Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form 1 ay by cy Gx where a, b, and c are constants and G is a continuous function. The related homogeneous equation 2 ay by cy 0 is called the complementary equation and plays an important role in the solution of the original nonhomogeneous equation (1). 3 Theorem The general solution of the nonhomogeneous differential equation (1) can be written as yx ypx ycx where yp is a particular solution of Equation 1 and yc is the general solution of the complementary Equation 2. Proof All we have to do is verify that if y is any solution of Equation 1, then y yp is a solution of the complementary Equation 2. Indeed ay yp by yp cy yp ay ayp by byp cy cyp ay by cy ayp byp cyp tx tx 0 We know from Additional Topics: Second-Order Linear Differential Equations how to solve the complementary equation. (Recall that the solution is yc c1 y1 c2 y2 , where y1 and y2 are linearly independent solutions of Equation 2.) Therefore, Theorem 3 says that we know the general solution of the nonhomogeneous equation as soon as we know a particular solution yp . There are two methods for finding a particular solution: The method of undetermined coefficients is straightforward but works only for a restricted class of functions G. The method of variation of parameters works for every function G but i0s usually more difficult to apply in practice. The Method of Undetermined Coefficients We first illustrate the method of undetermined coefficients for the equation ay by cy Gx where Gx) is a polynomial. It is reasonable to guess that there is a particular solution yp that is a polynomial of the same degree as G because if y is a polynomial, then ay by cy is also a polynomial. We therefore substitute ypx a polynomial (of the same degree as G ) into the differential equation and determine the coefficients. EXAMPLE 1 Solve the equation y y 2y x 2. SOLUTION The auxiliary equation of y y 2y 0 is r 2 r 2 r 1r 2 0 1 2 ■ NONHOMOGENEOUS LINEAR EQUATIONS with roots r 1, 2. So the solution of the complementary equation is yc c1 e x c2 e2x Since Gx x 2 is a polynomial of degree 2, we seek a particular solution of the form ypx Ax 2 Bx C Then yp 2Ax B and yp 2A so, substituting into the given differential equation, we have 2A 2Ax B 2Ax 2 Bx C x 2 or 2Ax 2 2A 2Bx 2A B 2C x 2 Polynomials are equal when their coefficients are equal. Thus Figure 1 shows four solutions of the differential equation in Example 1 in terms of the particular solution yp and the functions f x e x and tx e2 x. ■ ■ 8 2A 2B 0 2A B 2C 0 The solution of this system of equations is A 12 yp+2f+3g yp+3g 2A 1 B 12 C 34 A particular solution is therefore yp+2f _3 3 ypx 12 x 2 12 x 34 yp and, by Theorem 3, the general solution is _5 y yc yp c1 e x c2 e2x 12 x 2 12 x 34 FIGURE 1 If Gx (the right side of Equation 1) is of the form Ce k x, where C and k are constants, then we take as a trial solution a function of the same form, ypx Ae k x, because the derivatives of e k x are constant multiples of e k x. EXAMPLE 2 Solve y 4y e 3x. SOLUTION The auxiliary equation is r 2 4 0 with roots 2i, so the solution of the Figure 2 shows solutions of the differential equation in Example 2 in terms of yp and the functions f x