19.3. Second Order ODEs. Introduction. Prerequisites. Learning Outcomes

Transkript

1 Second Order ODEs 19.3 Introduction In this Section we start to learn how to solve second-order differential equations of a particular type: those that are linear and that have constant coefficients. Such equations are used widely in the modelling of physical phenomena, for example, in the analysis of vibrating systems, and the analysis of electrical circuits. The solution of these equations is achieved in stages. The first stage is to find what is called a complementary function. The second stage is to find a particular integral. Finally, the complementary function and the particular integral are combined to form the general solution of a second-order linear ODE Prerequisites Before starting this Section you should... Learning Outcomes After completing this Section you should be able to... 1 understand what is meant by a differential equation; (section 19.1) 2 understand complex numbers (Workbook 10) recognise a linear, constant coefficient equation understand what is meant by the terms auxiliary equation and complementary function find the complementary functions when the auxiliary equation has real, equal or complex roots

2 1. Constant coefficient equations We now proceed to study those second-order linear equations which have constant coefficients. The general form of such an equation is: a d2 y dx + bdy + cy = f(x) (1) where a, b, c are constants. The homogeneous form of (1) is a d2 y dx + bdy + cy =0 (2) The homogeneous form is found by ignoring the term which is independent of y,orits derivatives. To find the general solution of (1), it is first necessary to solve (2). The general solution of (2) is called the complementary function and will always contain two arbitrary constants. We will denote this solution by y cf. The technique for finding the complementary function is described in this section. Which of the following are constant coefficient equations? Which are homogeneous? a) dx +4dy +3y =e 2x, b) x d2 y +2y =0, dx2 c) d 2 x dt +3dx 2 dt +7x =0 d) dx +4dy +4y =0 a) is constant coefficient and is not homogeneous. b) is not constant coefficient because the coefficient of d2 y is x, avariable. The equation is homogeneous. c) is constant coefficient dx 2 and homogeneous. In this example the dependent variable is x. d) isconstant coefficient and homogeneous. What is a complementary function? A complementary function is the general solution of a homogeneous, linear differential equation HELM (VERSION 1: March 18, 2004): Workbook Level 1 2

3 2. Finding the complementary function To find the complementary function we must make use of the following property. If y 1 (x) and y 2 (x) are any two (linearly independent) solutions of a linear, homogeneous secondorder differential equation then the general solution y cf (x), is y cf (x) =Ay 1 (x)+by 2 (x) where A, B are constants. We see that the second-order linear ordinary differential equation has two arbitrary constants in its general solution. The functions y 1 (x) and y 2 (x) are linearly independent if one is not a multiple of the other. Example Verify that y 1 =e 4x and y 2 =e 2x both satisfy the constant coefficient homogeneous equation: dx 6dy +8y =0 (3) Write down the general solution of this equation. If y 1 =e 4x, differentiation yields: dy 1 dx =4e4x and similarly, 1 dx = 2 16e4x Substitution into the left hand side of (3) gives 16e 4x 6(4e 4x )+8e 4x, which equals 0, so that y 1 =e 4x is indeed a solution. Similarly if y 2 =e 2x, then dy 2 dx =2e2x and 2 dx 2 =4e2x. Substitution into the left hand side of (3) gives 4e 2x 6(2e 2x )+8e 2x, which equals 0, so that y 2 =e 2x is also a solution of quation (3). Now e 2x and e 4x are linearly independent functions. So, from the property stated above we have: as the general solution of (3). y cf (x) =Ae 4x + Be 2x Example Find values of k so that y =e kx is a solution of: dx dy 6y =0 Hence state the general solution. 3 HELM (VERSION 1: March 18, 2004): Workbook Level 1

