Nonhomogeneous Linear Equations

Transcription

1 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 ay by cy G x where a, b, and c are constants and G is a continuous function. The related homogeneous equation ay by cy 0 is called the complementary equation and plays an important role in the solution of the original nonhomogeneous equation (). 3 Theorem The general solution of the nonhomogeneous differential equation () can be written as y x x y c x where is a particular solution of Equation and is the general solution of the complementary Equation. y c Proof All we have to do is verify that if y is any solution of Equation, then y is a solution of the complementary Equation. Indeed a y b y c y ay a by b cy c ay by cy a b c t x t x 0 We know from Additional Topics: Second-Order Linear Differential Equations how to solve the complementary equation. (Recall that the solution is y c c y c y, where y and y are linearly independent solutions of Equation.) Therefore, Theorem 3 says that we know the general solution of the nonhomogeneous equation as soon as we know a particular solution. 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 G x where G x) is a polynomial. It is reasonable to guess that there is a particular solution 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 x a polynomial (of the same degree as G) into the differential equation and determine the coefficients. EXAMPLE Solve the equation y y y x. SOLUTION The auxiliary equation of y y y 0 is r r r r 0

2 NONHOMOGENEOUS LINEAR EQUATIONS with roots r,. So the solution of the complementary equation is y c c e x c e x Since G x x is a polynomial of degree, we seek a particular solution of the form x Ax Bx C Then Ax B and A so, substituting into the given differential equation, we have A Ax B Ax Bx C x or Ax A B x A B C x Figure shows four solutions of the differential equation in Example in terms of the particular solution y and the functions f x e x p and t x e x. +f+3g +3g +f _3 3 FIGURE 8 _5 Polynomials are equal when their coefficients are equal. Thus A A B 0 A B C 0 The solution of this system of equations is A B C 3 4 A particular solution is therefore x x x 3 4 and, by Theorem 3, the general solution is y y c c e x c e x x x 3 4 If G x (the right side of Equation ) is of the form Ce kx, where C and k are constants, then we take as a trial solution a function of the same form, x Ae kx, because the derivatives of e kx are constant multiples of e kx. Figure shows solutions of the differential equation in Example in terms of and the functions f x cos x and t x sin x. Notice that all solutions approach as x l and all solutions resemble sine functions when x is negative. 4 +f+g +g _4 +f FIGURE _ EXAMPLE Solve y 4y e 3x. SOLUTION The auxiliary equation is r 4 0 with roots i, so the solution of the complementary equation is y c x c cos x c sin x For a particular solution we try y. Then y and 9Ae 3x p 3Ae 3x p x Ae 3x. Substituting into the differential equation, we have 9Ae 3x 4 Ae 3x e 3x so 3Ae 3x e 3x and A 3. Thus, a particular solution is x 3e 3x and the general solution is y x c cos x c sin x 3e 3x If G x 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 x A cos kx B sin kx

3 NONHOMOGENEOUS LINEAR EQUATIONS 3 EXAMPLE 3 Solve y y y sin x. SOLUTION We try a particular solution x A cos x B sin x Then A sin x B cos x 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 A cos x B sin x sin x or 3A B cos x A 3B sin x sin x This is true if The solution of this system is 3A B 0 and A 3B so a particular solution is A 0 B 3 0 In Example we determined that the solution of the complementary equation is y c c e x c e x. Thus, the general solution of the given equation is If G x 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 we would try y x c e x c e x 0 cos x 3sin x x Ax B cos 3x Cx D sin 3x If G x is a sum of functions of these types, we use the easily verified principle of superposition, which says that if and are solutions of x 0 cos x 3 0 sin x y y 4y x cos 3x ay by cy G x ay by cy G x respectively, then is a solution of ay by cy G x G x EXAMPLE 4 Solve y 4y xe x cos x. SOLUTION The auxiliary equation is r 4 0 with roots, so the solution of the complementary equation is y. For the equation y 4y xe x c x c e x c e x we try x Ax B e x Then y, Ax A B e x p Ax A B e x, so substitution in the equation gives Ax A B e x 4 Ax B e x xe x or 3Ax A 3B e x xe x

