HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS

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1 CHAPTER 5 HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS SECTION 5. INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS In this section the central ideas of the theory of linear differential equations are introduced and illustrated concretely in the contet of second-order equations. These key concepts include superposition of solutions (Theorem ), eistence and uniqueness of solutions (Theorem ), linear independence, the Wronskian (Theorem 3), and general solutions (Theorem 4). This discussion of second-order equations serves as preparation for the treatment of nth order linear equations in Section 5.. Although the concepts in this section may seem somewhat abstract to students, the problems set is quite tangible and largely computational. In each of Problems 6 the verification that y and y satisfy the given differential equation is a routine matter. As in Eample, we then impose the given initial conditions on the general solution y = c y + c y. This yields two linear equations that determine the values of the constants c and c.. Imposition of the initial conditions y(0) = 0, y (0) = 5 on the general solution y ce ce c + c = 0, c c = 0 with solution = + yields the two equations c = 5/, c = 5/. Hence the desired particular solution is y() = 5(e - e - )/.. Imposition of the initial conditions y(0) =, y (0) = 5 on the general solution 3 3 y = ce + ce yields the two equations c+ c =, 3c 3c = 5 with solution c =, c = 3. Hence the desired particular solution is y() = e 3-3e Imposition of the initial conditions y(0) = 3, y (0) = 8 on the general solution y = ccos+ csinyields the two equations c = 3, c = 8 with solution c = 3, c = 4. Hence the desired particular solution is y() = 3 cos + 4 sin. 4. Imposition of the initial conditions y(0) = 0, y (0) = 0 on the general solution y = ccos5+ csin5yields the two equations c = 0, 5c = 0 with solution c = 3, c = 4. Hence the desired particular solution is y() = 0 cos 5 - sin 5. Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

2 5. Imposition of the initial conditions y(0) =, y (0) = 0 on the general solution y = ce + ce yields the two equations c+ c =, c+ c = 0 with solution c =, c =. Hence the desired particular solution is y() = e - e. 6. Imposition of the initial conditions y(0) = 7, y (0) = on the general solution 3 y = ce + ce yields the two equations c+ c = 7, c 3c = with solution c = 4, c = 3. Hence the desired particular solution is y() = 4e + 3e Imposition of the initial conditions y(0) =, y (0) = 8 on the general solution y c ce c + c =, c = 8 with solution = + yields the two equations c = 6, c = 8. Hence the desired particular solution is y() = 6 8e Imposition of the initial conditions y(0) = 4, y (0) = on the general solution 3 y = c+ ce yields the two equations c+ c = 4, 3c = with solution c = 4 / 3, c = / 3. Hence the desired particular solution is y() = (4 - e 3 )/3. 9. Imposition of the initial conditions y(0) =, y (0) = on the general solution y ce ce = + yields the two equations c =, c + c = with solution c =, c =. Hence the desired particular solution is y() = e - + e Imposition of the initial conditions y(0) = 3, y (0) = 3 on the general solution 5 5 y = ce + ce yields the two equations c = 3, 5c+ c = 3 with solution c = 3, c =. Hence the desired particular solution is y() = 3e 5 - e 5.. Imposition of the initial conditions y(0) = 0, y (0) = 5 on the general solution y cecos cesin = + yields the two equations c = 0, c + c = 5 with solution c = 0, c = 5. Hence the desired particular solution is y() = 5e sin.. Imposition of the initial conditions y(0) =, y (0) = 0 on the general solution 3 3 y = ce cos+ ce sinyields the two equations c =, 3c+ c = 5 with solution c =, c = 3. Hence the desired particular solution is y() = e -3 ( cos + 3 sin ). 3. Imposition of the initial conditions y() = 3, y () = on the general solution y = c + c yields the two equations c+ c = 3, c+ c = with solution c = 5, c =. Hence the desired particular solution is y() = Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

3 4. Imposition of the initial conditions y() = 0, y () = 5 on the general solution 3 y = c + c yields the two equations 4 c+ c/8= 0, 4c 3 c/6= 5 with solution c = 3, c = 6. Hence the desired particular solution is y() = 3-6/ Imposition of the initial conditions y() = 7, y () = on the general solution y = c + c lnyields the two equations c = 7, c+ c = with solution c = 7, c = 5. Hence the desired particular solution is y() = 7-5 ln. 6. Imposition of the initial conditions y() =, y () = 3 on the general solution y = ccos(ln ) + csin(ln ) yields the two equations c =, c = 3. Hence the desired particular solution is y() = cos(ln ) + 3 sin(ln ). 7. If y = c/ then or c =. + = / + / = ( )/ 0 unless either c = 0 y y c c c c 8. If y 3 = c then = 6 = 6 6 unless yy c c c c =. 9. If y = + then yy + y = + + = 3/ / 3/ ( /4) ( /) / Linearly dependent, because f() = π = π(cos + sin ) = π g(). Linearly independent, because 3 =+ if > 0, whereas 3 = if < 0.. Linearly independent, because + = c( + ) would require that c = with = 0, but c = 0 with = -. Thus there is no such constant c. 3. Linearly independent, because f() = +g() if > 0, whereas f() = -g() if < Linearly dependent, because g() = f(). 5. f() = e sin and g() = e cos are linearly independent, because f() = k g() would imply that sin = k cos, whereas sin and cos are linearly independent. 6. To see that f() and g() are linearly independent, assume that f() = c g(), and then substitute both = 0 and = π/. 7. Let L[y] = y + py + qy. Then L[y c ] = 0 and L[y p ] = f, so L[y c + y p ] = L[y c ] + [y p ] = 0 + f = f. Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

4 8. If y() = + c cos + c sin then y () = -c sin + c cos, so the initial conditions y(0) = y (0) = - yield c = -, c = -. Hence y = - cos - sin. 9. There is no contradiction because if the given differential equation is divided by to get the form in Equation (8) in the tet, then the resulting functions p() = -4/ and q() = 6/ are not continuous at = (a) y = and y = are linearly independent because require that c = with =, but c = - with = -. = c would 3 3 (b) The fact that W(y, y ) = 0 everywhere does not contradict Theorem 3, because when the given equation is written in the required form y - (3/)y + (3/ )y = 0, the coefficient functions p() = -3/ and q() = 3/ are not continuous at = W(y, y ) = - vanishes at = 0, whereas if y and y were (linearly independent) solutions of an equation y + py + qy = 0 with p and q both continuous on an open interval I containing = 0, then Theorem 3 would imply that W 0 on I. 3. (a) W = y y - y y, so AW' = A(y y + y y - y y - y y ) = y (Ay ) - y (Ay ) = y (-By - Cy ) - y (-By - Cy ) = -B(y y - y y ) (b) (c) and thus AW' = -BW. Just separate the variables. Because the eponential factor is never zero. In Problems 33 4 we give the characteristic equation, its roots, and the corresponding general solution r r r 3 + = 0; =,; y() = c e + c e r r r + 5 = 0; = 3, 5; y() = c e -5 +c e 3 r r r + 5 = 0; = 0, 5; y() = c + c e Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