cos 2x and tx sin 2x. Notice that all solutions approach as x l and all solutions resemble sine functions when x is negative. ■ ■ 4 complementary equation is ycx c1 cos 2x c2 sin 2x For a particular solution we try ypx Ae 3x. Then yp 3Ae 3x and yp 9Ae 3x. Substituting into the differential equation, we have 9Ae 3x 4Ae 3x e 3x yp+f+g yp+g yp _4 yp+f _2 FIGURE 2 so 13Ae 3x e 3x and A 131 . Thus, a particular solution is 2 ypx 131 e 3x and the general solution is yx c1 cos 2x c2 sin 2x 131 e 3x If Gx is either C cos kx or C sin kx, then, because of the rules for differentiating the sine and cosine functions, we take as a trial particular solution a function of the form ypx A cos kx B sin kx NONHOMOGENEOUS LINEAR EQUATIONS ■ 3 EXAMPLE 3 Solve y y 2y sin x. SOLUTION We try a particular solution ypx A cos x B sin x yp A sin x B cos x Then yp A cos x B sin x so substitution in the differential equation gives A cos x B sin x A sin x B cos x 2A cos x B sin x sin x 3A B cos x A 3B sin x sin x or This is true if 3A B 0 and A 3B 1 The solution of this system is A 101 B 103 so a particular solution is ypx 101 cos x 103 sin x In Example 1 we determined that the solution of the complementary equation is yc c1 e x c2 e2x. Thus, the general solution of the given equation is yx c1 e x c2 e2x 101 cos x 3 sin x If Gx is a product of functions of the preceding types, then we take the trial solution to be a product of functions of the same type. For instance, in solving the differential equation y 2y 4y x cos 3x we would try ypx Ax B cos 3x Cx D sin 3x If Gx is a sum of functions of these types, we use the easily verified principle of superposition, which says that if yp1 and yp2 are solutions of ay by cy G1x ay by cy G2x respectively, then yp1 yp2 is a solution of ay by cy G1x G2x EXAMPLE 4 Solve y 4y xe x cos 2x. SOLUTION The auxiliary equation is r 2 4 0 with roots 2, so the solution of the com- plementary equation is ycx c1 e 2x c2 e2x. For the equation y 4y xe x we try yp1x Ax Be x Then yp1 Ax A Be x, yp1 Ax 2A Be x, so substitution in the equation gives Ax 2A Be x 4Ax Be x xe x or 3Ax 2A 3Be x xe x 4 ■ NONHOMOGENEOUS LINEAR EQUATIONS Thus, 3A 1 and 2A 3B 0, so A 13 , B 29 , and yp1x ( 13 x 29 )e x For the equation y 4y cos 2x, we try yp2x C cos 2x D sin 2x In Figure 3 we show the particular solution yp yp1 yp 2 of the differential equation in Example 4. The other solutions are given in terms of f x e 2 x and tx e2 x. ■ ■ 5 Substitution gives 4C cos 2x 4D sin 2x 4C cos 2x D sin 2x cos 2x 8C cos 2x 8D sin 2x cos 2x or yp+2f+g Therefore, 8C 1, 8D 0, and yp+g yp+f _4 yp2x 18 cos 2x 1 yp By the superposition principle, the general solution is _2 y yc yp1 yp2 c1 e 2x c2 e2x ( 13 x 29 )e x 18 cos 2x FIGURE 3 Finally we note that the recommended trial solution yp sometimes turns out to be a solution of the complementary equation and therefore can’t be a solution of the nonhomogeneous equation. In such cases we multiply the recommended trial solution by x (or by x 2 if necessary) so that no term in ypx is a solution of the complementary equation. EXAMPLE 5 Solve y y sin x. SOLUTION The auxiliary equation is r 2 1 0 with roots i, so the solution of the com- plementary equation is ycx c1 cos x c2 sin x