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# Note for MATH 3321 at UH

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Date Created: 02/06/15
3 Equation 2 has complex conjugate roots7 r1 04 2 3 r2 Oz 7 i B B 7 0 Case I The characteristic equation has two7 distinct real roots7 r1 a r2 B In this case7 24190 5 and 242W 5 are solutions of Since 04 7 yl and y2 are not constant multiples of each other7 the pair y17y2 forms a fundamental set of solutions of equation 1 and y Cl em 025 is the general solution Note We can use the Wronskian to verify the independence of y1 and y2 60m e m a 604m B e m 50m 519m 751 a 50m 0473 ea z 7g 039 I Example 1 Find the general solution of the differential equation yH2y78y 0 SOLUTION The characteristic equation is r2 2r 7 8 0 74772 0 The characteristic roots are 71 74 r2 2 The functions y1z 5 y2z 52m form a fundamental set of solutions of the differential equation and y Cl 574m CZ 521 is the general solution of the equation I Example 2 Find a linearly independent pair of solutions of y 33 0 and give the general solution of the equation SOLUTION The characteristic equation is r2 3r rr 3 O7 and the characteristic roots are 71 O7 r2 73 Therefore the functions y1z 50m E 1 and y2z 5 31 are linearly independent solutions of the differential equation The general solution is y 011 025 01 025 l 77 Case II The characteristic equation has only one real root r 041 Then yl e and y2m m e are linearly independent solutions of equation 1 and y 01 e 0260m is the general solution Proof We know that y1z e is one solution of the differential equa tion we need to find another solution which is independent of y1 Since the characteristic equation has only one real root 4 the equation must be 2 rzarbria 7272ozr0420 and the differential equation 1 must have the form yH7204y042y 0 Now 2 Ce C any constant is also a solution of but 2 is not independent of y1 since it is simply a multiple of y1 We replace C by a function it which is to be determined if possible so that y us is a solution of Calculating the derivatives of y we have y ueu m y Que u 5 H 2 H y auew2auewu em Substitution into gives H azueam2au em u 50m7204 auemu 5M a2uem 0 This reduces to H az0 u e which implies u 0 since 5 7 0 Now u 0 is the simplest second order linear differential equation with constant coefficients the general solution is u 01 ng 01 1 Cg z and u1z 1 and u2z z form a fundamental set of solutions Since y u e we conclude that yl 1 5 em and 292 m 5 1In this case a is said to be a double root of the Characteristic equation 2This is an application of a general method called mutation of parameters We will use the method several times in the work that follows 78 are solutions of In particular7 yg me is a solution of which is independent of y1 e That is7 M and yg form a fundamental set of solutions of This can also be checked by using the Wronskian 60m eam aem 5M Imam e 6 ameml 7 axe 52 7 0 Finally7 the general solution of is y 01 em I 02x5 Note The solution y2z me can also be obtained by using Problem 15 in Exercises 32 I Example 3 Find the general solution of the differential equation yH76y9y0 SOLUTION The characteristic equation is r2 7 ST 9 0 r 7 32 0 There is only one characteristic root r1 r2 3 The functions y1z 53m y2z x 53m are linearly independent solutions of the differential equation and y0153m02m53m is the general solution I Case III The characteristic equation has complex conjugate roots Tlai 72047i Bio In this case y1z 5 cos x and y2z 5 sin Bx are linearly independent solutions of equation 1 and y 01 5 cos x I 02 5 sin Bx e 01 cos x I 02 sin 3m is the general solution Proof It is true that the functions 21z 40 erm and 22z awhile are linearly independent solutions of 17 but these are complex valued functions 79 and we are not equipped to handle such functions in this course We want real valued solutions of The characteristic equation in this case is rzarb T7 ai ri Oti 7 272047 04232 0 and the differential equation 1 has the form yH72aya232y039 We ll proceed in a manner similar to Case ll Set y us where u is to be determined if possible so that y is a solution of Calculating the derivatives of y we have y u e y 04 u em u cm H H y 04214 cm 204 u em u em Substitution