 (c) = 0, u = w (t ) = fi u( x, t) = 0, t - < v ds 17 (d) 18 (a) s = t, t = x - vt, u( x, t) = 2 du ∂u = -cu , u s= = e -t = -cds fi - u -1 = -cs + C, C = -et , u = u ∂s cs + et ct + e( x - vt ) 18 (b) As in (16 b), the peak value reaches 20 miles downstream at t=4 hours At t=4 and x=20 mi, 19 u(20, ) = = 0.05 fi c = = 4.75hr -1 4c + Chapter 13 Linear Two-Point Boundary Value Problems Section 13.1 Note: Part (a) of Exercises 1-7 are identical (a) (b) (b) (b) (b) (b) (b) (b) Ê tˆ Ê tˆ y ( t) = c1 cosÁ ˜ + c sinÁ ˜ + Ë 2¯ Ë 2¯ Applying the boundary conditions, we have y (0) = c1 + = 0, y ( p) = c + = Solving these simultaneous equations yields c1 = -4 and c = -2 Thus there is a unique solution: Ê tˆ Ê tˆ y ( t) = -4 cosÁ ˜ - sinÁ ˜ + Ë 2¯ Ë 2¯ c c y ¢ (0) = = 0, y ¢ ( p) = - = c1 = c = Thus there is a unique solution: y ( t) = 2 c Ê tˆ c Ê tˆ Differentiation gives us y ¢ = - sinÁ ˜ + cosÁ ˜ Applying the boundary conditions, we Ë 2¯ Ë 2¯ c2 have y ¢ (0) = = -2, y ( p) = c + = Solving these simultaneous equations yields c1 arbitrary and c = -4 Thus there are infinitely many solutions: Ê tˆ Ê tˆ y ( t) = c1 cosÁ ˜ - sinÁ ˜ + Ë 2¯ Ë 2¯ c y (0) = c1 + = 0, y ¢ ( p) = - = There is no solution c Ê tˆ c Ê tˆ Differentiation gives us y ¢ = - sinÁ ˜ + cosÁ ˜ Applying the boundary conditions, we Ë 2¯ Ë 2¯ have y (0) + y ¢ (0) = c1 + + c = 0, y ( p) + y ¢ ( p) = c + - c1 = Solving these simultaneous equations yields c1 = and c = -4 Thus there is a unique solution: Ê tˆ y ( t) = -4 sinÁ ˜ + Ë 2¯ -c y (0) + y ¢ (0) = c1 + + c = 0, y (p ) - y ¢ ( p) = c + - (2) = There are infinitely many t t solutions: y = c1 cos - (c1 + ) sin + 2 c1 Ê t ˆ c Ê tˆ Differentiation gives us y ¢ = - sinÁ ˜ + cosÁ ˜ Applying the boundary conditions, we Ë 2¯ Ë 2¯ have y (0) + y ¢ (0) = c1 + + c = 4, y ( p) - y ¢ ( p) = c + + c1 = These simultaneous equations cannot be solved, and so there are no solutions y ¢¢ - y = 0, y (0) = 0, y (2) = g = -4, a = 0, b = Differentiation gives us y ¢ = 2, y ¢¢ = Thus g = 0, a = 2, b = 346 • Chapter 13 Linear Two-Point Boundary Value Problems 10 11 14 (a) 14 (b) 14 (c) 14 (d) 15 (a) 15 (b) 15 (c) 15 (e) 16 (a) 16 (b) 16 (c) 16 (e) 17 (a) 17 (b) 17 (c) 17 (d) 18 (a) 18 (b) 18 (c) 18 (e) p p Ê pˆ y ¢¢ + gy = e t - sin t + g (e t + sin t) = 2e t , y (0) = 1, y Á ˜ = e + g = 1, a = 1, b = + e Ë 2¯ The general solution is y = c1 cos t + c sin t + Differentiation gives us y ¢ = -c1 sin t + c cos t From the boundary conditions, Ê pˆ Ê pˆ y (0) + a1 y ¢ (0) = = c1 + + a1c , y Á ˜ + y ¢ Á ˜ = b = c + - c1 From the graph, we can see Ë 2¯ Ë 2¯ Ê pˆ that y (0) = 1, y Á ˜ = , and so c1 + = fi c1 = 0, c + = fi c = Finally, Ë 2¯ + a1 ◊ = fi a1 = and + = b - fi b = t z¢¢ - tz¢ + z = 0, z(1) + z¢ (1) = 0, z(2) - z¢ (2) = z = c1t + c t , z¢ = c1 + 2c t [c1 + c ] + [c1 + 2c ] = 0, [2c1 + c ] - [c1 + c ] = c1 = c = 0, z = The given problem has a unique solution È2 3˘È c1 ˘ È9 ˘ y = c1t + c t , y ¢ = c1 + 2c t Í ˙Íc ˙ = Í3˙ y = 3t + t ẻ ẻ ẻ t z - tz¢ + z = 0, z(1) - z¢ (1) = 0, z(2) - z¢ (2) = z = c1t + c t , z¢ = c1 + 2c t Applying the boundary conditions gives us 2[c1 + c ] - [c1 + 2c ] = 0, [2c1 + c ] - [c1 + c ] = Solving these simultaneous equations gives us z = c t , c arbitrary By the Fredholm Alternative Theorem, the given problem does not have a unique solution y = c1t + c t , y ¢ = c1 + 2c t Applying the boundary conditions gives us y (1) - y ¢ (1) = c1 = 