Ordinarily, we would use the trial solution ypx A cos x B sin x but we observe that it is a solution of the complementary equation, so instead we try ypx Ax cos x Bx sin x Then ypx A cos x Ax sin x B sin x Bx cos x ypx 2A sin x Ax cos x 2B cos x Bx sin x The graphs of four solutions of the differential equation in Example 5 are shown in Figure 4. ■ ■ Substitution in the differential equation gives 4 yp yp 2A sin x 2B cos x sin x so A 12 , B 0, and _2π 2π ypx 12 x cos x yp _4 FIGURE 4 The general solution is yx c1 cos x c2 sin x 12 x cos x NONHOMOGENEOUS LINEAR EQUATIONS ■ 5 We summarize the method of undetermined coefficients as follows: 1. If Gx e kxPx, where P is a polynomial of degree n, then try ypx e kxQx, where Qx is an nth-degree polynomial (whose coefficients are determined by substituting in the differential equation.) 2. If Gx e kxPx cos mx or Gx e kxPx sin mx, where P is an nth-degree polynomial, then try ypx e kxQx cos mx e kxRx sin mx where Q and R are nth-degree polynomials. Modification: If any term of yp is a solution of the complementary equation, multiply yp by x (or by x 2 if necessary). EXAMPLE 6 Determine the form of the trial solution for the differential equation y 4y 13y e 2x cos 3x. SOLUTION Here Gx has the form of part 2 of the summary, where k 2, m 3, and Px 1. So, at first glance, the form of the trial solution would be ypx e 2xA cos 3x B sin 3x But the auxiliary equation is r 2 4r 13 0, with roots r 2 3i, so the solution of the complementary equation is ycx e 2xc1 cos 3x c2 sin 3x This means that we have to multiply the suggested trial solution by x. So, instead, we use ypx xe 2xA cos 3x B sin 3x The Method of Variation of Parameters Suppose we have already solved the homogeneous equation ay by cy 0 and written the solution as 4 yx c1 y1x c2 y2x where y1 and y2 are linearly independent solutions. Let’s replace the constants (or parameters) c1 and c2 in Equation 4 by arbitrary functions u1x and u2x. We look for a particular solution of the nonhomogeneous equation ay by cy Gx of the form 5 ypx u1xy1x u2xy2x (This method is called variation of parameters because we have varied the parameters c1 and c2 to make them functions.) Differentiating Equation 5, we get 6 yp u1 y1 u2 y2 u1 y1 u2 y2 Since u1 and u2 are arbitrary functions, we can impose two conditions on them. One condition is that yp is a solution of the differential equation; we can choose the other condition so as to simplify our calculations. In view of the expression in Equation 6, let’s impose the condition that 7 u1 y1 u2 y2 0 6 ■ NONHOMOGENEOUS LINEAR EQUATIONS yp u1 y1 u2 y2 u1 y1 u2 y2 Then Substituting in the differential equation, we get au1 y1 u2 y2 u1 y1 u2 y2 bu1 y1 u2 y2 cu1 y1 u2 y2 G or 8 u1ay1 by1 cy1 u2ay2 by2 cy2 au1 y1 u2 y2 G But y1 and y2 are solutions of the complementary equation, so ay1 by1 cy1 0 and ay2 by2 cy2 0 and Equation 8 simplifies to au1 y1 u2 y2 G 9 Equations 7 and 9 form a system of two equations in the unknown functions u1 and u2 . After solving this system we may be able to integrate to find u1 and u2 and then the particular solution is given by Equation 5. EXAMPLE 7 Solve the equation y y tan x, 0 x 2. SOLUTION The auxiliary equation is r 2 1 0 with roots