5 r + r = r = y() = c + c e -3/ 3 0; 0, 3/ ; r r = r = y() = c e -/ + c e 0;, /; r + r+ = r = y() = c e -/ +c e -3/ ; /, 3/; r + r+ = r = y() = (c + c )e -/ 4 4 0; /, /; r r+ = r = y() = (c + c )e / ; /3, /3; r r = r = y() = c e -4/3 + c e 5/ ; 4/3,5/; r r = r = y() = c e -4/7 + c e 3/5 35 0; 4 / 7, 3/ 5; In Problems we first write and simplify the equation with the indicated characteristic roots, and then write the corresponding differential equation r r+ = r + r = y + 0y = 0 ( 0)( 0) 0 0; r r+ = r = y 00y = 0 ( 0)( 0) 00 0; r+ r+ = r + r+ = y + 0y + 00y = 0 ( 0)( 0) ; r r = r r+ = y 0y + 000y = 0 ( 0)( 00) ; r r = r = y = 0 ( 0)( 0) 0; r r + = r r = y y y = 0 ( )( ) 0; 49. The solution curve with y(0) =, y (0) = 6 is 8 y = e 7e. We find that y = 0 when = ln(7/4) so e = 4 / 7 and e = 6 / 49. It follows that y (ln(7 / 4)) = 6/ 7, so the high point on the curve is (ln(7 / 4)),6 / 7) (0.56,.9), which looks consistent with Fig The two solution curves with y(0) = a and y(0) = b (as well as y (0) = ) are y = a+ e a+ e, y = b+ e b+ e. Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

6 Subtraction and then division by a - b gives e = e, so it follows that = -ln. Now substitution in either formula gives y = -, so the common point of intersection is (-ln, -). 5. (a) The substitution v = ln gives dy dy dv dy y = = = d dv d dv Then another differentiation using the chain rule gives d y d dy d dy y = = = d d d d dv = + = + dv dv dv d dv dv dy d dy dv dy d y. Substitution of these epressions for immediately the desired Eq. (): y and y into Eq. () in the tet then yields d y dy a + ( b a) + cy = 0. dv dv (b) If the roots r and r of the characteristic equation of Eq. () are real and distinct, then a general solution of the original Euler equation is v r v rv rv r r r y = ce + ce = c e + c e = c + c. 5. The substitution v = ln yields the converted equation d y/ dv y = 0 whose v characteristic equation r = 0 has roots r = and r =. Because e =, the corresponding general solution is y = ce + c e = c + v v c. 53. The substitution v = ln yields the converted equation d y / dv + dy / dv y = 0 whose characteristic equation r + r = 0 has roots r = 4 and r = 3. Because v e =, the corresponding general solution is y = ce + c e = c + c 4v 3v The substitution v = ln yields the converted equation whose characteristic equation 4r 4r 3 0 v Because e =, the corresponding general solution is 4 d y / dv + 4 dy / dv 3y = 0 + = has roots r = 3/ and r = /. 38 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

7 y = ce + c e = c + c 3 v/ v/ 3/ /. 55. The substitution v = ln yields the converted equation d y/ dv = 0 whose characteristic equation r = 0 has repeated roots r, r = 0. Because v = ln, the corresponding general solution is y = c + c v = c + c ln. 56. The substitution v = ln yields the converted equation d y / dv 4 dy / dv + 4y = 0 v whose characteristic equation r 4r+ 4= 0 has roots r, r =. Because e =, the corresponding general solution is y = ce + c ve = ( c + c ln v). v v SECTION 5. GENERAL SOLUTIONS OF LINEAR EQUATIONS Students should check each of Theorems through 4 in this section to see that, in the case n =, it reduces to the corresponding theorem in Section 5.. Similarly, the computational problems for this section largely parallel those for the previous section. By the end of Section 5. students should understand that, although we do not prove the eistence-uniqueness theorem now, it provides the basis for everything we do with linear differential equations. The linear combinations listed in Problems 6 were discovered "by inspection" that is, by trial and error.. (5/)() + (-8/3)(3 ) + (-)(5-8 ) = 0. (-4)(5) + (5)( - 3 ) + ()(0 + 5 ) = 0 3. ()(0) + (0)(sin ) + (0)(e ) = 0 4. ()(7) + (-7/)( sin ) + (-7/3)(3 cos ) = 0, because sin + cos =. 5. ()(7) + (-34)(cos ) + (7)(cos ) = 0, because cos = + cos. 6. (-)(e ) + ()(cosh ) + ()(sinh ) = 0, because cosh = (e + e - )/ and sinh = (e - e - )/. Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

8 7. W = 0 = is nonzero everywhere e e e = = is never zero. W e e 3e e e 4e 9e 3 9. W = e (cos + sin ) = e 0 0. W = -7 e ( + )( + 4) is nonzero for > 0.. W = 3 e is nonzero if 0.. W = - [ cos (ln ) + sin (ln )] = - is nonzero for > 0. In each of Problems 3-0 we first form the general solution y() = c y () + c y () + c 3 y 3 (), then calculate y () and y (), and finally impose the given initial conditions to determine the values of the coefficients c, c, c Imposition of the initial conditions y(0) =, y (0) =, y (0) = 0 on the general solution y = ce + ce + ce yields the three equations 3 c + c + c =, c c c =, c + c + 4c = with solution c = 4/3, c = 0, c3 = /3. Hence the desired particular solution is given by y() = (4e - e - )/3. 4. Imposition of the initial conditions y(0) = 0, y (0) = 0, y (0) = 3 on the general solution y = ce + ce + ce yields the three equations 3 3 c + c + c =, c + c + 3c =, c + 4c + 9c = with solution c = 3/, c = 3, c3 = 3/. Hence the desired particular solution is given by y() = (3e - 6e + 3e 3 )/. 5. Imposition of the initial conditions y(0) =, y (0) = 0, y (0) = 0 on the general solution y = ce + ce + ce yields the three equations Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