4 As suggested we try a solution of the form y =e kx. Differentiating we find dy dx = kekx Substitution into the given equation yields: and dx 2 = k2 e kx. k 2 e kx ke kx 6e kx =0 that is (k 2 k 6)e kx =0 The only way this equation can be satisfied for all values of x is if k 2 k 6=0 that is, (k 3)(k +2)=0sothat k =3ork = 2. That is to say, if y =e kx is to be a solution of the differential equation k must be either 3 or 2. We therefore have found two solutions. y 1 (x) =e 3x and y 2 (x) =e 2x These two functions are linearly independent and therefore the general solution is y cf (x) =Ae 3x + Be 2x The equation k 2 k 6=0for determining k is called the auxiliary equation. By substituting y =e kx, find values of k so that y is a solution of dx 3dy +2y =0 Hence, write down two solutions, and the general solution of this equation. Hint: substitute y =e kx to get the auxiliary equation k 2 3k + 2=0 The auxiliary equation can be factorised as (k 1)(k 2) = 0 and so the required values of k are 1 and 2. The two solutions are y =e x and y =e 2x. The general solution is y cf (x) =Ae x + Be 2x HELM (VERSION 1: March 18, 2004): Workbook Level 1 4

5 Example Find the auxiliary equation of the differential equation: a d2 y dx + bdy + cy =0 We try a solution of the form y =e kx so that dy dx = kekx and Substitution into the given differential equation yields: dx 2 = k2 e kx. ak 2 e kx + bke kx + ce kx =0 that is (ak 2 + bk + c)e kx =0 Since this equation is to be satisfied for all values of x, then is the required auxiliary equation. ak 2 + bk + c =0 Key Point The auxiliary equation of a d2 y dx 2 + bdy dx + cy =0 is ak2 + bk + c =0 Write down, but do not solve the auxiliary equations of the following: a) dx + dy + y =0, b) y 2d +7dy 3y =0 c) 4 d2 y dx +7y =0, d) + dy =0 (a) (b) (c) (d) k 2 + k + 1=0 b)2k 2 +7k 3=0 c)4k 2 + 7=0 d)k 2 + k =0 5 HELM (VERSION 1: March 18, 2004): Workbook Level 1

6 Solving the auxiliary equation gives the values of k which we seek. Clearly the nature of the roots will depend upon the values of a, b and c. Ifb 2 > 4ac the roots will be real and distinct. The two values of k thus obtained, k 1 and k 2, will allow us to write down two independent solutions: y 1 (x) =e k 1x and y 2 (x) =e k2x, and so the general solution of the differential equation will be: y(x) =Ae k 1x + Be k 2x Key Point If the auxiliary equation has real, distinct roots k 1 and k 2, the complementary function will be: y cf (x) =Ae k 1x + Be k 2x On the other hand, if b 2 =4ac the two roots of the auxiliary equation will be equal and this method will therefore only yield one independent solution. In this case, special treatment is required. If b 2 < 4ac the two roots of the auxiliary equation will be complex, that is, k 1 and k 2 will be complex numbers. The procedure for dealing with such cases will become apparent in the following examples. Example Find the general solution of: dx +3dy 10y =0 By letting y =e kx,sothat dy dx = kekx the auxiliary equation is found to be: and dx 2 = k2 e kx k 2 +3k 10 = 0 and so (k 2)(k +5)=0 so that k =2and k = 5. Thus there exist two solutions: y 1 =e 2x and y 2 =e 5x. We can write the general solution as: y = Ae 2x + Be 5x HELM (VERSION 1: March 18, 2004): Workbook Level 1 6

