Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Signal-flow graph wikipedia , lookup
Linear algebra wikipedia , lookup
Eigenvalues and eigenvectors wikipedia , lookup
Quadratic equation wikipedia , lookup
Cubic function wikipedia , lookup
Quartic function wikipedia , lookup
System of polynomial equations wikipedia , lookup
Elementary algebra wikipedia , lookup
History of algebra wikipedia , lookup
Differential Chapter 15 equations 15.1 Basic concepts, separable and homogeneous equations 15.2 First-order linear equations 15.3 Exact equations 15.4 Strategy for solving first-order equations 机动 目录 上页 下页 返回 结束 15.5 Second-Order Linear Equations A second-order linear differential equation has the form 2 d y dy (1) P( x) 2 Q( x) R( x) y G( x) dx dx where P, Q, R, and G are continuous functions. If G(x) = 0 for all x, such equations are called secondorder homogeneous linear equations. (This use of the word homogeneous has nothing to do with the meaning given in Section 15.1.) (2) d2y dy P ( x ) 2 Q( x ) R ( x ) y 0 dx dx If G ( x) 0 for some x, Equation 1 is nonhomogeneous. Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions y1 and y 2 of such an equation, then the linear combination 2y 2c 1y 1c y is also a solution. (3)Theorem If y1 ( x ) and y 2 ( x ) are both solutions of the linear equation (2) and c1 and c2 are any constants, then the function )x ( 2y 2c )x ( 1y 1c )x (y is also a solution of Equation 2. Proof have Since y1 and y 2 are solutions of Equation 2, we 0 1y )x (R 1y )x (Q 1y )x (P and 0 2y )x (R 2y )x (Q 2y )x (P Therefore y )x (R y )x (Q y )x (P ) 2y 2c 1y 1c()x (R ) 2y 2c 1y 1c()x (Q ) 2y 2c 1y 1c()x (P ) 2y 2c 1y 1c()x (R ) 2y 2c 1y 1c()x (Q )2y 2c 1y 1c()x (P ] 2y )x (R 2y )x (Q 2y )x (P [ 2c ] 1y )x (R 1y )x (Q 1y )x (P [1c 0 2c 0 1c 0 Thus 2 y 2c 1y 1c y is a solution of Equation 2. Let x and y are two variables, if neither x nor y is a constant multiple of the other, we say x and y are two linearly independent variables. For instance, the function f ( x) x 2 and g ( x) 5 x 2 are linearly dependent, but f ( x) e x and g ( x ) xe x are linearly independent. The second theorem says that the general solution of a homogeneous linear equation is a linear combination of two linearly independent solutions. (4)Theorem If y1 and y 2 are linearly independent solutions of Equation 2 , then the general solution is given by )x ( 2y 2c )x ( 1y 1c )x ( y where c1 and c2 are arbitrary constants. In general, it is not easy to discover particular solutions to a second-order linear equation. But it is always possible to do so if the coefficient functions P, Q and R are constant functions, that is, if the differential equation has the form (5) 0 yc yb ya It is not hard to think of some likely candidates for particular solutions of Equation 5. For example, the exponential function y erx because its derivatives are rx 2 rx constants multiple of itself: y re , y r e . Substitute these expression into Equation 5 ar 2 e rx brerx cerx 0 (ar 2 br c)e rx 0 Notice erx is never 0 so y e rx is a solution of Equation 5 if r is a root of the equation (6) ar 2 br c 0 which is called the auxiliary equation(or characteristic equation) of Equation 5. Using the quadratic formula, the root r1 and r2 of the auxiliary equation can be found: (7) b b 2 4ac r1 2a b b 2 4ac r2 2a We distinguish three cases according to the sign of the discriminant b 2 4ac . Case 1 b 2 4ac 0 In this case the roots r1 and r2 of the auxiliary equation are real and distinct, so y1 e r1 x and y 2 e r2 x are two