i, so the solution of y y 0 is c1 sin x c2 cos x. Using variation of parameters, we seek a solution of the form ypx u1x sin x u2x cos x Then yp u1 sin x u2 cos x u1 cos x u2 sin x Set u1 sin x u2 cos x 0 10 Then yp u1 cos x u2 sin x u1 sin x u2 cos x For yp to be a solution we must have yp yp u1 cos x u2 sin x tan x 11 Solving Equations 10 and 11, we get u1sin 2x cos 2x cos x tan x ■ ■ Figure 5 shows four solutions of the differential equation in Example 7. u1 sin x 2.5 u1x cos x (We seek a particular solution, so we don’t need a constant of integration here.) Then, from Equation 10, we obtain 0 yp π 2 u2 sin x sin 2x cos 2x 1 u1 cos x sec x cos x cos x cos x _1 FIGURE 5 So u2x sin x lnsec x tan x NONHOMOGENEOUS LINEAR EQUATIONS ■ 7 (Note that sec x tan x 0 for 0 x 2.) Therefore ypx cos x sin x sin x lnsec x tan x cos x cos x lnsec x tan x and the general solution is yx c1 sin x c2 cos x cos x lnsec x tan x Exercises A Click here for answers. S 16. y 3y 4 y x 3 xe x Click here for solutions. 17. y 2 y 10 y x 2ex cos 3x 18. y 4y e 3x x sin 2x 1–10 Solve the differential equation or initial-value problem using the method of undetermined coefficients. 1. y 3y 2y x 2. y 9y e 2 3. y 2y sin 4x 7. y y e x x 3, y0 2, 8. y 4y e x cos x, y0 2, ■ ■ ■ ■ ■ ■ y0 0 ■ ■ ■ ■ ■ ; 11–12 Graph the particular solution and several other solutions. What characteristics do these solutions have in common? ■ ■ ■ ■ 13 –18 ■ 13. y 9 y e x sin x 2 x 14. y 9 y xe cos x 15. y 9 y 1 xe 9x ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 23–28 Solve the differential equation using the method of variation of parameters. 1 1 ex 25. y 3y 2y ■ ■ ■ ■ ■ Write a trial solution for the method of undetermined coefficients. Do not determine the coefficients. 2x ■ 24. y y cot x, 0 x 2 12. 2y 3y y 1 cos 2x ■ ■ 23. y y sec x, 0 x 2 11. 4y 5y y e x ■ ■ 22. y y e x y0 1, ■ ■ 21. y 2y y e 2x y0 2 ■ ■ ■ 20. y 3y 2y sin x y0 1 10. y y 2y x sin 2x, ■ 19. y 4y x y0 0 y0 1, ■ Solve the differential equation using (a) undetermined coefficients and (b) variation of parameters. 6. y 2y y xex 5. y 4y 5y e ■ 19–22 4. y 6y 9y 1 x x 9. y y xe x, ■ 3x ■ 26. y 3y 2y sine x 27. y y 1 x e2x x3 28. y 4y 4y ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 8 ■ NONHOMOGENEOUS LINEAR EQUATIONS Answers S Click here for solutions. 1. y c1 e2x c2 ex 2 x 2 2 x 1 3. y c1 c2 e 2x 3 7 4 cos 4x 201 sin 4x 1 5. y e c1 cos x c2 sin x 10 ex 3 11 1 x 7. y 2 cos x 2 sin x 2 e x 3 6x 1 9. y e x ( 2 x 2 x 2) 5 11. The solutions are all asymptotic to yp e x10 as x l . Except for yp , all solutions approach _2 4 yp either or as x l . 1 40 2x _4 13. 15. 17. 19. 21. 23. 25. 27. yp Ae 2x Bx 2 Cx D cos x Ex 2 Fx G sin x yp Ax Bx C e 9x yp xex Ax 2 Bx C cos 3x Dx 2 Ex F sin 3x y c1 cos 2x c2 sin 2x 14 x y c1e x c2 xe x e 2x y c1 x sin x c2 ln cos x cos x y c1 ln1 ex e x c2 ex ln1 ex e 2x y [c1 12 x e xx dx]ex [c2 12 x exx dx]e x NONHOMOGENEOUS LINEAR EQUATIONS ■ 9 Solutions: Nonhomogeneous Linear Equations 1. The auxiliary equation is r 2 + 3r + 2 = (r + 2)(r + 1) = 0, so the complementary solution is yc (x) = c1 e−2x + c2 e−x . We try the particular solution yp (x) = Ax2 + Bx + C, so yp0 = 2Ax + B and yp00 = 2A. Substituting into the differential equation, we have (2A) + 3(2Ax + B) + 2(Ax2 + Bx + C) = x2 or 2Ax2 + (6A + 2B)x + (2A + 3B + 2C) = x2 . Comparing coefficients gives 2A = 1, 6A + 2B = 0, and 2A + 3B + 2C = 0, so A = 12 , B = − 32 , and C = −2x y(x) = yc (x) + yp (x) = c1 e + c2 e −x + 1 2 x 2 − 7 . 