4 4 NONHOMOGENEOUS LINEAR EQUATIONS Thus, 3A and A 3B 0, so A, B 3 9, and x ( 3 x 9)e x For the equation y 4y cos x, we try In Figure 3 we show the particular solution of the differential equation in Example 4. The other solutions are given in terms of f x e x and t x e x. FIGURE 3 +f+g +g +f _4 5 _ x C cos x D sin x Substitution gives 4C cos x 4D sin x 4 C cos x D sin x cos x or 8C cos x 8D sin x cos x Therefore, 8C, 8D 0, and x 8 cos x By the superposition principle, the general solution is y y c c e x c e x ( 3x 9)e x 8 cos x Finally we note that the recommended trial solution 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 if necessary) so that no term in x is a solution of the complementary equation. EXAMPLE 5 Solve y y sin x. SOLUTION The auxiliary equation is r 0 with roots i, so the solution of the complementary equation is y c x c cos x c sin x Ordinarily, we would use the trial solution x A cos x B sin x but we observe that it is a solution of the complementary equation, so instead we try x Ax cos x Bx sin x Then x A cos x Ax sin x B sin x Bx cos x x A sin x Ax cos x B cos x Bx sin x The graphs of four solutions of the differential equation in Example 5 are shown in Figure 4. 4 Substitution in the differential equation gives A sin x B cos x sin x _π π so A, B 0, and x x cos x _4 The general solution is FIGURE 4 y x c cos x c sin x x cos x

5 NONHOMOGENEOUS LINEAR EQUATIONS 5 We summarize the method of undetermined coefficients as follows:. If G x e kx P x, where P is a polynomial of degree n, then try x e kx Q x, where Q x is an nth-degree polynomial (whose coefficients are determined by substituting in the differential equation.). If G x e kx P x cos mx or G x e kx P x sin mx, where P is an nth-degree polynomial, then try x e kx Q x cos mx e kx R x sin mx where Q and R are nth-degree polynomials. Modification: If any term of by x (or by x if necessary). is a solution of the complementary equation, multiply EXAMPLE 6 Determine the form of the trial solution for the differential equation y 4y 3y e x cos 3x. SOLUTION Here G x has the form of part of the summary, where k, m 3, and P x. So, at first glance, the form of the trial solution would be x e x A cos 3x B sin 3x But the auxiliary equation is r 4r 3 0, with roots r 3i, so the solution of the complementary equation is y c x e x c cos 3x c sin 3x This means that we have to multiply the suggested trial solution by x. So, instead, we use x xe x A 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 y x c y x c y x y ters) c and c in Equation 4 by arbitrary functions u x and u x. We look for a particular where and y are linearly independent solutions. Let s replace the constants (or parame- solution of the nonhomogeneous equation ay by cy G x of the form 5 x u x y x u x y x (This method is called variation of parameters because we have varied the parameters c and c to make them functions.) Differentiating Equation 5, we get 6 u y u y u y u y u u Since and are arbitrary functions, we can impose two conditions on them. One condition is that 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 u y u y 0

6 6 NONHOMOGENEOUS LINEAR EQUATIONS Then u y u y u y u y Substituting in the differential equation, we get or a u y u y u y u y b u y u y c u y u y G 8 u ay by cy u ay by cy a u y u y G y y But and are solutions of the complementary equation, so ay by cy 0 and Equation 8 simplifies to and ay by cy 0 9 a u y u y G Equations 7 and 9 form a system of two equations in the unknown functions u and u. After solving this system we may be able to integrate to find u and u and then the particular solution is given by Equation 5. EXAMPLE 7 Solve the equation y y tan x, 0 x. SOLUTION The auxiliary equation is r 0 with roots i, so the solution of y y 0 is c sin x c cos x. Using variation of parameters, we seek a solution of the form x u x sin x u x cos x Then u sin x u cos x u cos x u sin x Set 0 u sin x u cos x 0 Then u cos x u sin x u sin x u cos x For to be a solution we must have u cos x u sin x tan x Solving Equations 0 and, we get Figure 5 shows four solutions of the differential equation in Example 7..5 u sin x cos x cos x tan x u sin x u x cos x (We seek a particular solution, so we don t need a constant of integration here.) Then, from Equation 0, we obtain 0 _ π u sin x cos x u sin x cos x cos x cos x sec x cos x FIGURE 5 So u x sin x ln sec x tan x

7 NONHOMOGENEOUS LINEAR EQUATIONS 7 (Note that sec x tan x 0 for 0 x.) Therefore x cos x sin x sin x ln sec x tan x cos x cos x ln sec x tan x and the general solution is y x c sin x c cos x cos x ln sec x tan x Exercises A Click here for answers. 0 Solve the differential equation or initial-value problem using the method of undetermined coefficients.. y 3y y x. y 9y e 3x 3. y y sin 4x 4. y 6y 9y x 5. y 4y 5y e x 6. x y y y xe 7. y y e x x 3, y 0, y y 4y e x cos x, y 0, y 0 9. y y xe x, y 0, y 0 0. y y y x sin x, y 0, y 0 0 ; Graph the particular solution and several other solutions. What characteristics do these solutions have in common?. 4y 5y y e x. y 3y y cos x 3 8 Write a trial solution for the method of undetermined coefficients. Do not determine the coefficients. 3. y 9y e x x sin x 4. y 9y xe x cos x 5. y 9y xe 9x S Click here for solutions. 6. y 3y 4y x 3 x e x 7. y y 0y x e x cos 3x 8. y 4y e 3x x sin x 9 Solve the differential equation using (a) undetermined coefficients and (b) variation of parameters. 9. y 4y x 0. y 3y y sin x. y y y e x. y y e x 3 8 Solve the differential equation using the method of variation of parameters. 3. y y sec x, 0 x 4. y y cot x, 0 x 5. y 3y y e x 6. y 3y y sin e x y y x y 4y 4y e x x 3