into gives 04214 e 204150 iiHeD m 7 204 auew ueu m 042 32 lieu 0 This reduces to Mew ozuew 0 which implies u 5 0 since 5 7g 0 Now7 uHBZuO is the equation of simple harmonic motion for example7 it models the oscillatory motion of a weight suspended on a spring The functions u1z cos x and u2z sin Bx form a fundamental set of solutions Verify this Since y u e we conclude that y1z 5 cos x and 242 em sin Bx are solutions of It s easy to see that M and yg form a fundamental set of solutions This can also be checked by using the Wronskian Finally7 we conclude that the general solution of equation 1 is y 01 5 cos Bx 025 sin Bx e 01 cos Bx 02 sin Bx I Example 4 Find the general solution of the differential equation y 7 43 13y 0 80 SOLUTION The characteristic equation is r2 7 4r 1 13 By the quadratic formula7 the roots are 2i32 774 742 74113 i 41 M16 7 52 7 41 M736 7 41 62 2 2 2 2 T17 T2 The characteristic roots are the complex numbers 71 2 32 r2 2 7 32 The functions y1z 52m cos 3m y2z 52m sin 3x are linearly independent solutions of the differential equation and y 01 52 cos 3x Cg 521 sin 3x 52 01 cos 3x Cg sin 3m is the general solution I Example 5 Find two linearly independent solutions of y 16y 0 SOLUTION The characteristic equation is r2 16 0 and the complex numbers 71 0 42 4i r2 0 7 4i 742 are the characteristic roots The functions 0 y1z e 1 cos 4x cos 4m y2z 501 sin 4x sin 4x are linearly independent solutions of the differential equation I In our next example we nd the solution of an initial value problem Example 6 Find the solution of the initial value problem 24 227151 07 240 27 20 6 SOLUTION The characteristic equation is 2 7 r 27 7 15 7 0 r5r73 0 51 31 The characteristic roots are 71 757 r2 3 The functions y1z e y2z e are linearly independent solutions of the differential equation and y 01 6751 CZ 53m is the general solution Before applying the initial conditions we need to calculate y ym 7501 575m 1 302 53m Now7 the conditions y0 27 y 0 76 are satisfied if and only if 0102 2 75013CZ 76 81 The solution of this pair of equations is 01 g 02 and the solution of the initial Value problem is ye 5m53m I Recovering a Differential Equation from Solutions You can also work backwards using the results above That is we can determine a second order linear homogeneous differential equation with constant coefficients that has given functions u and v as solutions Here are some examples Example 7 Find a second order linear homogeneous differential equation with constant coefficients that has the functions 52m vz 5 31 as solutions SOLUTION Since 52m is a solution 2 must be a root of the characteristic equation and 73m r 7 2 must be a factor of the characteristic polynomial Similarly e a solution means that 73 is a root and r 7 73 r 3 is a factor of the characteristic polynomial Thus the characteristic equation must be r 7 2r 3 0 which expands to r2 r 7 6 0 Therefore the differential equation is y y76y0 I Example 8 Find a second order linear homogeneous differential equation with constant coefficients that has y 01 e741 02 z e741 as its general solution SOLUTION Since 5 and ze are solutions 74 must be a double root of the characteristic equation Therefore the characteristic equation is r 7 742 r 42 0 which expands to r2 7 8r 7 16 0 and the differential equation is y 8yf16y 0 I Example 9 Find a second order linear homogeneous differential equation with constant coefficients that has em cos 2x as a solution SOLUTION Since 51 cos 2x is a solution the characteristic equation must have the complex numbers 1 2i and 1 7 2i as roots Although we didn t state it explicitly 51 sin 2x must also be a solution The characteristic equation must be r 7 1 2ir7 17 21 0 which expands to r2 7 2H 5 0 and the differential equation is yH72y5y0 I 82 Exercises 33 Find the general solution of the given differential equation 10 11 12 13 14 15 16 17 18 19 20 21 22 23 y 2378y0 yH713yl42y0 yH710325y0 y 235y0 3 4y 133 0 y 0 3 230 23 5373y0 y 7 123 0 y 123 0 y 7 23 1 2y 0 y 73y y 0 y 7 3 7 303 0 2y 1 33 0 2y 23 y 0 y 23 1 3y 0 y 7 83 163 0 5y y 7 y 0 Find the solution of the initial value problem y0 240 27 MO 1 y0 737 MO 1 y7r17 MW 1 ylt0gt 71 240 71 y 7 53 1 6y 0 17 30 1 y 4y 3y 0 y 2y y0 y 1y 0 y 