7 Example Find the general solution of: +4y =0 dx2 As before, let y =e kx so that dy dx = kekx and dx 2 = k2 e kx. The auxiliary equation is easily found to be: k 2 + 4=0that is, k 2 = 4 sothat k = ±2i, that is, we have complex roots. The two independent solutions of the equation are thus y 1 (x) =e 2ix y 2 (x) =e 2ix so that the general solution can be written in the form y(x) =Ae 2ix + Be 2ix However, in cases such as this, it is usual to rewrite the solution in the following way. Recall that Euler s relations give: so that e 2ix = cos 2x +isin 2x and e 2ix = cos 2x i sin 2x y(x) =A(cos 2x +isin 2x)+B(cos 2x i sin 2x) If we now relabel the constants such that A + B = C and Ai Bi =D we can write the general solution in the form: y(x) =C cos 2x + D sin 2x Example Given ay +by +cy =0, write down the auxiliary equation. If the roots of the auxiliary equation are complex (one root will always be the complex conjugate of the other) and are denoted by k 1 = α + βi and k 2 = α βi show that the general solution is: y(x) =e αx (A cos βx + B sin βx) 7 HELM (VERSION 1: March 18, 2004): Workbook Level 1

8 Substitution of y =e kx into the differential equation yields (ak 2 + bk + c)e kx =0and so the auxiliary equation is: ak 2 + bk + c =0 If k 1 = α + βi, k 2 = α βi then the general solution is y = Ce (α+βi)x + De (α βi)x where C and D are arbitrary constants. Using the laws of indices this is rewritten as: Then, using Euler s relations, we obtain: y = Ce αx e βix + De αx e βix =e αx (Ce βix + De βix ) y = e αx (C cos βx + Ci sin βx + D cos βx Di sin βx) = e αx {(C + D) cos βx +(Ci Di) sin βx} Writing A = C + D and B = Ci Di, we find the required solution: y =e αx (A cos βx + B sin βx) Key Point If the auxiliary equation has complex roots, α+βi and α βi, then the complementary function is: y cf =e αx (A cos βx + B sin βx) Find the general solution of y +2y +4y =0. The auxiliary equation is: k 2 +2k + 4=0 HELM (VERSION 1: March 18, 2004): Workbook Level 1 8

9 The auxiliary equation has complex roots given by: k = 1 ± 3i Using the keypoint above with α = 1 and β = 3 write down the general solution: y =e x (A cos 3x + B sin 3x) Example The auxiliary equation of ay + by + cy =0isak 2 + bk + c =0. Suppose this equation has equal roots k = k 1. Verify that y = xe k 1x is a solution of the differential equation. We have: y = xe k 1x y =e k 1x (1 + k 1 x) y =e k 1x (k 2 1x +2k 1 ) Substitution into the left-hand side of the differential equation yields: e k 1x {a(k 2 1x +2k 1 )+b(1 + k 1 x)+cx} =e k 1x {(ak bk 1 + c)x +2ak 1 + b} But ak bk 1 + c =0since k 1 satisfies the auxiliary equation. Also, k 1 = b ± b 2 4ac 2a but since the roots are equal, then b 2 4ac =0hence k 1 = b/2a. So 2ak 1 + b =0. Hence e k 1x {(ak bk 1 + c)x +2ak 1 + b} =e k 1x {(0)x +0} =0.Weconclude that y = xe k 1x is a solution of ay + by + cy =0when the roots of the auxiliary equation are equal. Key Point If the auxiliary equation has two equal roots, k 1, the complementary function is: y cf =(A + Bx)e k 1x 9 HELM (VERSION 1: March 18, 2004): Workbook Level 1

10 Example Obtain the general solution of the equation: dx +8dy +16y =0 As before, a trial solution of the form y =e kx yields an auxiliary equation k 2 +8k +16=0. This equation factorizes so that (k + 4)(k +4)=0and we obtain equal roots, that is, k = 4 (twice). If we proceed as before, writing y 1 (x) =e 4x y 2 (x) =e 4x, it is clear that the two solutions are not independent. We need to find a second independent solution. Using the result of the previous example we conclude that, because the roots of the auxiliary equation are equal, the second independent solution is y 2 = xe 4x. The general solution is then: y(x) =(A + Bx)e 4x Exercises 1. Obtain the general solutions, that is, the complementary functions, of the following homogeneous equations: (a) d2 y dx 3dy +2y =0 (b) (c) d2 x dt +5dx +6x =0 2 dt (d) (e) d2 y dx 4dy +4y =0 (f) (g) d2 y dx 2dy + y =0 (i) d2 y dx + dy 2y =0 (j) (k) d2 y dx 2dy =0 d2 y dx +7dy +6y =0 d2 y dt 2 +2dy dt + y =0 d2 y dt + dy +8y =0 2 dt (h) d2 y dt + dy +5y =0 2 dt d2 y +9y =0 dx2 (l) d2 x 16x =0 dt2 2. Find the auxiliary equation for the differential equation L d2 i dt + R di 2 dt + 1 C i =0 Hence write down the complementary function. 3. Find the complementary function of the equation d2 y dx 2 + dy dx + y =0 HELM (VERSION 1: March 18, 2004): Workbook Level 1 10