9 c =, c + c = 0, c + c + c = 0 3 with solution c =, c =, c3 =. Hence the desired particular solution is given by y() = ( - + )e. 6. Imposition of the initial conditions y(0) =, y (0) = 4, y (0) = 0 on the general solution y = ce + ce + ce yields the three equations 3 c + c =, c + c + c = 4, c + 4c + 4c = with solution c =, c = 3, c3 = 0. Hence the desired particular solution is given by y() = -e + 3e - 0e. 7. Imposition of the initial conditions y(0) = 3, y (0) =, y (0) = on the general solution y = c+ ccos3+ c3sin3yields the three equations c + c = 3, 3c =, 9c = 3 with solution c = 9/9, c = /9, c3 = /3. Hence the desired particular solution is given by y() = (9 - cos 3-3 sin 3)/9. 8. Imposition of the initial conditions y(0) =, y (0) = 0, y (0) = 0 on the general solution y = e c+ ccos+ csin yields the three equations 3 c + c =, c + c + c = 0, c + c = with solution c =, c =, c3 =. Hence the desired particular solution is given by y() = e ( - cos - sin ). 9. Imposition of the initial conditions y() = 6, y () = 4, y () = on the general solution y = c+ c + c yields the three equations 3 3 c + c + c = 6, c + c + 3c = 4, c + 6c = with solution c =, c =, c3 = 3. Hence the desired particular solution is given by y() = Imposition of the initial conditions y() =, y () = 5, y () = on the general solution y = c+ c + c lnyields the three equations 3 Section 5. 4 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

10 c + c =, c c + c = 5, 6c 5c = 3 3 with solution c =, c =, c3 =. Hence the desired particular solution is given by y() = ln. In each of Problems -4 we first form the general solution y() = y c () + y p () = c y () + c y () + y p (), then calculate y (), and finally impose the given initial conditions to determine the values of the coefficients c and c.. Imposition of the initial conditions y(0) =, y (0) = on the general solution y = ccos+ csin+ 3yields the two equations c =, c + 3 = with solution c =, c = 5. Hence the desired particular solution is given by y() = cos - 5 sin Imposition of the initial conditions y(0) = 0, y (0) = 0 on the general solution y = ce + ce 3 yields the two equations c+ c 3= 0, c c = 0 with solution c = 4, c =. Hence the desired particular solution is given by y() = 4e - e Imposition of the initial conditions y(0) = 3, y (0) = on the general solution 3 y = ce + ce yields the two equations c+ c = 3, c+ 3c = with solution c =, c = 4. Hence the desired particular solution is given by y() = e - + 4e Imposition of the initial conditions y(0) = 4, y (0) = 8 on the general solution y = ce cos+ ce cos+ + yields the two equations c+ = 4, c+ c + = 8 with solution c = 3, c = 4. Hence the desired particular solution is given by y() = e (3 cos + 4 sin ) L[y] = L[y + y ] = L[y ] + L[y ] = f + g 6. (a) y = and y = 3 (b) y = y+ y = The equations c + c + c 3 = 0, c + c 3 + 0, c 3 = 0 4 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

11 (the latter two obtained by successive differentiation of the first one) evidently imply by substituting = 0 that c = c = c 3 = 0. n 8. If you differentiate the equation c0 + c+ c + + cn = 0 repeatedly, n times in succession, the result is the system c c c c n n = 0 c c nc n n = 0 ( n )! c + n! c = 0 n n nc! n = 0 of n+ equations in the n+ coefficients c0, c, c,, c n. Upon substitution of = 0, the (k+)st of these equations reduces to kc! k = 0, so it follows that all these coefficients must vanish. 9. If c 0 e r + c e r + + c n n e r = 0, then division by e r yields c 0 + c + + c n n = 0, so the result of Problem 8 applies. 30. When the equation y - y + y = 0 is rewritten in standard form y + (-/)y + (/ )y = 0, the coefficient functions p () = -/ and p () = / are not continuous at = 0. Thus the hypotheses of Theorem 3 are not satisfied. 3. (a) Substitution of = a in the differential equation gives y ( a) = p y ( a) q( a). (b) If y(0) = and y (0) = 0, then the equation y - y - 5y = 0 implies that y (0) = y (0) + 5y(0) = Let the functions y, y,, y n be chosen as indicated. Then evaluation at = a of the (k - )st derivative of the equation c y + c y + c n y n = 0 yields c k = 0. Thus c = c = = c n = 0, so the functions are linearly independent. 33. This follows from the fact that Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

12 a b c = ( b a)( c b)( c a). a b c 34. W(f, f,, f n ) = V ep(r i ), and neither V nor ep(r i ) vanishes. 36. If y = vy then substitution of the derivatives y = vy + vy, y = vy + vy + v y in the differential equation y + py + qy = 0 gives [ vy vy v y] p[ vy vy ] q[ vy] v[ y py qy ] v y vy pvy = 0, = 0. But the terms within brackets vanish because y is a solution, and this leaves the equation yv + y + py v = 0 that we can solve by writing v y = = + v p ln v ln y p d ln C, y p d C p d e y y v = e v = C d+ K. With C = and K = 0this gives the second solution e y = y d. y p d When we substitute y = v in the given differential equation and simplify, we get the separable equation v + v = 0 that we solve by writing v = ln v = ln + ln A, v A v = v = Aln + B. With A= and B = 0we get v = lnand hence y = 3 ln. 44 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

13 3 38. When we substitute y = v in the given differential equation and simplify, we get the separable equation v + 7v = 0 that we solve by writing v 7 = ln v = 7ln + ln A, v A A v = v = + B With A= 6andB = 0we get v / 6 = and hence y = / 3. / 39. When we substitute y = ve in the given differential equation and simplify, we eventually get the simple equation v = 0 with general solution v = A+ B. / With A= and B = 0we get v = and hence y. = e 40. When we substitute y = v in the given differential equation and simplify, we get the separable equation v v = 0 that we solve by writing v = lnv = + ln A, v v = Ae v = Ae + B. With A= and B = 0we get v = e and hence y. = e 4. When we substitute y = ve in the given differential equation and simplify, we get the separable equation ( + v ) + v = 0 that we solve by writing v = = + lnv = + ln( + ) + ln A, v + + v = A( + e ) v = A ( + e ) d = A( + e ) + B. With A= and B = 0we get v = ( + e ) and hence y = When we substitute y = v in the given differential equation and simplify, we get the separable equation ( ) v = v that we solve by writing Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

14 v = = +, v ( ) + ln v = ln + ln( + ) + ln( ) + ln A, ( ) A v = = A v = A + B. With A= and B = 0we get v = + / and hence y = When we substitute y = v in the given differential equation and simplify, we get the separable equation ( ) v = ( 4 ) v that we solve by writing v 4 = = +, v ( ) + ln v = ln ln( + ) ln( ) + ln A, A v = = A + +, ( ) + v = A + ln( + ) ln( ) + B. With A= and B = 0we get + v = ln( + ) + ln( ) y = ln. 44. When we substitute y = v / cos in the given differential equation and simplify, we eventually get the separable equation (cos ) v = (sin ) v that we solve by writing v sin = v = + A = + A v cos v = Asec v = Atan + B. ln ln cos ln lnsec ln, With A= and B = 0we get v = tanand hence y / / (tan )( = cos ) = sin. 46 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