72y 2y 0 83 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 y 42 4y 0 274 2 Mil 1 Find a differential equation y ay 1 by 0 that is satis ed by the given functions y1m 52m y2m 575m y1m 363m y2m 2x55 y1z cos 2x y2z 7 2 sin 2x 2m 2m cos 4m y2z 5 sin 4m Find a differential equation y ay 1 by 0 Whose general solution is the given expression 34 Clem2 0252 y C163m 025 y C1157m COS 3m 025 m sin 3x 34 Clem2 szemZ y 01 cos 4m 02 sin 495 Find the solution y of the initial value problem y 7 y 7 2y O y0 a y 0 2 Then nd 04 such that 7 0 as x 7 00 Find the solution y ofthe initial value problem 4y 7 O y0 27 y 0 3 Then nd 3 such that 7 0 as x 7 00 Given the differential equation y 7 2a 7 1y aa 7 1y O a Determine the values of a if any for which all solutions have limit 0 as HOO 0 Determine the values of a if any for which all solutions are unbounded as HOO Exercises 37 39 are concerned with the differential equation 1 y ay 1 by 0 Where a and b are constants Give a condition on a and b which will imply that a 1 has solutions of the form yl 5W yg em a 3 distinct real numbers b 1 has solutions of the form yl e yg me 04 a real number c 1 has solutions of the form yl 5 cos 3m yg 5 sin 3m 04 3 real numbers Prove that if a and b are both positive7 then all solutions have limit 0 as x 7 oo 84 39 Prove a If a 0 and b gt 0 then all solutions of the equation are bounded b If a gt 0 and b 0 and y is a solution then lim k for some constant k 700 Determine k for the solution that satis es the initial conditions y0 a MO B 40 Show that the general solution of the differential equation y 7 Lazy 0 w a positive constant can be written y 01 cosh um 02 sinh um 41 Suppose that the roots r1 r2 of the characteristic equation 2 are real and distinct Then they can be written as T1 13 r2 04 7B where 04 and B are real Show that the general solution of equation 1 in this case can be expressed in the form y e 01 cosh Bx 02 sinh 3m Euler Equations A second order linear homogeneous equation of the form dzy dy 2 z ax yiio where 04 and B are constants is called an Euler equation 42 Prove that the Euler equation can be transformed into the second order equation with constant coefficients 2 d y dy 7 7 b 0 122 adz y Where a and b are constants by means of the change of independent variable 2 ln m Find the general solution of the Euler equations 43 xzy 7 my 7 8y 0 44 xzy 7 me 2y 0 45 xzy 7 3zy 4y 0 46 xzy 7 my 5y 0 85 33 Homogeneous Equations with Constant Coef cients We emphasized in Sections 31 and 32 that there are no general methods for solving second or higher order linear differential equations However there are some special cases for which solution methods do exist In this and the following sections we consider such a case linear equations with constant coefficients In this section we treat homogeneous equations nonhomogeneous equations will be treated in the next two sections A second order linear homogeneous di erential equation with constant coe icients is an equation which can be written in the form y ay by0 1 where a and b are real numbers You have seen that the function y 5quot is a solution of the rst order linear equation 3 ay 0 the model for exponential growth and decay This suggests the possibility that equation 1 may also have an exponential function y em as a solution If y 5 then 3 re and y r2 e Substitution into 1 gives r2 em 1 a r e b 5 em r2 1 ar 1 b 0 rm Since em 7 0 for all x we conclude that y e is a solution of 1 if and only if r2arb0 2 Thus if r is a root of the quadratic equation 2 then y em is a solution of equation 1 we can nd solutions of 1 by nding the roots of the quadratic equation DEFINITION 1 Given the differential equation The corresponding quadratic equa tion r2 1 ar 1 b 0 is called the characteristic equation of 1 the quadratic polynomial r2 1 ar 1 b is called the characteristic polynomial The roots of the characteristic equation are called the characteristic roots The nature of the solutions of the differential equation 1 depends on the nature of the roots of its characteristic equation There are three cases to consider 1 Equation 2 has two distinct real roots r1 4 r2 2 Equation 2 has only one real root r Oz 76

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