linearly independent solutions of Equation 5. (8) If the roots r1 and r2 of the auxiliary equation ` ar 2 br c 0 are real and unequal, then the general solution of ay by cy 0 is y c1e r1 x c2 e r2 x Example 1 Solve the equation y y 6 y 0 d 2 y dy Example 2 Solve the equation 3 2 y 0. dx dx Case 2 b 2 4ac 0 In this case r1 r2 ; that is, the root of the auxiliary equation are real and equal. Denote r as the common value of r1 and r2 , we have (9) b r 2a so 2ar b 0 We know that y1 e rx is one solution of Equation 5. We now verify that y2 xerx is also a solution: ay2 by2 cy2 a ( 2re rx r 2 xerx ) b(e rx rxerx ) cxerx ( 2ar b)e rx ( ar 2 br c) xerx 0(e rx ) 0( xerx ) 0 Since y1 e and y2 xe are linearly independent solutions, Theorem 4 provides us with the general solution: rx rx (10) If the auxiliary equation ar 2 br c 0 has only one real root r, then the general solution of ` ay by cy 0 is y c1e rx c2 xerx (c1 c2 x)e rx Example 3 Solve the equation 4 y 12 y 9 y 0 Case 3 b 2 4ac 0 In this case the roots r1 and r2 of the auxiliary equation are complex numbers, we can write r1 i r2 i where b (2a ), 4ac b 2 (2a) Using Euler’s equation ei cos i sin we write the solution of the differential equation as y C1e r1 x C2 e r2 x C1e ( i ) x C2 e ( i ) x C1ex (cos x i sin x) C2 ex (cos x i sin x) ex [(C1 C2 ) cos x i (C1 C2 ) sin x] ex (c1 cos x c2 sin x) where c1 C1 C2 , c2 C1 C2 We summarize the discussion as follows: (11) If the roots of auxiliary equation ar 2 br c 0 are the complex numbers r1 i , r2 i , then the general solution of ay by cy 0 is y ex (c1 cos x c2 sin x) Example 4 Solve the equation y 6 y 13 y 0. Initial-value and boundary-value problems An initial-value problem for the second-order Equation 1 or 2 consists of finding a solution y of the differential equation that also satisfies initial conditions of the form y( x0 ) y0 y( x0 ) y1 where y 0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and P ( x ) 0 there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem. A boundary-value problem for Equation 1 consists of finding a solution of the differential equation that also satisfies boundary conditions of the form y( x0 ) y0 y( x1 ) y1 In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution. Example 5 Solve the initial-value problem y y 6 y 0 y (0) 1 y(0) 0 Example 6 Solve the initial-value problem y y 0 y (0) 2 y(0) 3 Example 7 Solve the boundary-value problem y 2 y y 0 y (0) 1 y (1) 3 15.1 Basic concepts, separable and homogeneous equations 机动 目录 上页 下页 返回 结束 Basic concepts In Sections 8, We have known the concept of the differential equations. 10 Two kinds of equations Def: An ordinary differential equation: involves an unknown function of a single variable and some of its derivatives. Such as dy x e xy dx 机动 目录 上页 下页 返回 结束 A partial differential equation: involves an unknown function of two or more variables and some of its partial derivatives. Such as 2u 2u 2u 2 xy 2 2 0 2 x x y 20 The order of the equation The order of a differential equation is the order of the highest derivative that appears in the equation. dy x e xy dx Is an ordinary differential equation of order 1 y 2 xy y sin x Is a third-order differential equation 2u 2u 2u 2 xy 2 2 0 2 x x y Is a second-order partial differential equation 机动 目录 上页 下页 返回 结束 12.1.2 Separable equations Note: We study only ordinary differential equations mainly In general,a first-order differential equation has dy the form : F ( x, y ) dx where F is some function of the two variables x and y 10 Define the separable equations