15 SECTION 5.3 HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS This is a purely computational section devoted to the single most widely applicable type of higher order differential equations linear ones with constant coefficients. In Problems 0, we write first the characteristic equation and its list of roots, then the corresponding general solution of the given differential equation. Eplanatory comments are included only when the solution of the characteristic equation is not routine. r 4 = r r+ = 0; r =,; y = c e + c e -. r 3r = r r 3 = 0; r = 0, 3/ ; y = c + c e 3/. r + 3r 0 = r+ 5 r = 0; r = 5,; y = c e + c e r 7r+ 3 = r r 3 = 0; r = /,3; y = c e / + c e r + 6r+ 9 = r+ 3 = 0; r = 3, 3; y = c e -3 + c e r + r+ = r = ( ± ) 5 5 0; 5 5 / y() = e -5/ [c ep( 5/) + c ep(- 5/)] 7. 4r r+ 9 = r 3 = 0; r = 3/, 3/ ; y = c e 3/ + c e 3/ r 6r+ 3 = 0; r = 6± 6 / = 3± i; y = e 3 (c cos + c sin ) 8. r + 8r+ 5 = 0; r = 8± 36 / = 4± 3 i; y = e -4 (c cos 3 + c sin 3) 9. 5r + 3r = r 5r+ 3 = 0; r = 0, 0, 0, 3/ 5; y = c + c + c 3 + c 4 e -3/ r 8r + 6r = r r 4 = 0; r = 0, 0, 4, 4; y = c + c + c 3 e 4 + c 4 e r 3r + 3r r = r r = 0; r = 0,,,; y = c + c e + c 3 e + c 4 e r + r + 4r = r 3r+ = 0; r = 0, /3, /3 y() = c + c e -/3 + c 3 e -/3 Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

16 4 4. r + 3r 4 = r r + 4 = 0; r =,, ± i y() = c e + c e - + c 3 cos + c 4 sin 5. r r + = r = r r+ = r = ;,,, y() = c e + c e + c 3 e - + c 4 e r + 8r + 8 = r + 9 = 0; r = ± 3 i, ± 3i y() = (c + c )cos 3 + (c 3 + c 4 )sin 3 7. r + r + = r + r + = r = ± i ± i (3 4) 0; /, / 3, y() = c cos(/ ) + c sin(/ ) + c 3 cos(/ 3) + c 4 sin(/ 3) 4 8. r 6 = r 4 r + 4 = 0; r =,, ± i y() = c e + c e - + c 3 cos + c 4 sin r + r r = r r + r = r r+ = 0; r =,, ; 3 9. y() = c e + c e - + c 3 e - 0. r 4 + r 3 + 3r + r + = (r + r + ) = 0; ( ± i) ( ± i) y = e -/ (c + c )cos( 3/) + e -/ (c 3 + c 4 )sin( 3/) 3 /, 3 /. Imposition of the initial conditions y(0) = 7, y (0) = on the general solution 3 y = ce + ce yields the two equations c+ c = 7, c+ 3c = with solution c = 5, c =. Hence the desired particular solution is y() = 5e + e 3.. Imposition of the initial conditions y(0) = 3, y (0) = 4 on the general solution /3 y = e ccos ( / 3) + csin ( / 3) yields the two equations c = 3, c/ 3 + c/ 3 = 4 with solution c = 3, c = 5 3. Hence the desired particular /3 solution is y = e 3cos ( / 3) + 5 3sin ( / 3 ). 3. Imposition of the initial conditions y(0) = 3, y (0) = on the general solution 3 y = e ( ccos4+ csin4) yields the two equations c = 3, 3c+ 4c = with solution c = 3, c =. Hence the desired particular solution is y() = e 3 (3 cos 4 - sin 4). 48 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

17 4. Imposition of the initial conditions y(0) =, y (0) =, y (0) = 3 on the general solution y = c+ ce + ce yields the three equations / 3 c + c + c =, c c / =, 4 c + c / 4 = with solution c = 7 /, c = /, c3 = 4. Hence the desired particular solution is y() = (-7 + e + 8e -/ )/. 5. Imposition of the initial conditions y(0) =, y (0) = 0, y (0) = on the general solution y c c ce /3 = yields the three equations c + c =, c c / 3 = 0, 4 c / 9 = with solution c = 3/4, c = 3/, c3 = 9/4. Hence the desired particular solution is y() = ( e -/3 )/4. 6. Imposition of the initial conditions y(0) =, y (0) =, y (0) = 3 on the general solution y = c+ ce + ce yields the three equations c + c = 3, 5c + c = 4, 5c 0c = with solution c = 4/5, c = 9/5, c3 = 5. Hence the desired particular solution is y() = (4-9e -5-5e -5 )/ First we spot the root r =. Then long division of the polynomial r + 3r 4 by r - yields the quadratic factor r + 4r+ 4 = ( r+ ) with roots r = -, -. Hence the general solution is y() = c e + c e - + c 3 e First we spot the root r =. Then long division of the polynomial r 3 - r - 5r - by the factor r - yields the quadratic factor r + 3r + = (r + )(r + ) with roots r = -, -/. Hence the general solution is y() = c e + c e - + c 3 e -/ First we spot the root r = 3. Then long division of the polynomial r + 7 by r + 3 yields the quadratic factor r 3r 9 r = 3± i 3 /. Hence the + with roots general solution is y() = c e -3 + e 3/ [c cos(3 3/) + c 3 sin(3 3/)]. 30. First we spot the root r = -. Then long division of the polynomial r 4 - r 3 + r - 3r - 6 Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

18 by r + yields the cubic factor r 3 - r + 3r - 6. Net we spot the root r =, and another long division yields the quadratic factor r + 3 with roots r = ± i 3. Hence the general solution is y() = c e - + c e + c 3 cos 3 + c 4 sin The characteristic equation r 3 + 3r + 4r - 8 = 0 has the evident root r =, and long division then yields the quadratic factor r + 4r + 8 = (r + ) + 4 corresponding to the comple conjugate roots - ± i. Hence the general solution is y() = c e + e - (c cos + c 3 sin ). 3. The characteristic equation r 4 + r 3-3r - 5r - = 0 has root r = that is readily found by trial and error, and long division then yields the factorization (r - )(r + ) 3 = 0. Thus we obtain the general solution y() = c e + (c + c 3 + c 4 )e Knowing that y = e 3 is one solution, we divide the characteristic polynomial r 3 + 3r - 54 by r - 3 and get the quadratic factor r + 6r + 8 = (r + 3) + 9. Hence the general solution is y() = c e 3 + e -3 (c cos 3 + c 3 sin 3). 34. Knowing that y = e /3 is one solution, we divide the characteristic polynomial 3r 3 - r + r - 8 by 3r - and get the quadratic factor r + 4. Hence the general solution is y() = c e /3 + c cos + c 3 sin. 35. The fact that y = cos is one solution tells us that r + 4 is a factor of the characteristic polynomial 6r 4 + 5r 3 + 5r + 0r + 4. Then long division yields the quadratic factor 6r + 5r+ = (3r+ )(r+ ) with roots r = /, /3. Hence the general solution is y() = c e -/ + c e -/3 + c 3 cos + c 4 sin 36. The fact that y = e - sin is one solution tells us that (r + ) + = r + r + is a factor of the characteristic polynomial 9r 3 + r + 4r Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