1, y (2) - y ¢ (2) = c1 = Solving these simultaneous equations gives us y = t + c t , c arbitrary t z¢¢ - tz¢ + z = 0, 3z(1) - z¢ (1) = 0, z(2) - z¢ (2) = z = c1t + c t , z¢ = c1 + 2c t 3[c1 + c ] - 2[c1 + 2c ] = 0, 5[2c1 + c ] - 6[c1 + c ] = c1 = c By the Fredholm Alternative Theorem, the given problem does not have a unique solution y = c1t + c t , y ¢ = c1 + 2c t y (1) - y ¢ (1) = c1 - c = 2, y (2) - y ¢ (2) = c1 - c = There is no solution t z¢¢ - tz¢ + z = 0, z(1) - z¢ (1) = 0, z(2) - z¢ (2) = z = c1t + c t , z¢ = c1 + 2c t Applying the boundary conditions gives us [c1 + c ] - 2[c1 + 2c2 ] = 0, 2[2c1 + 4c ] - [c1 + 4c2 ] = Solving these simultaneous equations gives us z = 0, c1 = c = By the Fredholm Alternative Theorem, the given problem has a unique solution È-1 -3˘È c1 ˘ È-5 ˘ y = c1t + c t , y ¢ = c1 + 2c t Applying the boundary conditions gives us Í ˙Í ˙ = Í ˙ Î ˚Îc ˚ Î ˚ Solving this equation gives us y = t + t 5 t z¢¢ - tz¢ + z = 0, z(1) - z¢ (1) = 0, z(2) - z¢ (2) = z = c1t + c t , z¢ = c1 + 2c t [c1 + c ] - [c1 + 2c ] = 0, [2c1 + c ] - 2[c1 + c ] = c = 0, c1 arbitrary By the Fredholm Alternative Theorem, the given problem does not have a unique solution y = c1t + c t , y ¢ = c1 + 2c t y (1) - y ¢ (1) = -c = 1, y (2) - y ¢ (2) = -4 c = y = c1t - t , c1 arbitrary Chapter 13 Linear Two-Point Boundary Value Problems • 347 19 (a) t z¢¢ - tz¢ + z = 0, z(1) - 3z¢ (1) = 0, 3z(2) - z¢ (2) = 19 (b) z = c1t + c t , z¢ = c1 + 2c t Applying the boundary conditions gives us 4[c1 + c ] - 3[c1 + 2c ] = 0, 3[2c1 + c ] - 4[c1 + c ] = Solving these simultaneous equations gives us z = c (2 t + t ), c arbitrary 19 (c) By the Fredholm Alternative Theorem, the given problem does not have a unique solution 19 (e) y = c1t + c t , y ¢ = c1 + 2c t Applying the boundary conditions gives us y (1) - y ¢ (1) = c1 - 2c = 1, y (2) - y ¢ (2) = 2c1 - c = These simultaneous equations cannot be solved, and thus there is no solution 20 (a) Both Theorems guarantee a unique solution 20 (b) c1 + c + = 7, c1 + 2c + = fi y = 2e - t + e t + 21 (a) Since a0 a1 > , Theorem 13.2 is not applicable Likewise, the form of the boundary conditions makes Theorem 13.3 not applicable 21 (c) z¢¢ - z = 0, z(0) + z¢ (0) = 0, z(ln 2) + z¢ (ln 2) = Thus z = c1e - t + c 2e t , z¢ = -c1e - t + c 2e t From 1 the boundary conditions, we have c1 + c - c1 + c = and c1 + 2c - c1 + 2c = Solving 2 these simultaneous equations yields c = 0, c1 arbitrary Therefore, there is no unique solution to the given problem 21 (e) y = c1e - t + c 2e t + 4, y ¢ = -c1e - t + c 2e t From the boundary conditions, we have 1 c1 + c + - c1 + c = and c1 + 2c + - c1 + 2c = There is no solution to these 2 simultaneous equations, and so the problem has no solution 22 (a) Theorem 13.2 guarantees a unique solution 1 22 (b) c1 + c + + c1 - c = 0, c1 + 2c + - c1 + 2c = 12 fi y = -2e - t + 2e t + 2 23 (a) Theorem 13.2 guarantees a unique solution 23 (b) y = c1e - t + c 2e t + 4, y ¢ = -c1e - t + c 2e t From the boundary conditions, we have c1 + c + = 11 and - c1 + 2c = Solving these simultaneous equations yields c1 = 4, c = , and so y ( t) = e - t + 3e t + 24 (a) Since q( t) > , neither theorem can guarantee a unique solution 24 (c) z¢¢ + z = 0, z(0) + z¢ (0) = 0, z( p) + z¢ ( p) = z = c1 cos t + c sin t, z¢ = -c1 sin t + c cos t c1 + c = -c1 - c = c1 = -c , c arbitrary There is no unique solution to the given problem 24 (e) y = c1 cos t + c sin t + 2, y ¢ = - c1 sin t + c cos t c1 + + c = 7, - c1 + - c = -3 y = (5 - c ) cos t + c sin t + 25 (a) Since q( t) > , neither theorem can guarantee a unique solution 25 (c) z¢¢ + z = 0, z(0) + z¢ (0) = 0, z( p) + z¢ ( p) = Thus z = c1 cos t + c sin t, z¢ = -c1 sin t + c cos t From the boundary conditions, we have c1 + c = and -c1 - c = Solving these simultaneous equations yields c1 = -c , c arbitrary Therefore, there is no unique solution to the given problem 25 (e) y = c1 cos t + c sin t + 2, y ¢ = - c1 sin t + c cos t From the boundary conditions, we have c1 + + c = 7, - c1 + - c = There is no solution to these simultaneous equations, and so the problem has no solution 26 (a) Since q( t) > , neither theorem can guarantee a unique solution 348 • Chapter 13 Linear Two-Point Boundary Value Problems 26 (c) z¢¢ + z = 0, z(0) = 0, z( p) = z = c1 cos t + c sin t, z¢ = -c1 sin t + c cos t c1 = -c1 = c1 = 0, c arbitrary There is no unique solution to the given problem 26 (e) y = c1 cos t + c sin t + c1 + = 7, - c1 + = There is no solution to the given problem 27 (a) Since q( t) > , neither theorem can guarantee a unique solution 27 (c) z¢¢ + z = 0, z(0) = 0, z( p) + z¢ ( p) = Thus z = c1 cos t + c sin t, z¢ = -c1 sin t + c cos t From the boundary conditions, we have c1 = and -c1 - c = Thus c1 = c = , and so the problem has a unique solution 27 (d) y = c1 cos t + c sin t + 2, y ¢ = - c1 sin t + c cos t From the boundary conditions, we have c1 + = 8, - c1 + - c = Therefore, y ( t) = cos t - sin t + 28 (a) and (c) same as (26) 28 (e) y = c1 cos t + c sin t + c1 + = 8, - c1 + = -4 y = cos t + c sin t + 29 (a) y1 ( t) is a nonzero solution if g( t) π and/or a π Since c and c1 are not both zero, a π ensures nontrivial initial conditions y ( t) is a nontrivial solution since a0 and a1 are not both zero 31 (b) Choose c1 = 1, c = Then a0c1 - a1c = Then we have t y1¢¢ - ty1¢ + y1 = 2, y1 (1) = 3, y1¢ (1) = Thus y1 = c1t + c t ln t + , and with the boundary conditions we have c1 + = 3, c1 + c [ln t + 1] = Therefore, c1 = 1, c = -1, and so y1 ( t) = t - t ln t + Then, t y 2¢¢ - ty 2¢ + y = 0, y (1) = 0, y 2¢ (1) = -1 Thus y = c1t + c t ln t , and with the boundary conditions we have c1 = 0, c1 + c = -1 Therefore, c1 = 0, c = -1, and so y ( t) = - t ln t - ln Finally, y s = y1 + sy = t - t ln t + - st ln t , y s¢ (2) = - ln - + s(- ln - 1) = fi s = + ln t ln t ◊ ln y ( t) = t - t ln t + + + ln 32 (b) Choose c1 = 1, c = y1¢¢ + y1¢ = sin t, y1 (0) = 3, y1¢ (0) = y1 = c1 cos t + c sin t + sin t , c1 = 3, 2c + = y1 ( t) = cos t - sin t + sin t y = c1 cos t + c sin t c1 = 1, 2c = -1 y ( t) = cos t - sin t y s = y1 + sy 2 Ê 29 Ê pˆ ˆ Ê pˆ Ê pˆ Ê pˆ y1 Á ˜ + sy Á ˜ + y1¢ Á ˜ + sy 2¢ Á ˜ = fi s = - Á - ˜ Ë 4¯ ¯ Ë 4¯ Ë 4¯ Ë 4¯ 5Ë Ê -29 + 2 ˆ Ê 1 ˆ y = cos t - sin t + sin t + Á Á cos t - sin t˜ ˜ ¯ Ë ¯Ë 33 (b) Choose c1 = 1, c = Then we have y1¢¢ - t y1 = 1, y1 (0) = 0, y1¢ (0) = ; - y1 (1) finally, y 2¢¢ - t y = 0, y (0) = 0, y 2¢ (0) = -1 Thus y1 (1) + sy (1) = fi s = y (1) - y1 (1) y r( t) = y1 + sy = y1 + y (1) 34 (b) g = 0, a = 0, b = y1¢¢ + ty1¢ - y1 = 0, y1 (0) = 0, y1¢ (0) = fi y1 = ; y 2¢¢ + ty 2¢ - y = 0, y (0) = 0, y 2¢ (0) = -1 y ( t) y1 + sy = sy , y (1) = sy (1) = fi s = fi y ( t) = y (1) y (1) Chapter 13 Linear Two-Point Boundary Value Problems • 349 Section 13.2 (b) x y ¢¢ = - f ( x ) fi y ¢ = - Ú f (l ) dl + c1 Using the second boundary