4 3 x 2 Thus the general solution is + 74 . 3. The auxiliary equation is r 2 − 2r = r(r − 2) = 0, so the complementary solution is yc (x) = c1 + c2 e2x . Try the particular solution yp (x) = A cos 4x + B sin 4x, so yp0 = −4A sin 4x + 4B cos 4x and yp00 = −16A cos 4x − 16B sin 4x. Substitution into the differential equation gives (−16A cos 4x − 16B sin 4x) − 2(−4A sin 4x + 4B cos 4x) = sin 4x ⇒ (−16A − 8B) cos 4x + (8A − 16B) sin 4x = sin 4x. Then −16A − 8B = 0 and 8A − 16B = 1 1 and B = − 20 . Thus the general solution is y(x) = yc (x) + yp (x) = c1 + c2 e2x + 1 40 cos 4x − 1 20 ⇒ A= 1 40 sin 4x. 5. The auxiliary equation is r 2 − 4r + 5 = 0 with roots r = 2 ± i, so the complementary solution is yc (x) = e2x (c1 cos x + c2 sin x). Try yp (x) = Ae−x , so yp0 = −Ae−x and yp00 = Ae−x . Substitution gives Ae−x − 4(−Ae−x ) + 5(Ae−x ) = e−x 2x y(x) = e (c1 cos x + c2 sin x) + ⇒ 10Ae−x = e−x ⇒ A= 1 . 10 Thus the general solution is 1 −x e . 10 7. The auxiliary equation is r 2 + 1 = 0 with roots r = ±i, so the complementary solution is yc (x) = c1 cos x + c2 sin x. For y00 + y = ex try yp1 (x) = Aex . Then yp0 1 = yp001 = Aex and substitution gives Aex + Aex = ex A = 12 , so yp1 (x) = 12 ex . For y00 + y = x3 try yp2 (x) = Ax3 + Bx2 + Cx + D. ⇒ Then yp0 2 = 3Ax2 + 2Bx + C and yp002 = 6Ax + 2B. Substituting, we have 6Ax + 2B + Ax3 + Bx2 + Cx + D = x3 , so A = 1, B = 0, 6A + C = 0 ⇒ C = −6, and 2B + D = 0 ⇒ D = 0. Thus yp2 (x) = x3 − 6x and the general solution is y(x) = yc (x) + yp1 (x) + yp2 (x) = c1 cos x + c2 sin x + 12 ex + x3 − 6x. But 2 = y(0) = c1 + and 0 = y0 (0) = c2 + y(x) = 3 2 cos x + 11 2 1 2 −6 ⇒ c2 = 11 . 2 1 2 ⇒ c1 = 3 2 Thus the solution to the initial-value problem is sin x + 12 ex + x3 − 6x. 9. The auxiliary equation is r 2 − r = 0 with roots r = 0, r = 1 so the complementary solution is yc (x) = c1 + c2 ex . Try yp (x) = x(Ax + B)ex so that no term in yp is a solution of the complementary equation. Then yp0 = (Ax2 + (2A + B)x + B)ex and yp00 = (Ax2 + (4A + B)x + (2A + 2B))ex . Substitution into the differential equation gives (Ax2 + (4A + B)x + (2A + 2B))ex − (Ax2 + (2A + B)x + B)ex = xex ⇒ ¡ ¢ (2Ax + (2A + B))ex = xex ⇒ A = 12 , B = −1. Thus yp (x) = 12 x2 − x ex and the general solution is ¡ ¢ y(x) = c1 + c2 ex + 12 x2 − x ex . But 2 = y(0) = c1 + c2 and 1 = y0 (0) = c2 − 1, so c2 = 2 and c1 = 0. The ¡ ¡ ¢ ¢ solution to the initial-value problem is y(x) = 2ex + 12 x2 − x ex = ex 12 x2 − x + 2 . 10 ■ NONHOMOGENEOUS LINEAR EQUATIONS 11. yc (x) = c1 e−x/4 + c2 e−x . Try yp (x) = Aex . Then 10Aex = ex , so A = 1 10 and the general solution is y(x) = c1 e−x/4 + c2 e−x + 1 x e . 10 The solutions are all composed of exponential curves and with the exception of the particular solution (which approaches 0 as x → −∞), they all approach either ∞ or −∞ as x → −∞. As x → ∞, all solutions are asymptotic to yp = 1 x e . 