19 Then long division yields the linear factor 9r - 7. Hence the general solution is y() = c e 7/9 + e - (c cos + c 3 sin ) The characteristic equation is r r = r 3 ( r ) = 0, so the general solution is y = A+ B+ C + De. Imposition of the given initial conditions yields the equations A+ D = 8, B+ D =, C+ D = 3, D = 7 with solution A=, B = 5, C = 3, D = 7. Hence the desired particular solution is y = e. 38. Given that r = 5 is one characteristic root, we divide (r 5) into the characteristic 3 polynomial r 5r + 00r 500 and get the remaining factor r Thus the general solution is 5 y = Ae + Bcos0+ Csin0. Imposition of the given initial conditions yields the equations A+ B = 0, 5A+ 0C = 0, 5A 00B = 50 with solution A=, B =, C = 0. Hence the desired particular solution is 5 y = e cos r = r r + r, so the differential equation is y 6y + y 8y = ( r )( r 4) r r 4r 8 + = +, so the differential equation is 4. 4 ( r 4)( r 4) r 6 y y + 4y 8y = 0. + =, so the differential equation is y (4) 6 y = ( r 4) r r 48r 64 + = + + +, so the differential equation is = 0. (6) (4) y y y y 44. (a) = i, -i (b) = -i, 3i Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

20 45. The characteristic polynomial is the quadratic polynomial of Problem 44(b). Hence the general solution is y = ce + ce = c(cos isin ) + c(cos3+ isin 3 ). i 3i 46. The characteristic polynomial is r ir r i r i + 6 = ( + )( 3 ) so the general solution is y = ce + ce = c(cos3+ isin 3 ) + c(cos isin ). 3i i 47. The characteristic roots are r = ± + i 3 = ± (+ i 3) so the general solution is y = ce + ce = ce cos 3 + isin 3 + ce cos 3 isin 3 (+ i 3) (+ i 3) 48. The general solution is y() = Ae + Be α + Ce ß where α = (- + i 3 )/ and β = (- - i 3 )/. Imposition of the given initial conditions yields the equations A+ B+ C = A+ αb+ βc = 0 A+ α B+ β C = 0 that we solve for A = B = C = /3. Thus the desired particular solution is given ( + i 3) / ( 3) / by y = 3 ( e + e + e ), which (using Euler's relation) reduces to the given real-valued solution. 49. The general solution is y = Ae + Be - + C cos + D sin. Imposition of the given initial conditions yields the equations A+ B+ C = 0 A B + D = 0 4A+ B C = 0 8A B D = 30 that we solve for A =, B = -5, C = 3, and D = -9. Thus y() = e - 5e cos - 9 sin. 50. If > 0 then the differential equation is y + y = 0 with general solution y = Acos + Bsin.. But if < 0 it is y y = 0 with general solution y = C cosh + Dsin. To satisfy the initial conditions y(0) =, y (0) = 0 we choose 5 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

21 A= C = and B = D = 0. But to satisfy the initial conditions y(0) = 0, y (0) = we choose A= C = 0 and B = D =. The corresponding solutions are defined by cos if 0, sin if 0, y = y = cosh if 0; sinh if In the solution of Problem 5 in Section 3. we showed that the substitution v = ln gives dy dy d y dy d y y = = and y = = +. d dv d dv dv A further differentiation using the chain rule gives 3 d y dy 3 d y 3 d y y = = + d dv dv dv Substitution of these epressions for y, y, and y into the third-order Euler equation 3 a y + b y + c y + d y = 0 and collection of coefficients quickly yields the desired constant-coefficient equation 3 d y d y dy 3 a + ( b 3 a) + ( c b+ a) + d y = 0. dv dv dv In Problems 5 through 58 we list first the transformed constant-coefficient equation, then its characteristic equation and roots, and finally the corresponding general solution with v = ln v and e = d y 9 0; 9 0; 3 y r r i dv + = + = = ± y = ccos(3 v) + csin(3 v) = ccos(3ln ) + csin(3ln ) d y dy = 0; = 0; = 3± 4 y r r r i dv dv 3v 3 y e = ccos(4 v) + csin(4 v) = ccos(4ln ) + csin(4ln ) [ ] [ ] d y d y dv dv y = v c+ cv+ ce = c+ cln+ c = 0; r + 3r = 0; r = 0, 0, d y d y dy = 0; r 4r + 4r = 0; r = 0,, 3 dv dv dv v v y = c+ ce + cve = c+ c+ cln 3 3 Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

22 d y 3 = 0; r = 0; r = 0, 0, 0 3 dv y = c+ cv+ cv = c+ cln + c(ln ) d y d y dy = 0; r 4r + 4r = 0; r = 0, 3± 3 3 dv dv dv y = c+ ce + cve = c+ c + c (3 3) v (3+ 3) v d y d y dy y = 0; r + 3r + 3r+ = 0; r =,, 3 dv dv dv v v v y = ce + cve + cve 3 = c+ cln + c3(ln ) SECTION 5.4 Mechanical Vibrations In this section we discuss four types of free motion of a mass on a spring undamped, underdamped, critically damped, and overdamped. However, the undamped and underdamped cases in which actual oscillations occur are emphasized because they are both the most interesting and the most important cases for applications.. Frequency: ω0 = k/ m = 6/4 = rad/sec= / π Hz Period: P = π / ω0 = π /= π sec. Frequency ω0 = k/ m = 48/ 0.75 = 8 rad / sec = 4 / π Hz Period: P = π / ω0 = π /8 = π /4 sec 3. The spring constant is k = 5 N/0.0 m = 75 N/m. The solution of = 0 with (0) = 0 and (0) = -0 is (t) = - sin 5t. Thus the amplitude is m; the frequency is ω0 = k/ m = 75/3 = 5 rad/sec=.5/ π Hz; and the period is π/5 sec. 4. (a) With m = /4 kg and k = (9 N)/(0.5 m) = 36 N/m we find that ω 0 = rad/sec. The solution of + 44 = 0 with (0) = and (0) = -5 is (t) = cos t - (5/)sin t = (3/)[(/3)cos t - (5/3)sin t] (t) = (3/)cos(t - α) 54 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