11 Answers 1. (a) y = Ae x + Be 2x (b) y = Ae x + Be 6x (c) x = Ae 2t + Be 3t (d) y = Ae t + Bte t (e) y = Ae 2x + Bxe 2x (f) y =e 0.5t (A cos 2.78t + B sin 2.78t) (g) y = Ae x + Bxe x (h) x =e 0.5t (A cos 2.18t + B sin 2.18t) (i) y = Ae 2x + Be x (j) y = A cos 3x + B sin 3x (k) y = A + Be 2x (l) x = Ae 4t + Be 4t 2. Lk 2 + Rk + 1 C =0 i(t) =Aek 1t + Be k 2t k 1,k 2 = 1 2L ( R ± R 2 C 4L C ) 3. e (A x/2 cos 3 x + B sin ) 3 x What is meant by a particular integral? Given a second order o.d.e. a d2 y dx + bdy + cy= f(x) a particular integral is any function, y p (x), which satisfies the equation. That is, any function which when substituted into the left hand side and simplified, results in the function on the right. We denote a particular integral by y p (x). Show that is a particular integral of y = 1 4 e2x dx dy 6y =e2x (4) Starting with y = 1 4 e2x, find dy dx and d2 y dx 2 : dy = 1 dx 2 e2x, = e 2x dx 2 11 HELM (VERSION 1: March 18, 2004): Workbook Level 1

12 Now substitute these into (4): Substitution into (4) yields e 2x ( 1e2x) 6 ( 1e2x) which simplifies to e 2x, the same as 2 4 the right hand side. Therefore y = 1 4 e2x is a particular integral and we write (attaching a subscript p) y p (x) = 1 4 e2x What is a particular integral? A particular integral is any solution of an inhomogeneous differential equation. 4. Finding a particular integral In the previous section we explained what is meant by a particular integral. Now we look at how one is actually found. In fact our method is rather crude. It involves trial and error and educated guesswork. We try solutions which are of the same general form as the f(x) onthe right hand side. As a guide, use Table 1. Table 1. Trial solutions to find the particular integral f(x) constant term c Trial solution constant term γ polynomial in x polynomial in x of degree r: of degree r: ax r + + bx + c αx r + + βx + γ a cos kx a sin kx ae kx α cos kx + β sin kx α cos kx + β sin kx αe kx HELM (VERSION 1: March 18, 2004): Workbook Level 1 12

13 Example Find a particular integral of the equation dx dy 6y =e2x (5) We shall attempt to find a solution of the inhomogeneous problem by trying a function of the same form as that on the right-hand side. In particular, let us try y(x) =αe 2x, where α is a constant that we shall now determine. If y(x) =αe 2x then Substitution in (5) gives: that is, dy dx =2αe2x and dx 2 =4αe2x. 4αe 2x 2αe 2x 6αe 2x =e 2x 4αe 2x =e 2x so that y will be a solution if α is chosen so that 4α =1,that is, α = 1. Therefore the 4 particular integral is y p (x) = 1 4 e2x. By trying a solution of the form y = αe x find a particular integral of the equation d2 y dx + dy 2y 2 =3e x dx Substitute y = αe x into the given equation to find α, and hence the particular integral. α = 3 2 ; y p(x) = 3 2 e x Example Obtain a particular integral of the equation: dx 2 6dy dx +8y = x 13 HELM (VERSION 1: March 18, 2004): Workbook Level 1