8 8 NONHOMOGENEOUS LINEAR EQUATIONS Answers S Click here for solutions y c e x c e x x 3 x 7 4 y c c e x 40 cos 4x 0 sin 4x y e x c cos x c sin x 0 e x y 3 cos x sin x e x x 3 6x y e x ( x x ). 5 The solutions are all asymptotic to e x 0 as x l. Except for, _ 4 all solutions approach either or as x l _4 Ae x Bx Cx D cos x Ex Fx G sin x Ax Bx C e 9x xe x Ax Bx C cos 3x Dx Ex F sin 3x y c cos x c sin x 4 x y c e x c xe x e x y c x sin x c ln cos x cos x y c ln e x e x c e x ln e x e x 7. y [c x e x x dx]e x [c x e x x dx]e x

9 NONHOMOGENEOUS LINEAR EQUATIONS 9 Solutions: Nonhomogeneous Linear Equations. The auxiliary equation is r +3r +=(r +)(r +)=0,sothecomplementarysolutionis y c (x) =c e x + c e x. We try the particular solution (x) =Ax + Bx + C, soy 0 p =Ax + B and y 00 p =A. Substituting into the differential equation, we have (A)+3(Ax + B)+(Ax + Bx + C) =x or Ax +(6A +B)x +(A +3B +C) =x. Comparing coefficients gives A =, 6A +B =0,and A +3B +C =0,soA =, B = 3,andC = 7. Thus the general solution is 4 y(x) =y c(x)+(x) =c e x + c e x + x 3 x The auxiliary equation is r r = r(r ) = 0,sothecomplementarysolutionisy c(x) =c + c e x.trythe particular solution (x) =A cos 4x + B sin 4x, soy 0 p = 4A sin 4x +4B cos 4x and y 00 p = 6A cos 4x 6B sin 4x. Substitution into the differential equation gives ( 6A cos 4x 6B sin4x) ( 4A sin4x + 4B cos 4x) = sin4x ( 6A 8B)cos4x +(8A 6B)sin4x =sin4x. Then 6A 8B =0and 8A 6B = A = 40 and B =. Thus the general solution is y(x) =y 0 c(x)+ (x) =c + c e x + cos 4x sin 4x The auxiliary equation is r 4r +5=0with roots r =± i, sothecomplementarysolutionis y c(x) =e x (c cos x + c sin x). Try (x) =Ae x,soy 0 p = Ae x and y 00 p = Ae x. Substitution gives Ae x 4( Ae x )+5(Ae x )=e x 0Ae x = e x A =. Thus the general solution is 0 y(x) =e x (c cos x + c sin x)+ 0 e x. 7. The auxiliary equation is r +=0with roots r = ±i, sothecomplementarysolutionis y c(x) =c cos x + c sin x. Fory 00 + y = e x try (x) =Ae x.theny 0 p = y 00 p = Ae x and substitution gives Ae x + Ae x = e x A =,soyp (x) = ex.fory 00 + y = x 3 try (x) =Ax 3 + Bx + Cx + D. Then y 0 p =3Ax +Bx + C and y 00 p =6Ax +B. Substituting, we have 6Ax +B + Ax 3 + Bx + Cx + D = x 3,soA =, B =0, 6A + C =0 C = 6, andb + D =0 D =0.Thus (x) =x 3 6x and the general solution is y(x) =y c(x)+ (x)+ (x) =c cos x + c sin x + ex + x 3 6x. But=y(0) = c + c = 3 and 0=y 0 (0) = c + y(x) = 3 cos x + sin x + ex + x 3 6x. 6 c =. Thus the solution to the initial-value problem is 9. The auxiliary equation is r r =0with roots r =0, r =so the complementary solution is y c(x) =c + c e x. Try (x) =x(ax + B)e x so that no term in is a solution of the complementary equation. Then y 0 p =(Ax +(A + B)x + B)e x and y 00 p =(Ax +(4A + B)x +(A +B))e x. Substitution into the differential equation gives (Ax +(4A + B)x +(A +B))e x (Ax +(A + B)x + B)e x = xe x (Ax +(A + B))e x = xe x A =, B =. Thus (x) = x x e x and the general solution is y(x) =c + c e x + x x e x.but=y(0) = c + c and =y 0 (0) = c,soc =and c =0.The solution to the initial-value problem is y(x) =e x + x x e x = e x x x +.