23 where α = π - tan - (5/) (b) C = 3/.0833 m and T = π/ sec. 5. The gravitational acceleration at distance R from the center of the earth is g = GM/R. According to Equation (6) in the tet the (circular) frequency ω of a pendulum is given by ω = g/l = GM/R L, so its period is p = π/ω = πr L / GM. 6. If the pendulum in the clock eecutes n cycles per day (86400 sec) at Paris, then its period is p = 86400/n sec. At the equatorial location it takes 4 hr min 40 sec = sec for the same number of cycles, so its period there is p = 86560/n sec. Now let R = 3956 mi be the Earth s "radius" at Paris, and R its "radius" at the equator. Then substitution in the equation p /p = R /R of Problem 5 (with L = L ) yields R = mi. Thus this (rather simplistic) calculation gives 7.33 mi as the thickness of the Earth's equatorial bulge. 7. The period equation p = = ( ) 00 yields.9795 mi 0,450 ft for the altitude of the mountain. 8. Let n be the number of cycles required for a correct clock with unknown pendulum length L and period p to register 4 hrs = sec, so np = The given clock with length L = 30 in and period p loses 0 min = 600 sec per day, so np = Then the formula of Problem 5 yields L p np = = = L p np 86400, so L = (30)(86400 /87000) 9.59 in. 0. The F = ma equation ρπr h = ρπr hg - πr g simplifies to + (g/ρh) = g. The solution of this equation with (0) = (0) = 0 is (t) = ρh( - cos ω 0 t) where ω 0 = g / ρ h. With the given numerical values of ρ, h, and g, the amplitude of oscillation is ρh = 00 cm and the period is p = π ρ h/ g.0 sec. Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

24 . The fact that the buoy weighs 00 lb means that mg = 00 so m = 00/3 slugs. The weight of water is 6.4 lb/ft 3, so the F = ma equation of Problem 0 is (00/3)'' = πr. It follows that the buoy's circular frequency ω is given by ω = (3)(6.4π)r /00. But the fact that the buoy s period is p =.5 sec means that ω = π/.5. Equating these two results yields r ft 3.8 in.. (a) Substitution of M r = (r/r) 3 M in F r = -GM r m/r yields F r = -(GMm/R 3 )r. (b) Because GM/R 3 = g/r, the equation mr'' = F r yields the differential equation r'' + (g/r)r = 0. (c) The solution of this equation with r(0) = R and r'(0) = 0 is r(t) = Rcos ω 0 t where ω 0 = g / R. Hence, with g = 3. ft/sec and R = (3960)(580) ft, we find that the period of the particle s simple harmonic motion is p = π/ω 0 = π R / g sec min. 3. (a) The characteristic equation 0r + 9r+ = (5r+ )(r+ ) = 0 has roots r = /5, /. When we impose the initial conditions (0) = 0, (0) = 5 on the t/5 t/ general solution () t = ce + c e we get the particular solution ( t/5 t/) t () = 50 e e. = 5 0 = = 0 t/ t/5 t/5 t/0 (b) The derivative t e e e ( e ) when t = 0 ln(5/ 4).344. Hence the mass's farthest distance to the right is given by (0ln(5/ 4)) = 5 /5 = (a) The characteristic equation 5r 0r 6 (5r ) = + + = has roots r = ± 5 i /5 = /5± 3 i. When we impose the initial conditions t /5 (0) 0, (0) 4 () t = e Acos3t+ Bsin3t we get A = 0, B = 5. The corresponding particular solution is given by = = on the general solution 56 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

25 t/5 t/5 t () = e (0cos3t+ 5sin3) t = 5e cos(3 t α) where α = tan (3/ 4) (b) Thus the oscillations are "bounded" by the curves /5 = ± 5e t and the pseudoperiod of oscillation is T = π /3 (because ω = 3). 5. With damping The characteristic equation (/ ) r + 3r+ 4 = 0 has roots r =, 4. When we impose the initial conditions (0) =, (0) = 0 on the general solution t 4t () t = ce + ce we get the particular solution (t) = 4e -t - e -4t that describes overdamped motion. Without damping The characteristic equation (/ ) r + 4 = 0 has roots r =± i. When we impose the initial conditions (0) =, (0) = 0 on the general solution ut () = Acos( t) + Bsin( t) we get the particular solution ut () = cos( t). The graphs of () t and u() t are shown in the following figure. 3 t u - 6. With damping The characteristic equation 3r + 30r+ 63= 0 has roots r = 3, 7. When we impose the initial conditions (0) =, (0) = on the general solution 3t 7t () t = ce + ce we get the particular solution (t) = 4e -3t - e -7t that describes overdamped motion. Without damping The characteristic equation 3r + 63= 0 has roots r =± i. When we impose the initial conditions (0) =, (0) = on the general solution ut () = Acos t + Bsin t we get the particular solution ut = cos( t ) + sin( ) cos( 0.49) t t. The graphs of () t and u() t are shown in the figure at the top of the net page. Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

26 u t - 7. With damping The characteristic equation r + 8r+ 6= 0 has roots r = 4, 4. When we impose the initial conditions (0) = 5, (0) = 0 on the general solution 4t () t = ( c+ ct) e 4t we get the particular solution t () = 5 e (t+ ) that describes critically damped motion. Without damping The characteristic equation r + 6 = 0 has roots r =± 4. i When we impose the initial conditions (0) = 5, (0) = 0 on the general solution ut = Acos(4 t) + Bsin(4 t) we get the particular solution 5 5 ut = 5cos(4 t) + sin(4 t) 5 cos(4t 5.895). The graphs of () t and u() t are shown in the following figure. 5 u t With damping The characteristic equation r + r+ 50= 0 has roots r = 3± 4 i. When we impose the initial conditions (0) = 0, (0) = 8 on the general solution 3t () t e Acos4t Bsin4t = + we get the particular solution 3t 3t t () = e sin4t = e cos(4t 3 π /) that describes underdamped motion. 58 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

27 Without damping The characteristic equation r + 50= 0 has roots r =± 5 i. When we impose the initial conditions (0) = 0, (0) = 8 on the general solution ut = Acos(5 t) + Bsin(5 t) we get the particular solution 8 8 3π ut () = sin(5) t = cos 5t 5 5. The graphs of () t and u() t are shown in the following figure. - u t 9. The characteristic equation 4r + 0r+ 69 = 0 has roots r = 5/± 6 i. When we impose the initial conditions (0) = 4, (0) = 6 on the general solution 5 t / () t = e Acos6t+ Bsin6t we get the particular solution (t) = 5 t / e [ 4 cos 6t sin 6t ] t / e cos(6t ) that describes underdamped motion. Without damping The characteristic equation 4r + 69 = 0 has roots r =± 3 i/. When we impose the initial conditions (0) = 4, (0) = 6 on the general solution u() t = Acos(3/) t + Bsin(3 t/) we get the particular solution u(t) = 3t 3 3t 4cos + sin cos t. 4 t -4 u Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