14 In the last example, we found that a fruitful approach was to assume a solution in the same form as that on the right-hand side. Suppose we assume a solution y(x) =αx and proceed to determine α. This approach will actually fail, but let us see why. If y(x) =αx then dy = α dx and d2 y =0. Substitution into the differential equation yields 0 6α +8αx = x and α ought dx 2 now to be chosen so that this expression is true for all x. Ifweequate the coefficients of x we find 8α =1so that α = 1, but with this value of α the constant terms are inconsistent (that 8 is 6 on the left, but zero on the right). Clearly a particular integral of the form αx is not 8 possible. The problem arises because differentiation of the term αx produces constant terms which are unbalanced on the right-hand side. So, we try a solution of the form y(x) =αx + β with α, β constants. This is consistent with the recommendation in Table 1. Proceeding as before dy dx = α, dx 2 =0. Substitution in the differential equation now gives: 0 6α +8(αx + β) =x Equating coefficients of x and then equating constant terms we find: 8α =1 ( ) 6α +8β =0 ( ) From ( ), α = 1 8 and then from ( ) 6 ( 1 8) +8β =0 so that, 8β = 3 4 that is, β = The required particular integral is y p(x) = 1 8 x Find a particular integral for the equation: dx 6dy +8y =3cos x First try to decide on an appropriate form for the trial solution. Refer to Table 1 if necessary HELM (VERSION 1: March 18, 2004): Workbook Level 1 14

15 y = α cos x + β sin x in which α, β are constants to be found. We shall try a solution of the form y(x) =α cos x + β sin x. Differentiating, we find: dy dx = α sin x + β cos x = α cos x β sin x dx2 Substitution into the differential equation gives: ( α cos x β sin x) 6( α sin x + β cos x)+8(α cos x + β sin x) =3cosx Equate coefficients of cos x in your previous answer: 7α 6β =3 Also, equate coefficients of sin x in your previous answer: 7β +6α =0 Solve these simultaneously to find α and β, and hence the particular integral: α = ,β=,y 85 p(x) = cos x sin x Finding the general solution of a second-order inhomogeneous equation The general solution of a second-order linear inhomogeneous equation is the sum of its particular integral and the complementary function. In section 19.5 you learned how to find a complementary function, and in the previous section you learnt how to find a particular integral. We now put these together to find the general solution. 15 HELM (VERSION 1: March 18, 2004): Workbook Level 1

16 Find the general solution of dx +3dy 10y 2 =3x2 dx The complementary function was found in section 19.5 page 6 to be y cf = Ae 2x + Be 5x. The particular integral is found by trying a solution of the form y = ax 2 + bx + c. Substitute into the homogeneous equation to find a, b and c, and hence y p (x). a = 3, b = 9, c = 57, y p(x) = 3 10 x2 9 x Thus the general solution is y = y p (x)+y cf (x) = 3 10 x2 9 x Ae2x + Be 5x Key Point The general solution of a constant coefficient ordinary differential equation a d2 y dx 2 + bdy dx + cy = f(x) is y = y p + y cf being the sum of the particular integral and the complementary function. y p contains no arbitrary constants; y cf contains two arbitrary constants. HELM (VERSION 1: March 18, 2004): Workbook Level 1 16

17 Example An LC circuit with sinusoidal input. The differential equation governing the flow of current in a series LC circuit when subject to an applied voltage v(t) = V 0 sin ωt is v L d2 i dt C i = ωv 0 cos ωt L C Obtain its general solution. i The homogeneous equation is L d2 i cf dt + i cf 2 C =0. Letting i cf =e kt we find the auxiliary equation is Lk =0so that k = ±i/ LC. Therefore, C the complementary function is: i cf = A cos t t + B sin LC LC where A and B arbitrary constants To find a particular integral try i p = E cos ωt + F sin ωt, where E, F are constants. We find: di p dt = ωe sin ωt + ωf cos ωt d 2 i p dt 2 = ω2 E cos ωt ω 2 F sin ωt 17 HELM (VERSION 1: March 18, 2004): Workbook Level 1