condition, we have x l 0 c1 = Ú f (l ) dl fi y ( x ) = - Ú Ú f (s )dsdl + x Ú f (l )dl Integration by parts and some y ( x ) = Ú xf (l ) dl + Ú lf (l ) dl = Ú G( x, l ) f (l ) dl with simplification give us x x 0 Ïl, £ l £ x G ( x, l ) = Ì Ó x, x < l £ 1 (c) (b) Ú x x x 6ldl + Ú 6l2 dl = x - x x y ¢ = - Ú f (l ) dl + c1 c1 = y ¢ (0) = fi y ( x ) = - Ú x l 0 y (1) + y ¢ (1) = fi c = Ú l Ú f (s )dsdl + c Ú f (s )ds dl + Ú f (l )dl 1 x Ï2 - x, £ l £ x y ( x ) = Ú (2 - x ) f (l ) dl + Ú (2 - l ) f (l ) dl = Ú G( x, l ) f (l ) dl G( x, l ) = Ì 0 x Ĩ2 - l , x < l £ (c) (b) x x Ú (2 - l )6ldl + Ú (2 - x )6ldl = - x y ¢ = - Ú f (l ) dl + c fi y ( x ) = - Ú Ú f (s ) dsdl + c x + c Using the boundary conditions, we have c - 2c = and - Ú Ú f (s ) dsdl + c + c = Thus c = Ú Ú f (s ) dsdl, c = 2c and 3 x 1 so y ( x ) = - Ú x l 0 l l 0 x 2 0 x +2 Ú f (s )dsdl + Ú Ú f (s )dsdl Integration by parts and some l l 0 simplification give us ( x + 2) x (l + 2) - l ) f (l ) dl = Ú G( x, l ) f (l ) dl with - x ) f (l ) dl + Ú y( x) = Ú ( ( 0 x 3 Ï (l + 2)(1 - x ) , 0£l £ x Ô G ( x, l ) = Ì Ơ ( x + 2)(1 - l ) , x < l £ Ó 1- x x x +2 (c) - l )6ldl = ( x + 2) - x ( l + 2)6ldl + ( Ú Ú 3 x x x l (b) y ¢ = - Ú f (l ) dl + c1 y ( x ) = - Ú Ú f (s ) dsdl + c1 x + c 0 y (0) - y ¢ (0) = c - c1 = fi c1 = c = Ú l Ú f (s )ds dl - Ú f (l )dl y ( x ) = Ú (- x )(l + 1) f (l ) dl - Ú ( x + 1)lf (l ) dl = Ú G( x, l ) f (l ) dl 0 (c) (c) x 1 x Ï- x (l + 1), £ l £ x G ( x, l ) = Ì Ó -l ( x + 1), x < l £ x -( x + 1) Ú l 6ldl - x Ú (l + 1)6ldl = -2 - x - x x 1 1 y = cos t + sin t, y1 = cos t - sin t, y = cos t + sin t 4 2 350 • Chapter 13 Linear Two-Point Boundary Value Problems 10 11 y + y1 - y = cos t - sin t + sin t 18 1 y = -2 y + y1 = - cos t - sin t - sin t y = -4 y1 - y = -3 cos t + sin t 2 t t y = -18 t -1 - + , y1 = - t -1, y = t -1 t y = y - y1 + y = -45 t -1 - - + t 2 33 -1 t t y = y - y1 + y = t - + m = e t , e t y ¢ ¢ + e t y = e t sin t 11 m=e (i) (ii) (iii) (b) (i) (ii) 12 y= ( ) ¢ - 3t y¢ = e 2 2 ¢ t2 -t -t -t -t m = e , Ê e y ¢ˆ + 2e y = e Ë ¯ -2 -2 3 t m = t , t y¢ ¢ + t y = t e - t2 ( , e - t2 ) 14 ( ) m = t , ( t y ¢ )¢ - sin t( y ) = 15 y ¢¢ + tan ty ¢ + sec ty = t sec t fi m = sec t, (sec ty ¢ )¢ + sec ty = t sec t 13 16 17 18 19 20 2 -1 m = t, ( ty ¢ )¢ + t 2e t y = t First, we rewrite the equation as follows: y ¢¢ - y ¢ - y = Then we have (l - 2)(l + 1) = fi y (t) = c1e - t + c 2e 2t l2 - 3l + = (l - 2)(l - 1) = fi y ( t) = c1e t + c 2e t l (l - 1) + 2l - = (l + 2)(l - 1) = fi y ( t) = c1t + c t -2 , t > l + l - = (l - 2)(l + 3) = fi y ( t) = c1t2 + c t -3 , t > Section 13.3 (b) p( t) = 1, f ( t) = - sin t (c) f ( t) = c1t + c , f (0) - f ¢ (0) = c - c1 = fi f = t + 1, y ( t) = c1t + c , y (2) - y ¢ (2) = c1 + c = fi y = t - 1, W = p( s)W ( s) = ◊ (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = -2 From (8), then, Ï Ô- ( t + 1)( s - 1), £ t £ s G( t, s) = Ì Ơ- ( s + 1)( t - 1), s < t £ Ó 2 t 1 (d) y ( t) = Ú G( t, s) f ( s) ds = ( t - 1) Ú ( s + 1) sin sds + ( t + 1) Ú ( s - 1) sin sds 0 t 2 -t (b) p( t) = 1, f ( t) = -e Chapter 13 Linear Two-Point Boundary Value Problems • 351 (c) f ( t) = c1 cos t + c sin t, f (0) = c1 = fi f ( t) = sin t , y ( t) = cos t , Ï sin t cos s, £ t £ s W = p( s)W ( s) = ◊ (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = G( t, s) = Ì p Ĩsin s cos t, s < t £ p (d) t p t y ( t) = Ú G( t, s) f ( s) ds = - cos t Ú e - s