10 13. Here yc (x) = c1 cos 3x + c2 sin 3x. For y00 + 9y = e2x try yp1 (x) = Ae2x and for y 00 + 9y = x2 sin x try yp2 (x) = (Bx2 + Cx + D) cos x + (Ex2 + F x + G) sin x. Thus a trial solution is yp (x) = yp1 (x) + yp2 (x) = Ae2x + (Bx2 + Cx + D) cos x + (Ex2 + F x + G) sin x. 15. Here yc (x) = c1 + c2 e−9x . For y 00 + 9y0 = 1 try yp1 (x) = Ax (since y = A is a solution to the complementary equation) and for y 00 + 9y0 = xe9x try yp2 (x) = (Bx + C)e9x . 17. Since yc (x) = e−x (c1 cos 3x + c2 sin 3x) we try yp (x) = x(Ax2 + Bx + C)e−x cos 3x + x(Dx2 + Ex + F )e−x sin 3x (so that no term of yp is a solution of the complementary equation). Note: Solving Equations (7) and (9) in The Method of Variation of Parameters gives u01 = − Gy2 a (y1 y20 − y2 y10 ) u02 = and Gy1 a (y1 y20 − y2 y10 ) We will use these equations rather than resolving the system in each of the remaining exercises in this section. 19. (a) The complementary solution is yc (x) = c1 cos 2x + c2 sin 2x. A particular solution is of the form yp (x) = Ax + B. Thus, 4Ax + 4B = x ⇒ A= 1 4 solution is y = yc + yp = c1 cos 2x + c2 sin 2x + 14 x. and B = 0 ⇒ yp (x) = 14 x. Thus, the general (b) In (a), yc (x) = c1 cos 2x + c2 sin 2x, so set y1 = cos 2x, y2 = sin 2x. Then y1 y20 − y2 y10 = 2 cos2 2x + 2 sin2 2x = 2 so u01 = − 12 x sin 2x ⇒ R ¡ ¢ u1 (x) = − 12 x sin 2x dx = − 14 −x cos 2x + 12 sin 2x [by parts] and u02 = 12 x cos 2x R ¡ ¢ ⇒ u2 (x) = 12 x cos 2xdx = 14 x sin 2x + 12 cos 2x [by parts]. Hence ¢ ¢ ¡ ¡ yp (x) = − 14 −x cos 2x + 12 sin 2x cos 2x + 14 x sin 2x + 12 cos 2x sin 2x = 14 x. Thus y(x) = yc (x) + yp (x) = c1 cos 2x + c2 sin 2x + 14 x. 21. (a) r 2 − r = r(r − 1) = 0 ⇒ r = 0, 1, so the complementary solution is yc (x) = c1 ex + c2 xex . A particular solution is of the form yp (x) = Ae2x . Thus 4Ae2x − 4Ae2x + Ae2x = e2x ⇒ 2x x ⇒ Ae2x = e2x x yp (x) = e . So a general solution is y(x) = yc (x) + yp (x) = c1 e + c2 xe + e2x . ⇒ A=1 (b) From (a), yc (x) = c1 ex + c2 xex , so set y1 = ex , y2 = xex . Then, y1 y20 − y2 y10 = e2x (1 + x) − xe2x = e2x R and so u01 = −xex ⇒ u1 (x) = − xex dx = −(x − 1)ex [by parts] and u02 = ex ⇒ R u2 (x) = ex dx = ex . Hence yp (x) = (1 − x)e2x + xe2x = e2x and the general solution is y(x) = yc (x) + yp (x) = c1 ex + c2 xex + e2x . NONHOMOGENEOUS LINEAR EQUATIONS ■ 11 23. As in Example 6, yc (x) = c1 sin x + c2 cos x, so set y1 = sin x, y2 = cos x. Then sec x cos x y1 y20 − y2 y10 = − sin2 x − cos2 x = −1, so u01 = − = 1 ⇒ u1 (x) = x and −1 R sec x sin x u02 = = − tan x ⇒ u2 (x) = − tan xdx = ln |cos x| = ln(cos x) on 0 < x < π2 . Hence −1 yp (x) = x sin x + cos x ln(cos x) and the general solution is y(x) = (c1 + x) sin x + [c2 + ln(cos x)] cos x. 25. y1 = ex , y2 = e2x and y1 y20 − y2 y10 = e3x . So u01 = Z −e2x e−x =− and −x 3x (1 + e )e 1 + e−x e−x ex ex dx = ln(1 + e−x ). u02 = = 3x so 1 + e−x (1 + e−x )e3x e + e2x ¶ µ Z ex ex + 1 u2 (x) = − e−x = ln(1 + e−x ) − e−x . Hence dx = ln 3x 2x e +e ex u1 (x) = − yp (x) = ex ln(1 + e−x ) + e2x [ln(1 + e−x ) − e−x ] and the general solution is y(x) = [c1 + ln(1 + e−x )]ex + [c2 − e−x + ln(1 + e−x )]e2x . ex e−x and 27. y1 = e−x , y2 = ex and y1 y20 − y2 y10 = 2. So u01 = − , u02 = 2x 2x Z x Z −x e e dx + ex dx. Hence the general solution is yp (x) = −e−x 2x 2x µ µ ¶ Z x Z −x ¶ e e y(x) = c1 − dx e−x + c2 + dx ex . 2x 2x

© Copyright 2018