10 0 NONHOMOGENEOUS LINEAR EQUATIONS. y c(x) =c e x/4 + c e x.try(x) =Ae x.then 0Ae x = e x,soa = and the general solution is 0 y(x) =c e x/4 + c e x + 0 ex. 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.asx, all solutions are asymptotic to = 0 ex. 3. Here y c(x) =c cos 3x + c sin 3x. Fory 00 +9y = e x try (x) =Ae x and for y 00 +9y = x sin x try (x) =(Bx + Cx + D)cosx +(Ex + Fx+ G)sinx. Thus a trial solution is (x) = (x)+ (x) =Ae x +(Bx + Cx + D)cosx +(Ex + Fx+ G)sinx. 5. Here y c(x) =c + c e 9x.Fory 00 +9y 0 =try (x) =Ax (since y = A is a solution to the complementary equation) and for y 00 +9y 0 = xe 9x try (x) =(Bx + C)e 9x. 7. Since y c(x) =e x (c cos 3x + c sin 3x) we try (x) =x(ax + Bx + C)e x cos 3x + x(dx + Ex + F )e x sin 3x (sothatnotermof is a solution of the complementary equation). Note: Solving Equations (7) and (9) in The Method of Variation of Parameters gives u 0 Gy = a (y y 0 y y 0 ) and u 0 Gy = a (y y 0 y y 0 ) We will use these equations rather than resolving the system in each of the remaining exercises in this section. 9. (a) The complementary solution is y c (x) =c cos x + c sin x. A particular solution is of the form (x) =Ax + B. Thus,4Ax +4B = x A = and B =0 yp(x) = x. Thus, the general 4 4 solution is y = y c + = c cos x + c sin x + 4 x. (b) In (a), y c(x) =c cos x + c sin x, sosety =cosx, y =sinx. Then y y 0 y y 0 =cos x +sin x =so u 0 = x sin x R u (x) = x sin xdx = 4 x cos x + sin x [barts] and u 0 = x cos x R u (x) = x cos xdx = 4 x sin x + cos x [barts]. Hence (x) = 4 x cos x + sin x cos x + 4 x sin x + cos x sin x = x.thus 4 y(x) =y c (x)+ (x) =c cos x + c sin x + x. 4. (a) r r = r(r ) = 0 r =0,, so the complementary solution is y c (x) =c e x + c xe x. A particular solution is of the form (x) =Ae x. Thus 4Ae x 4Ae x + Ae x = e x Ae x = e x A = (x) =e x. So a general solution is y(x) =y c(x)+(x) =c e x + c xe x + e x. (b) From (a), y c (x) =c e x + c xe x,sosety = e x, y = xe x. Then, y y 0 y y 0 = e x ( + x) xe x = e x and so u 0 = xe x u (x) = R xe x dx = (x )e x [barts] and u 0 = e x u (x) = R e x dx = e x. Hence (x) =( x)e x + xe x = e x and the general solution is y(x) =y c (x)+ (x) =c e x + c xe x + e x.

11 NONHOMOGENEOUS LINEAR EQUATIONS 3. As in Example 6, y c(x) =c sin x + c cos x, sosety =sinx, y =cosx. Then y y 0 y y 0 = sin x cos x =, sou 0 sec x cos x = = u (x) =x and u 0 sec x sin x = = tan x u (x) = R tan xdx =ln cos x =ln(cosx) on 0 <x< π.hence (x) =x sin x +cosx ln(cos x) and the general solution is y(x) =(c + x)sinx +[c +ln(cosx)] cos x. 5. y = e x, y = e x and y y 0 y y 0 = e 3x.Sou 0 = Z u (x) = e x +e dx x =ln(+e x ). u 0 = Z e x µ e x + u (x) = dx =ln e 3x + ex e x e x e x = and ( + e x )e3x +e x e x ( + e x )e = e x so 3x e 3x + ex e x =ln(+e x ) e x.hence (x) =e x ln( + e x )+e x [ln( + e x ) e x ] and the general solution is y(x) =[c +ln(+e x )]e x +[c e x +ln(+e x )]e x. 7. y = e x, y = e x and y y 0 y y 0 =.Sou 0 = ex x, u0 = e x Z e (x) = e x x y(x) = µ Z c e x Z dx + ex x e x x dx e x + x µ c + x and dx. Hence the general solution is Z e x x dx e x.