sin sds - sin t Ú e - s cos sds (b) p( t) = 1, f ( t) = -e t (c) f ( t) = c1 cos t + c sin t, f (0) + f ¢ (0) = c1 + c = fi f = cos t - sin t , y ( t) = c1 cos t + c sin t, y ¢ (1) = -c1 sin1 + c cos1 = fi y = cos t cos1 + sin t sin1 = cos( t - 1) , W = p( s)W ( s) = ◊ (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = - cos1 - sin1 From (8), then, Ï -(cos t - sin t) cos( s - 1) , 0£ t£ s Ô cos1 + sin1 G( t, s) = Ì Ô -(cos s - sin s) cos( t - 1) , s < t £ Ó cos1 + sin1 cos t - sin t cos( t - 1) t (d) y ( t) = Ú G( t, s) f ( s) ds = cos( s - 1)e sds cos s - sin s)e sds + ( Ú Ú t 0 cos1 + sin1 cos1 + sin1 (b) p( t) = 1, f ( t) = -e t (c) f ( t) = cosh t, y ( t) = sinh2( t - 1) , W = p( s)W ( s) = ◊ (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = -2 cosh Ï cosh t sinh 2( s - 1) , 0£ t£ s Ô2 cosh G( t, s) = Ì Ô - cosh s sinh 2( t - 1) , s < t £ Ó cosh cosh t sinh 2( t - 1) t (d) y ( t) = Ú G( t, s) f ( s) ds = sinh 2( s - 1)e sds cosh se sds + Ú cosh Út cosh 2 (b) p( t) = 1, f ( t) = e - t (c) f = sinh t , y = sinh( t - 1) , W = p( s)W ( s) = ◊ (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = - sinh1 From (8), then, Ï - sinh t sinh( s - 1) , 0£ t£ s Ô sinh1 G( t, s) = Ì Ơ - sinh s sinh( t - 1) , s < t £ Ó sinh1 sinh t sinh( t - 1) t (d) y ( t) = Ú G( t, s) f ( s) ds = sinh( s - 1) s2 ds sinh( s) s2 ds + Ú Ú 0 t sinh1 sinh1 (b) p( t) = e t , f ( t) = -e t sin t (c) f ( t) = e -2 t sin t , y ( t) = e -2( t - 2) sin( t - 2) , W = p( s)W ( s) = ◊ (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = -e t + s sin Ï -e -2( t + s) sin t sin( s - 2) , 0£ t£ s ÔÔ G( t, s) = Ì -2( t + s) sin sin s sin( t - 2) Ô -e , s< t£2 ÔÓ sin 2 e -2 t sin( t - 2) t s e -2 t sin t 2 s (d) y ( t) = Ú G( t, s) f ( s) ds = Ú0 e sin sds + sin Út e sin(s - 2) sin sds sin (b) p( t) = e t , f ( t) = - t 3e t (c) f ( t) = c1 + c 2e -2 t , f (-1) = c1 + c 2e = fi f = - e -2( t +1) , y ( t) = c1 + c 2e -2 t , y ¢ (1) = -2c = fi y = 1, 352 • Chapter 13 Linear Two-Point Boundary Value Problems W = p( s)W ( s) = e t ◊ (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = e t (2e -2( s+1) - 0) From (8), then, Ï1 -2( t +1) , -1£ t £ s Ô2 e - e G( t, s) = Ì Ơ e - e -2( s+1) , s < t £ Ó2 t 1 1 (d) y ( t) = Ú G( t, s) f ( s) ds = - e Ú - e -2( s+1) ( s3e s ) ds - e - e -2( t +1) Ú s3e sds t 2 -1 -1 (b) p( t) = t , f ( t) = -1 (c) f ( t) = t - y ( t) = t - , W = p( s)W ( s) = t -1 ◊ (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = -6 Ï ( t - 1)( s2 - ) , 1£ t £ s ÔÔ G( t, s) = Ì 2 Ô- ( s - 1)( t - ) , s < t £ ƠĨ ( ( ) ) ( (d) y ( t) = Ú (t G( t, s) f ( s) ds = -3 t ) ( ) - 4) t t - 1) 2 ( (s - 1)ds + Út (s - 4)ds Ú1 (b) p( t) = e , f ( t) = - te -3 t (c) f ( t) = c1e t + c 2e t , f (0) = c1 + c = fi f = e t - e t , y ( t) = c1e t -1 + c 2e t - , y (1) = c1 + c = fi y = e t -1 - e t - , W = p( s)W ( s) = e -3 t (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = -e -1 + e -2 From (8), then, Ï -(e t - e t )(e s-1 - e s- ) , 0£ t£ s ÔÔ e -1 - e -2 G( t, s) = Ì 2s 2t - s t -1 e e e e ( ) ( ) Ơ , s< t £1 ƠĨ e -1 - e -2 e t - e t s-1 s- e t -1 - e t - t s s -3 s (d) y ( t) = Ú G( t, s) f ( s) ds = -1 -2 Ú (e - e )( se ) ds + -1 -2 Ú (e - e )( se -3 s ) ds 0 e -e t e -e -3 t -3 t 10 (b) p( t) = e , f ( t) = - te 1 10 (c) f ( t) = e t - e t , y ( t) = e t -1 - e 2( t -1) , 2 e -1 - e -2 W = p( s)W ( s) = e -3 t ◊ (y ( s)f ¢ ( s) - f ( s)y ¢ ( s)) = Ï e t - e t