28 0. With damping The characteristic equation r + 6r+ 40= 0 has roots r = 4± i. When we impose the initial conditions (0) = 5, (0) = 4 on the general solution 4t () t = e Acost+ Bsint we get the particular solution (t) = e -4t (5 cos t + sin t) 3 e -4t cos(t -.760) that describes underdamped motion. Without damping The characteristic equation r + 40= 0 has roots r =± 5 i. When we impose the initial conditions (0) = 5, (0) = 4 on the general solution ut () = Acos( 5 t) + Bsin( 5 t) we get the particular solution 9 u(t) = 5cos( 5t) + sin( 5t) cos ( ) 5 The graphs of () t and u() t are shown in the following figure. 5 t. 5 t -5 u. With damping The characteristic equation r + 0r+ 5 = 0 has roots r = 5± 0 i. When we impose the initial conditions (0) = 6, (0) = 50 on the general solution 5t () t = e Acos0t+ Bsin0t we get the particular solution (t) = e -5t (6 cos 0t + 8 sin 0t) 0 e -5t cos(0t ) that describes underdamped motion. Without damping The characteristic equation r + 5 = 0 has roots r =± 5 5 i. When we impose the initial conditions (0) = 6, (0) = 50 on the general solution ut () = Acos(5 5 t) + Bsin(5 5 t) we get the particular solution u(t) = 6cos( 5 5t) + 5sin( 5 5t) 4 cos ( ) t. The graphs of () t and u() t are shown in the figure at the top of the net page. 60 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

29 6 t -6 u. (a) With m = /3 = 3/8 slug, c = 3 lb-sec/ft, and k = 4 lb/ft, the differential equation is equivalent to = 0. The characteristic equation 3r + 4r+ 9 = 0 has roots r = 4± 4 3 i. When we impose the initial conditions 4t = = on the general solution t () e ( Acos4t 3 Bsin4t 3) (0), (0) 0 the particular solution (t) = e -4t [cos 4t 3 + (/ 3)sin 4t 3] = + we get = (/ 3)e -4t [( 3/)cos 4t 3 + (/)sin 4t 3] (t) = (/ 3)e -4t cos(4t 3 π /6). (b) The time-varying amplitude is / 3.5 ft; the frequency is rad/sec; and the phase angle is π /6. 3. (a) With m = 00 slugs we get ω = k /00. But we are given that ω = (80 cycles/min)(π)( min/60 sec) = 8π/3, and equating the two values yields k 708 lb/ft. (b) With ω = π(78/60) sec -, Equation () in the tet yields c 37.3 lb/(ft/sec). Hence p = c/m.865. Finally e -pt = 0.0 gives t.47 sec. 30. In the underdamped case we have (t) = e -pt [A cos ω t + B sin ω t], (t) = -pe pt [A cos ω t + B sin ω t] + e -pt [-Aω sin ω t + Bω cos ω t]. The conditions (0) = 0, (0) = v 0 yield the equations A = 0 and -pa + Bω = v 0, whence B = (v 0 + p 0 )/ω. Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

30 3. The binomial series α αα ( ) αα ( )( α + = ) +! 3! 3 ( ) α converges if <. (See, for instance, Section 0.8 of Edwards and Penney, Calculus: Early Transcendentals, 7th edition, Prentice Hall, 008.) With α = / and = c /4mk in Eq. () of Section 5.4 in this tet, the binomial series gives ω k c k c m 4m m 4mk = ω0 p = = 4 ω 0 k c c c =. m 8mk 8m k 8mk 3. If (t) = Ce -pt cos(ω t - α) then (t) = -pce -pt cos(ω t - α) + Cω e -pt sin(ω t - α) = 0 yields tan(ω t - α) = -p/ω. 33. If = (t ) and = (t ) are two successive local maima, then ω t = ω t + π so = C ep(-pt ) cos(ω t - α), = C ep(-pt ) cos(ω t - α) = C ep(-pt ) cos(ω t - α). Hence / = ep[-p(t - t )], and therefore ln( / ) = -p(t - t ) = πp/ω. 34. With t = 0.34 and t =.7 we first use the equation ω t = ω t + π from Problem 3 to calculate ω = π/(0.83) 7.57 rad/sec. Net, with = 6.73 and =.46, the result of Problem 33 yields Then Equation (6) in this section gives and finally Equation () yields p = (/0.83) ln(6.73/.46).84. c = mp = (00/3)(.84).5 lb-sec/ft, k = (4m ω + c )/4m lb/ft. 6 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

31 35. The characteristic equation r + r+ = 0 has roots r =,. When we impose the initial conditions (0) = 0, () = 0 on the general solution () t = ( c+ ct) e t we get the particular solution (). t = te t n n 36. The characteristic equation r + r+ ( 0 ) = 0 has roots r = ± 0. When we impose the initial conditions (0) = 0, () = 0 on the general solution ( n n ) ( t) = cep + 0 t + cep 0 t we get the equations n n c + c = 0, + 0 c + 0 c = with solution c = 5, c = 5. This gives the particular solution n n n n n n n t ep(0 t) ep( 0 t) n t n ( t) = 0 e = 0 e sinh(0 t). n n 37. The characteristic equation r + r+ (+ 0 ) = 0 has roots r = ± 0 i. When we impose the initial conditions (0) = 0, () = 0 on the general solution t n n () t = e Acos0 t + Bsin0 t n n we get the equations c = 0, c+ 0 c = with solution c = 0, c = 0. This n t n gives the particular solution 3 ( t) = 0 e sin(0 t). n n t n t sinh(0 t) t 38. lim ( t) = lim 0 e sinh(0 t) = te lim = te and n n n n 0 t sin(0 t) = = =, t n n t n t t lim 3( t) lim 0 e sin(0 t) te lim te n n n 0 n using the fact that instance). lim(sin θ) / θ = lim(sinh θ) / θ = 0 (by L'Hôpital's rule, for θ 0 θ 0 Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

32 SECTION 5.5 NONHOMOGENEOUS EQUATIONS AND THE METHOD OF UNDETERMINED COEFFICIENTS The method of undetermined coefficients is based on "educated guessing". If we can guess correctly the form of a particular solution of a nonhomogeneous linear equation with constant coefficients, then we can determine the particular solution eplicitly by substitution in the given differential equation. It is pointed out at the end of Section 5.5 that this simple approach is not always successful in which case the method of variation of parameters is available if a complementary function is known. However, undetermined coefficients does turn out to work well with a surprisingly large number of the nonhomogeneous linear differential equations that arise in elementary scientific applications. In each of Problems 0 we give first the form of the trial solution y trial, then the equations in the coefficients we get when we substitute y trial into the differential equation and collect like terms, and finally the resulting particular solution y p.. y = Ae A = y p = (/5)e 3 3 trial ; 5 ;. ytrial = A+ B; A B = 4, B = 3; y p = -(5 + 6)/4 3. ytrial = Acos3+ Bsin3 ; 5A 3B = 0, 3A 5B = ; y p = (cos 3-5 sin 3)/39 4. ytrial = Ae + Be ; 9A+ B = 0, 9B = 3; y p = (-4e + 3e )/9 5. First we substitute sin = ( - cos )/ on the right-hand side of the differential equation. Then: ytrial = A+ Bcos+ Csin ; A= /, 3B+ C = /, B 3C = 0; y p = (3 + 3 cos - sin )/6 6. y A B C A B C B C C trial = + + ; = 0, = 0, 7 = ; y p = ( )/ First we substitute sinh = (e - e - )/ on the right-hand side of the differential equation. Then: ytrial = Ae + Be ; 3A= /, 3B = /; y p = (e - - e )/6 = -(/3)sinh 64 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