18 (contd.) Substitution into the inhomogeneous equation yields: L( ω 2 E cos ωt ω 2 F sin ωt)+ 1 C (E cos ωt + F sin ωt) =ωv 0 cos ωt Equating coefficients of sin ωt gives: ω 2 LF +(F/C)=0. Equating coefficients of cos ωt gives: ω 2 LE +(E/C)=ωV 0. Therefore F =0and E = CV 0 ω/(1 ω 2 LC). Hence the particular integral is i p = CV 0ω cos ωt. 1 ω 2 LC Finally, the general solution is: i = i cf + i p = A cos t LC + B sin t LC + CV 0ω cos ωt 1 ω 2 LC 6. Inhomogeneous term appearing in the complementary function Occasionally you will come across a differential equation a d2 y dx + bdy + cy = f(x) for which the inhomogeneous term, f(x), forms part of the complementary function. One such example is the equation dx dy 6y 2 =e3x dx It is straightforward to check that the complementary function is y cf = Ae 3x + Be 2x. Note that the first of these terms has the same form as the inhomogeneous term, e 3x,onthe right-hand side of the differential equation. You should verify for yourself that trying a particular integral of the form y p (x) =αe 3x will not work in a case like this. Can you see why? Instead, try a particular integral of the form y p (x) =αxe 3x.Verify that dy p dx = αe3x (3x +1) and p dx 2 = αe3x (9x +6). Substitute these expressions into the differential equation to find α = 1. Finally, the particular 5 integral is y p (x) = 1 5 xe3x and so the general solution to the differential equation is: y = Ae 3x + Be 2x xe3x HELM (VERSION 1: March 18, 2004): Workbook Level 1 18

19 Exercises 1. Find the general solution of the following equations: (a) d2 x dt 2dx 3x =6. 2 dt (b) (d) d2 x dt +11dx +30x =8t 2 dt (e) (g) d2 y +9y =4e8x (h) dx2 d2 y dx +5dy +4y =8 (c) d2 y dx +2dy +3y =2sin 2x (f) d2 x 16x =9e6t dt2 d2 y dt +5dy +6y =2t 2 dt 2. Find a particular integral for the equation d2 x dt 3dx +2x 2 =5e3t dt 3. Find a particular integral for the equation d2 x dt 2 x =4e 2t 4. Obtain the general solution of y y 2y =6. 5. Obtain the general solution of the equation dx +3dy +2y =10cos 2x. dy Find the particular solution satisfying y(0) = 1, (0) =0. dx d2 y dt + dy + y =4cos 3t 2 dt 6. Find a particular integral for the equation 7. Find the general solution of (a) d2 x dt 6dx +5x =3 2 dt (b) dx 2 + dy dx + y =1+x d2 x dt 2 2dx dt + x =et Answers 1. (a) x = Ae t + Be 3t 2 (b) y = Ae x + Be 4x +2 (c) y = Ae 2t + Be 3t + 1t (d) x = Ae 6t + Be 5t t (e) y =e x [A sin 2x + B cos 2x] cos 2x 17 sin 2x (f) y =e 0.5t (A cos 0.866t + B sin 0.866t) cos 3t sin 3t (g) y = A cos 3x + B sin 3x e 8x (h) x = Ae 4t + Be 4t e6t 2. x p =2.5e 3t 3. x p = 4 3 e 2t 4. y = Ae 2x + Be x 3 5. y = Ae 2x + Be x + 3 sin 2x 1 cos 2x, e 2x + 3 sin 2x y p = x cos 2x 7. (a) x = Ae t + Be 5t (b) x = Ae t + Bte t t2 e t 19 HELM (VERSION 1: March 18, 2004): Workbook Level 1