e s-1 - e 2( s-1) 2 Ô, 0£ t£ s -1 -2 Ô e e G( t, s) = Ì s 2s t -1 2( t -1) Ô e - 12 e e - 12 e , s< t £1 ÔÓ e -1 - e -2 ( )( ) ( )( ) 10 (d) y ( t) = Ú G( t, s) f ( s) ds ( 11 12 13 ) ( ( - 12 e t - t s e t - 12 e t 2s -3 s = Ú0 e - e se ds + e -1 - e -2 e -1 - e -2 p( t) = 1, q( t) = 0, a0 = 0, a1 = 1, b0 = 1, b1 = 1 p( t) = 1, q( t) = 0, a0 = 1, a1 = 1, b0 = 1, b1 = p( t) = 1, q( t) = 1, a0 = 1, a1 = 0, b0 = 0, b1 = 2e t -1 ) ) (e Ú t s -1 ) - e s- se -3 sds Chapter 13 Linear Two-Point Boundary Value Problems • 353 p( t) = 1, q( t) = -4, a0 = 1, a1 = 0, b0 = 1, b1 = Ïk cos t cos(1 - s), £ t £ s 16 (a) G( t, s) = Ì k cos s sin(1 - s) + k sin s cos(1 - s) = -1, k = sin1 Ó k cos s cos(1 - t), s < t £ 16 (b) g = 1, a0 = 0, a1 = 1, b0 = 0, b1 = 14 Section 13.4 r È 1˘ r È ˘ After differentiating and forming the matrices and vectors, we have y ¢ = Í ˙ y + Ícos t ˙ Ỵ-1 ˚ Ỵ ˚ È-1˘ È2 -1˘ r È0 ˘ r y (0) + Í y (1) = Í ˙ From the boundary conditions, we form Í ˙ ˙ Ỵ2˚ Ỵ0 ˚ Ỵ1 ˚ ˘ È1 ˘ r È0 ˘ r È0 ˘ r È0 ˘ r È ˙y + Í1 ˙, Í y (0) + Í y (2) = Í ˙ y¢ = Í ˙ ˙ ÍỴ0 - ˙˚ ÍỴ ( t + 1) ˙˚ Ỵ0 ˚ Ỵ1 -1˚ Ỵ0 ˚ After differentiating and forming the matrices and vectors, we have ˘r È ˘ r È y ¢ = Í -1 t -1 ˙ y + Í -1 ˙ From the boundary conditions, we form Ỵ- t e - t ˚ Ỵ2 t ˚ È-3˘ È0 ˘ r È0 ˘ r Í0 ˙ y (1) + Í1 ˙ y (2) = Í ˙ Ỵ ˚ Ỵ Ỵ ˚ ˚ ˘ È1 -1˘ È È0 0 ˘ È2˘ È 0˘ r Í ˙ Í ˙r ˙r ˙r Í Í Í ˙ y ¢ = Í 0 1˙ y + Í ˙ , Í2 ˙ y (0) + Í0 0 ˙ y (2) = Í-1˙ ÍỴe - t + sin t ˙˚ ÍỴ0 0 ˙˚ ÍỴ0 -3˙˚ ÍỴ ˙˚ ÍỴ-1 -3 3˙˚ After differentiating and forming the matrices and vectors, we have È ˘ È 0˘ r Í ˙ ˙r Í y¢ = Í 0 ˙ y + Í ˙ From the boundary conditions, we form ÍỴ3t -3 sin t ˙˚ ÍỴ-2 t -3 t -2 ˙˚ È0 0˘ È1 ˘ È1 1 ˘ ˙r ˙r Í Í ˙ Í Í0 0 ˙ y (-2) + Í-1 -1˙ y (-1) = Í4 ˙ ÍỴ ˙˚ ÍỴ2 ˙˚ ÍỴ0 0 ˙˚ È1 ˘ [1] È0 ˘ È1 ˘È 1˘ È0 ˘È e e ˘ È 1 ˘ P [ 0] = Í P D = = =Í , , ˙ Í0 ˙ Í0 ˙Í-1 1˙ + Í0 ˙Í ˙ The 3˙ Ỵ0 ˚ Ỵ ˚ Ỵ ˚Ỵ ˚ Ỵ ˚Ỵ-e e ˚ Ỵ-e e ˚ r r determinant of D is nonzero, and so there is a unique solution for every g( t) and a È1 -1˘ [1] È0 ˘ È1 -1˘È 1˘ È0 ˘È e e ˘ È ˘ P =Í =Í ˙, P = Í1 -1˙, D = Í0 ˙Í-1 1˙ + Í1 -1˙Í ˙ The 3˙ Ỵ0 ˚ Ỵ ˚ Ỵ ˚Ỵ ˚ Ỵ ˚Ỵ-e e ˚ Ỵ2e ˚ r r determinant of D is zero, and so there is not a unique solution for every g( t) and a [ 0] 354 • Chapter 13 Linear Two-Point Boundary Value Problems È1 -1˘ [1] È0 ˘ È1 -1˘È 1˘ È0 ˘È e e ˘ È2 ˘ P =Í =Í ˙, P = Í1 ˙, D = Í0 ˙Í-1 1˙ + Í1 ˙Í ˙ The 3˙ Ỵ0 ˚ Ỵ ˚ Ỵ ˚Ỵ ˚ Î ˚Î-e e ˚ Î0 2e ˚ r r determinant of D is nonzero, and so there is a unique solution for every g( t) and a [ 0] È1 ˘ [1] È0 ˘ È1 ˘È 1˘ È0 ˘È e e ˘ È1 ˘ P [ 0] = Í P D = = =Í , , ˙ Í1 ˙ Í0 ˙Í-1 1˙ + Í1 ˙Í ˙ The determinant 3˙ Ỵ0 ˚ Ỵ ˚ Ỵ ˚Ỵ ˚ Î ˚Î-e e ˚ Îe e ˚ r r of D is nonzero, and so there is a unique solution for every g( t) and a -t 3t È3 -1 ˘ r È ˘ r È(2e - 1)e - 3e ˘ r È2e - 1˘ r 10 (b) Í ( ) = Y = y t c c = c fi = Í ˙ Í ˙ Í-1˙ 3˙ 6e ÍỴ(2e - 1)e - t + 3e t ˙˚ 6e Î -3 ˚ Î0 2e ˚ Î ˚ ˘ r È1 ˘ È -e ˘ È1 r 11 (b) Í -1 c = Í ˙fic = Í ˙ Then we have -e ˙˚ -e - e -1 Ỵ-e -1 ˚ Ỵe Ỵ0 ˚ -t +3 -t e t ˘È e ˘ e t -1 ˘ r r Èe Èe = y ( t) = Yc = Í ˙Í ˙ Í ˙ e + e -1 Ỵe - t -e t ˚Ỵe -1 ˚ e + e -1 Ỵe - t + -e t -1 ˚ ˘ r È0 ˘Èe -1 e ˘È (e - 1) ˘ È 1 ˙ Í ˙ Í 12 (b) f = - Í = ˙ ˙Í -1 -e -3 ˚Í (1 - e -3 ) ˙ ÍỴ- + 2e + e ˙˚ Ỵ0 ˚Ỵe ˚ Ỵ ˘ r r È -e -1˘È -1 1 ˙ fi c = -c1 Í c =D f =- 3 ˙ -1 Í + e ˙ e + e Ỵ-e -1 ˚Í- + Ỵ 2e ˚ 3t 1 t È ˘ È ˘ -t 3t -t Èe - t e t ˘Í c1 + (e - 1) ˙ Íc1 (e - e ) + (1 - e ) + (e - 1) ˙ r y ( t) = Í - t ˙Í ˙= Í ˙ -e t ˚Í-c + 1 - e -3 t ˙ Íc e - t + e t + 1 - e - t - e t - ˙ Ỵe ( )˚ Î 1( ) 2( ) 6( )˚ Î È1 0 ˘È1 0 ˘ È0 0 ˘Èe -2 0 ˘ È1 0 ˘ r r ˙ Í ˙ ˙Í ˙ Í ˙Í Í 13 e e ˙ = Í0 1 ˙ (see exercises y ( t) = Yc ; D = Í0 ˙Í0 1 ˙ + Í0 0 ˙Í ÍỴ0 0 ˙˚ÍỴ0 -1 ˙˚ ÍỴ0 ˙˚ÍỴ -e e ˙˚ ÍỴ0 -e e ˙˚ È Í1 È1 ˘ Í r r Í ˙ r e3 -1 r 6-9) Dc = a = Í0 ˙ fi c = D a = Í0 Í e +e ÍỴ1 ˙˚ e Í0 ÍỴ e + e ˘ ˘ È È -2 t ˙ Í -2 t ˙ Í e Èe 0˘ t t Í r e -e ˙ Í ˙Í -1 ˙ ˙ Í = y ( t) = Í e t e t ˙Í ˙ e + e ˙ Í e3 + e ˙ t t Í t ÍỴ -e e ˙˚ 3t ˙ Íe + e ˙ Í ÍỴ e + e ˙˚ ÍỴ e + e ˙˚ ˘ È ˘ Í ˙ È1 ˘ ˙ Í ˙ -1 ˙ -1 Í ˙ = ˙ Therefore, Í ˙ e + e ˙Í ˙ Í e - e ˙ ˙ÍỴ1 ˙˚ Í ˙ ÍỴ e - e ˙˚ e + e ˙˚ Chapter 13 Linear Two-Point Boundary Value Problems • 355 x ˘ [ ] È1 ˘ [ l ] È 0 ˘ r ÈI inc ˘ È1 18 (a) Y = Í =Í -1 ˙ P ˙, P = Í-G ˙, a = Í ˙ Ỵ1 x + b ˚ Ỵ0 ˚ Î ˚ Î ˚ l ˘ˆ r È Ê [ ] È1 ˘ ˘r r [ l ] È1 P P c + = Í1 - G (1 - G)l + b -1 ˙c = a Í1 b -1 ˙ Í1 l + b -1 ˙˜ Á Ë Ỵ Ỵ ˚ ˚ Ỵ ˚¯ È + (1 - G)b (l - x ) ˘ È ˘ È +˘ Í I inc I r + (1 - G)bl ˙ inc inc ˙I fi c = Í -(1 - G) I ˙ fi Í - ˙ = Í G + (1 - G)b (l - x ) ˙ Í I -1 ˙ Í Ỵ ˚ Ỵ (1 - G)l + b ˚ ÍỴ + (1 - G)bl ˙˚ dI (1 + (1 - G)bl)(1 - bl) - (G + (1 - G)bl)(-bl) = 18 (b) (0) = > dG (1 + (1 - G)bl) (1 + (1 - G)bl)2 dR = -bx + C Imposing R(l) = , we have = -bl + C fi C = + bl 19 (b) Ú = R -1 (R - 1) b (l - x ) -1 = b (l - x ) + fi R = From (11), Therefore, R -1 + b (l - x ) È1 + b (l - x ) ˘ inc - È b (l - x ) ˘ inc I+ I =Í ˙I , I = Í bl + ˙I fi I + = R Ỵ bl + ˚ Ỵ ˚ Section 13.5 (a) (b) (a) (b) (a) (b) p = 1, q = 0, r = u = c1 cos l x + c sin l x, u¢ = - l c1 sin l x + l c cos l x From the boundary conditions, we have u¢ (0) = l c = and u(1) = c1 cos l + c sin l = From l c = 0, l = and / or c = If l = 0, then c1 = We thus conclude that zero is not an eigenvalue since u = in that case If l π 0, c = and c1 cos l = Thus (2n - 1)p 1ˆ ÊÊ 1ˆ ˆ Ê ln = fi l n = Á n - ˜ p , un = cosÁ Á n - ˜ px˜ Ë ËË 2¯ 2¯ ¯ p = 1, q = 0, r = u = c1 cos l x + c sin l x, u¢ = - l c1 sin l x + l c cos l x From the boundary conditions, we have u(0) = c1 = fi u = c sin l x From u¢ (1) = c l cos l = 0, l = and / or cos l = If l = 0, then u = We thus conclude that zero is not an eigenvalue cos l = fi 1ˆ 1ˆ ÊÊ 1ˆ ˆ Ê Ê l n = Á n - ˜ p fi l n = Á n - ˜ p , un = sinÁ Á n - ˜ px˜ Ë Ë ËË 2¯ 2¯ 2¯ ¯ p = 1, q = -1, r = u = c1 cos l + 1x + c sin l + 1x, u¢ = - l + 1c1 sin l + 1x + l + 1c cos l + 1x From the boundary conditions, we have u¢ (0) = c l + = fi c = and / or l + = Therefore, u = c1 cos l + 1x u¢ (1) = -c1 l + sin l + = fi l n + = np Thus l n = -1 + ( np) , un = cos( npx ) 356 • Chapter 13 Linear Two-Point Boundary Value Problems p = 1, q = 1, r = u = c1 cos l - 1x + c sin l - 1x, u¢ = - l - 1c1 sin l - 1x + l - 1c cos l - 1x From the boundary conditions, we have u(0) = c1 = fi u = c sin l - 1x From Ê npx ˆ Ê np ˆ u(2) = c sin l - ◊ = 0, l n - = np Thus l n = + Á ˜ , un = sinÁ ˜ Ë ¯ Ë 2¯ (a) p = 1, q = 0, r = (b) u = c1 cos l x + c sin l x, u¢ = - l c1 sin l x + l c cos l x From the boundary conditions, we have u(0) = c1 = fi u = c sin l x From (a) (b) ( ) [ ] u(1) + u¢ (1) = c sin l + l cos l = 0, l n = - tan l n , un = sin l n x (or u would be zero) From root-finding software, we obtain l1 = 4.11586, l = 24.1393, l = 63.6591 (a) p = 1, q = -4, r = (b) u = c1 cos 2l + x + c sin 2l + x From the boundary conditions, we have ( ) u(0) = c1 = fi u = c sin 2l + x From u( 3) = c sin 2l + ◊ = 0, 2l