33 8. First we note that cosh is part of the complementary function y = c cosh + c sinh.. Then: c ytrial = Acosh + Bsinh ; 4A= 0, 4B = ; y p = (/4) sinh 9. First we note that e is part of the complementary function y c = c e + c e -3. Then: ytrial = A+ ( B+ C) e ; 3A=, 4B+ C = 0, 8C = ; y p = -(/3) + ( - )e /6. 0. First we note the duplication with the complementary function yc = ccos3+ csin 3. Then: ytrial = Acos3+ Bsin3 ; 6B =, 6A= 3; y p = ( sin 3-3 cos 3)/6. First we note the duplication with the complementary function y = c + c cos+ c sin. Then: c 3 ytrial = A+ B ; 4A=, 8B = 3; y p = (3 - )/8. First we note the duplication with the complementary function y = c + c cos + c sin. Then: c 3 ytrial = A+ Bcos + Csin ; A=, B = 0, C = ; y p = + (/) sin ytrial = e Acos + Bsin ; 7A+ 4B = 0, 4A+ 7B = ; y p = e (7 sin - 4 cos )/ First we note the duplication with the complementary function y = c+ c e + c + c e Then: c 3 4. ytrial = A+ B e ; 8A+ 4B = 0, 4B = ; y p = (-3 e + 3 e )/4 5. This is something of a trick problem. We cannot solve the characteristic equation 5 4 r + 5r = 0 to find the complementary function, but we can see that it contains no constant term (why?). Hence the trial solution ytrial = A leads immediately to the particular solution y p = = + ( + + ) y A B C D e 3 trial ; 9A= 5,8B+ 6C+ D = 0,8C+ D = 0,8D = ; Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

34 y p = (45 + e 3-6e e 3 )/8 7. First we note the duplication with the complementary function yc = ccos + csin. Then: [ ] ytrial = ( A+ B)cos + ( C+ D)sin ; B+ C = 0, 4D =, A+ D =, 4B = 0; y p = ( sin - cos )/4 8. First we note the duplication with the complementary function y = ce + c e + c e + c e Then: c 3 4. y trial = (Ae ) + (B + C) e ; 6A=, B+ 38C = 0, 4C = ; y p = -(4e - 9e + 6 e )/44 9. First we note the duplication with the part c+ c of the complementary function (which corresponds to the factor r of the characteristic polynomial). Then: y trial = (A + B + C ); 4A+ B =, B+ 48C = 0, 4C = 3; y p = ( )/ First we note that the characteristic polynomial r r has the zero r = corresponding to the duplicating part e of the complementary function. Then: ytrial = A+ Be ; A= 7, 3B = ; y p = -7 + (/3)e In Problems 30 we list first the complementary function y c, then the initially proposed trial function y i, and finally the actual trial function y p in which duplication with the complementary function has been eliminated.. yc = e ( ccos + csin ) ; yi = e ( Acos + Bsin ) y = e ( A + B ) p cos sin. y = ( c + c + c ) + ( c e ) + ( c e ) yi = ( A+ B+ C ) + ( De ) 3 yp = ( A+ B+ C ) + ( De ) c ; y = c cos + c sin ; yi = ( A+ B)cos + ( C+ D)sin yp = ( A+ B)cos + ( C+ D)sin 3. c [ ] 66 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

35 4. 5. y c c e c e 3 4 c = ; 3 yi = ( A+ B) + ( C+ D) e yp = ( A+ B) + ( C+ D) e y ce c e c = + ; 3 yi ( A B) e = + + ( C+ D) e yp = ( A+ B) e + ( C+ D) e 3 6. y = e ( c + c ) cos sin ; c 3 3 yi = ( A+ B) e cos + ( C+ D) e sin 3 3 yp = ( A+ B) e cos + ( C+ D) e sin 7. yc = ( ccos + csin ) + ( c3cos+ c3sin ) yi = ( Acos + Bsin ) + ( Ccos+ Dsin ) y = ( A + B ) + ( C + D ) p cos sin cos sin 8. y = ( c + c ) + ( c cos3+ c sin3) c 3 3 yi = A+ B + C cos3 + D + E + F sin 3 yp = A+ B+ C cos3+ D+ E+ F sin3 9. y = c + c + c e + c e + c e ; c i y = A+ B e + Ce + De p y = A+ B e + Ce + De = ( + ) + ( + ) y c c e c c e c 3 4 yi = yp = A+ B+ C cos + D+ E+ F sin In Problems 3 40 we list first the complementary function y c, the trial solution y tr for the method of undetermined coefficients, and the corresponding general solution y g = y c + y p where y p results from determining the coefficients in y tr so as to satisfy the given nonhomogeneous differential equation. Then we list the linear equations obtained by imposing the given initial conditions, and finally the resulting particular solution y(). y = c cos+ c sin ; y = A+ B 3. c tr y = c cos+ c sin + / g Section Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.

36 c =, c + / = y = cos + (3/4)sin + / 3. y = ce + c e ; y = Ae c tr y ce c e e g = + + /6 c + c + /6 = 0, c c + /6 = 3 ( ) y = 5e 6 e + e / 6 y = c cos3+ c sin 3 ; y = Acos+ Bsin 33. c tr y = c cos3+ c sin 3 + (/5)sin c g =, 3c + / 5 = 0 y = 5cos3 sin3+ 3sin /5 34. y = c cos + c sin ; y = ( Acos + Bsin ) c tr y = c cos + c sin + sin g c =, c = ; y = cos sin + sin y = e c cos + c sin ; y = A+ B 35. c tr y = e c cos + c sin + + / g c + = 3, c + c + / = 0 y = e 4cos 5sin /+ + / 36. y ( c = c+ c+ c3e + c4e ; ytr = A+ B+ C ) y c c c e c e 4 g = /6 /48 c + c + c =, c c + c =, 4c + 4c /8=, 8c + 8c = y = e 33e 4 /9 37. = + + ; = + ( + ) y c c e c e y A B C e c 3 tr y c ce ce e e 3 g = / + /6 68 Chapter 5 Copyright 00 Pearson Education, Inc. Publishing as Prentice Hall.