n + = np Ê np ˆ Ê npx ˆ Thus l n = Á ˜ - 2, un = sinÁ ˜ Ë ¯ 2Ë ¯ 10 (a) p = e x , q = -2e x , r = e x -2 ± - (l + 2) = -1 ± i l - fi u = c1e - x cos l - 1x + c 2e - x sin l - 1x From the boundary conditions, we have 10 (b) r + r + (l + 2) = fi r = ( ) ( ) u(0) = c1 = 0, u(1) = c 2e -1 sin l - = fi l n - = np fi l n = + ( np) , un = e - x sin npx 11 (a) p = e x , q = -e x , r = e x -1 ± - (l + 1) =- ±i l + 2 x x fi u = c1e cosÊË l + xˆ¯ + c 2e sinÊË l + xˆ¯ From the boundary conditions, we have -x u(0) = c1 = fi u = c 2e sinÊË l + xˆ¯ , ˘ -x È u¢ = c 2e Í- sinÊË l + xˆ¯ + l + cosÊË l + xˆ¯ ˙ Î ˚ Also, ˘ -1 -1 È u(1) + u¢ (1) = c 2e sin l + + 2c 2e Í- sinÊË l + ˆ¯ + l + cosÊË l + ˆ¯ ˙ = Ỵ ˚ x ÊÊ 1ˆ ˆ 1ˆ Ê 1ˆ Ê Therefore, l n + = Á n - ˜ p fi l n = - + Á n - ˜ p un = e sinÁ Á n - ˜ px˜ Ë Ë ¯ Ë ¯ Ë 2¯ ¯ 12 (a) p = e - x , q = 0, r = e - x ± - 4l 12 (b) r - r + l = fi r = = ± i l - 14 2 x x fi u = c1e cosÊË l - xˆ¯ + c 2e sinÊË l - xˆ¯ 11 (b) r + r + (l + 1) = fi r = Chapter 13 Linear Two-Point Boundary Value Problems • 357 From the boundary conditions, we have np Ê np ˆ u(0) = c1 = 0, u(2) = c 2e sinÊË l - ˆ¯ = fi l n - = fi ln = + Á ˜ , Ë 2¯ x Ê npx ˆ un = e sinÁ ˜ Ë ¯ 13 (a) p = e x , q = 0, r = e x -1 ± - l 13 (b) r + r + l = fi r = = - ± i l - 14 2 -1 ( x -1) -1 ( x -1) Ê ˆ fi u = c1e cosË l - ( x - 1)¯ + c 2e sinÊË l - ( x - 1)ˆ¯ From the boundary conditions, we have -1 u(1) = c1 = 0, u(2) = c 2e sinÊË l - ˆ¯ = fi l n - = np fi l n = + ( np) , -1 ( x -1) sin( np( x - 1)) 14 (a) p = x, q = 0, r = x 14 (b) First, let u = x r Now, r( r - 1) + r + l = r + l = fi r = ±i l Thus u = c1 cos l ln x + c sin l ln x From the boundary conditions, we have un = e ( ) ( u(1) = c = 0, u( ) = c sin( ) l ln ) = fi Ê np ˆ Ê np ln x ˆ l n ln = np fi l n = Á ˜ , un = sinÁ ˜ Ë ln ¯ Ë ln ¯ x 15 (b) u = c1 cos l ln x + c sin l ln x From the boundary conditions, we have 15 (a) From Problem 14, p = x, q = 0, r = ( ) ( ) l Ê np ˆ u¢ (1) = c l = 0, u¢ ( 3) = -c1 sin l ln = fi l n ln = np fi l n = Á ˜ , Ë ln 3¯ Ê np ln x ˆ un = cosÁ ˜ Ë ln ¯ p = 1, q = -1, r = fi q ¢ = (l + 1) sin q + cos2 q = + l sin q R¢ = -lR sin q cosq u(0) = R(0) sin q (0) = 0, u(1) = R(1) sin q (1) = 0, sin q (0) = sin q (1) = p = 1, q = 2, r = fi q ¢ = ( 3l - 2) sin q + cos2 q R¢ = (1 - 3l + 2) R sin q cosq = -3(l - 1) R sin q cosq u¢ (0) = R(0) cosq (0) = 0, u(1) = R(1) sin q (1) = 0, cosq (0) = 0, sin q (1) = p(0) p = e -2 x , q = 0, r = e -2 x fi q ¢ = le -2 x sin q + e x cos2 q R¢ = (e x - le -2 x ) R sin q cosq u(0) = u(2) = Therefore, sin q (0) = 0, sin q (2) = 2 2 p = e - x , q = -e - x , r = e - x fi q ¢ = e - x (l + 1) sin q + e x cos2 q 2 R¢ = e x - (l + 1)e - x R sin q cosq u(0) = R(0) sin q (0) = 0, u¢ (1) = R(1) cosq (1) = p(1) Therefore, sin q (0) = 0, cosq (1) = ( 18 19 20 21 ( ) ) ... Linear Differential Equations Chapter 3: First Order Nonlinear Differential Equations 26 Chapter 4: Second Order Linear Differential Equations 51 Chapter 5: Higher Order Linear Differential Equations. .. Solutions of Linear Differential Equations 268 Chapter 11: Second Order Partial Differential Equations and Fourier Series ?? Chapter 12: First Order Partial Differential Equations and the Method... initial value problem is n +1 which can be rewritten as ln y = ln y - t n +1 n +1 10 • Chapter First Order Linear Differential Equations Substituting values from the table gives us the necessary equations