When F can be factored as a function of x times a function F ( x, y ) g ( x ) f ( y ) then dy g ( x) f ( y ) dx Is called a separable equation 机动 目录 上页 下页 返回 结束 Form (1) dy g ( x) f ( y ) dx Also can be written as: (2) dy g ( x) dx h( y ) h( y )dy g ( x)dx We integrate both sides: h( y)dy g ( x)dx H ( y ) G ( x) C This defines y implicitly as a function of x. Sometimes ,we can solve for y in terms of x , thus , y=f(x) are called general solutions of the equation 机动 目录 上页 下页 返回 结束 20 An initial-value problem The separable equation dy g ( x) dx h( y ) Has a solution satisfies an initial condition of the form (3) y( x0 ) y0 We say this is an initial problem。 In general, there is a unique solution to the initial problem given by equations (2),(3) 机动 目录 上页 下页 返回 结束 Example 1 (a) Solve the differential equation (b) Solve the initial-value problem dy e y cos x. dx y e y cos x y (0) 1 dy y e cos x. Solution (a) dx y e dy cos xdx. Write it as y e dy cos xdx. e y sin x C y ln(sin x C ) This is the general solution, involves an arbitrary constant C Solution (b) y(0) =1 tell us x=o,y=1 So y ln(sin x e) 1 ln(sin 0 C ) 1 ln C , C e 机动 目录 上页 下页 返回 结束 Example 2 Solution: Solve the differential equation ln x y xy xy 3 ln x ( y y )dy dx Write it as x We integrate both sides 3 ln x ( y y )dy x dx 3 So y2 y4 1 2 1 ln x C 2 4 2 2 机动 目录 上页 下页 返回 结束 Example 3 Solve the differential equation 2 x y Solution:Write it as 3x e 4 y3 ( y y 3 )dy ln x dx x We integrate both sides 3 ( y y )dy ln x dx x So y2 y4 1 2 1 ln x C 2 4 2 2 机动 目录 上页 下页 返回 结束 Homogeneous equations A first-order differential equation If F(x,y) can be written as The form (4) is called homogeneous equations For instance: Turn it as : dy F ( x, y ) dx y F ( x, y ) g ( ) x dy y g( ) dx x dy x 2 xy y 2 x y ln x ln y 2 2 dx x y x 2y y y 2 y 1 ( ) ( ) 1 ( ) dy y x x ln( ) x y 2 y dx x 1 ( ) 1 2( ) x x 机动 目录 上页 下页 返回 结束 y y g ( ) x We see make the change of variable then thus y xv v xv g (v) and so y x y 1v xv v xv v y xv g (v ) v v x or This is a separable differential equation , it’s general solution v ( x ) So, the original differential equation has the general solution y xv ( x ) 机动 目录 上页 下页 返回 结束 xy y 2 y x2 Example 5 Solve the differential equation Solution: write it as make the substitution v y y y ( ) ( ) 2 x x y x , which gives, y v xv 2 v so v xv v v 2 v x this is a separable differential equation, in a usual way solve it, 1 ln x C v x y ln x C aslo vo dv 2 v v yo v0 dx x 1 ln x C is another solution 机动 目录 上页 下页 返回 结束 Example 6 Solve the homogeneous differential equation y y xy sin y sin x x x 机动 目录 上页 下页 返回 结束 15.1Homework P 955 1.3.5. 15.2 First-order linear equations 机动 目录 上页 下页 返回 结束 The general solutions of the first-order linear equations dy P( x) y Q( x) dx (1) where P and Q are contunuous functions on agiven interval. It’s solution is ye P ( x )dx (2) Q( x)e P( x)dx dx C 机动 目录 上页 下页 返回 结束 When Q(x)=0, form (1) become dy P( x) y 0 dx (3) This is a separate equations ,solve it dy y P( x)dx, ln y P( x)dx C , y e P ( x ) dx C Thus y Ce P ( x ) dx (4) Is the solution of the equation form (3), C is an arbitrary constant 机动 目录 上页 下页 返回 结束 Equation (3) dy P( x) y 0 dx The solution is written as y Ce P ( x ) dx Let us replace the constant C in it by arbitrary We look for the solution of the equation of the form dy P( x) y Q(x) dx y( x) C ( x)e P ( x ) dx (5) 机动 目录 上页 下页 返回 结束 Differentiating Equation (5),we get P ( x ) dx P ( x ) dx dC(x) P ( x ) dx e C ( x)(1) P( x)e P( x)C ( x)e Q( x) dx dC(x) P ( x ) dx e Q( x) dx P ( x ) dx dC(x) Q ( x )e dx P ( x ) dx C(x) Q( x)e dx C y( x) C ( x)e P ( x ) dx Put it into (5), thus: The solution of the form (1) is ye P ( x ) dx Q( x)e P ( x ) dx dx C (6) Another method see page 997 机动 目录 上页 下页 返回 结束 Example 1 Solve the differential equation dy 3x 2 y 6 x 2 dx Solution: this is a linear equation By the general solution form (6) P( x) 3x 2 First step: The second step: e P ( x ) dx e 3 x 2 dx P ( x ) dx dx Q ( x ) e e 6x Q( x ) 6 x 2 x3 2 e dx x3 2 e dx 2e x3 Last step: ye x3 2e x C 3 3 2 Ce 机动 目录 x3 x3 上页 下页 返回 结束 Example 2 Find the solution of the initial-value problem x 2 y xy 1 x0 y (1) 2 Solution : We must first put the differential equation into standard form: 1 1 y y 2 x x First: Second: e P ( x ) dx e 1 dx x e x0 ln x e ln x1 1 x 1 1 x2 xdx xdx 1 y ln x C x P ( x ) dx dx Q( x)e Last: Since y(1)=2, We have 2 ln x 1 ln1 C C 1 Therefore, the solution to the initial-value problem is y 1 ln x 2 x 机动 目录 上页 下页 返回 结束 Example 3 find the solution of the initial-value problem dy ty 3t , dt t 2 1 y (2) 2 机动 目录 上页 下页 返回 结束 A Bernoulli differential equation dy P( x) y Q( x) y n dx Observe that ,if n=0 or 1,it is linear. n the substitution uy transforms the Bernoulli equation into the linear equation du (1 n) P( x)u Q( x) dx Example 1 Solve the differential equation dy y 6 xy 2 dx x 机动 目录 上页 下页 返回 结束 15.2 homework 15.3 Exact equations Definition of the exact f ( x, y ) C Suppose the equation (1) Then y=f(x) satisfies a first-order differential equation obtained by using the Chain Rule to differentiate both sides of equation with respect to x: f x( x, y) f y( x, y) y 0 (2) A differential equation of the form of Equation (2)is called exact。 机动 目录 上页 下页 返回 结束 Definition: A first –order differential equation of the form dy P ( x, y ) Q ( x, y ) 0 dx (3) is called exact if there is a function f(x,y) such that f x( x, y) P( x, y) f y( x, y ) Q( x, y ) If f(x,y) is known,thus the solution is given implicitly by f ( x, y ) C (4) We may be able to solve Equation 4 for y as an explicit function of x. 机动 目录 上页 下页 返回 结束 Theorem We have the following convenient method for the exactness of a differential equation. Theorem: Q ( x, y ) Suppose P ( x, y ) and have continuous partial derivatives on a simply –connected domain. The the differential equation dy P ( x, y ) Q ( x, y ) 0 dx is exact if and only if P Q Y x (5) 机动 目录 上页 下页 返回 结束 Find solutions of exact equation Example 1 Solve the differential equation 4 x 3 y 3( x y 2 ) y 0 Solution Here 2 Q ( x , y ) 3 x 3 y P( x, y) 4 x 3 y and have continuous partial derivatives on R2 Also P Q 3 y x So the differential equation is exact by theorem . Thus there exists a function f such that f x( x, y) 4 x 3 y f y( x, y ) 3 x 3 y (6) 2 (7) 机动 目录 上页 下页 返回 结束 To determine f we first integrate 6 with respect to x : f ( x, y) 2 x 3xy g ( y) 2 (8) Now we differentiate 8 with respect to y: f y( x, y) 0 3x g ( y) 3x g ( y) (9) Comparing 7 and 9,we see that g ( y ) 3 y 2 and so g ( y) y3 We do not need the arbitrary constant here Thus f ( x, y) 2 x 3xy y So 2 x2 3xy y3 C 2 3 is the solution. 机动 目录 上页 下页 返回 结束 Example 2 Solution Solve the differential equation sin y (1 x cos y) y 0 Here P( x, y) sin y and Q( x, y) 1 x cos y have continuous partial derivatives on R2 Also P Q cos y y x So the differential equation is exact by theorem . thus there exists a function f such that f x( x, y) sin y f y( x, y ) 1 x cos y 机动 目录 上页 下页 返回 结束 To determine f we first integrate f x( x, y) sin y f ( x, y) x sin y g ( y) with respect to x : Now we differentiate it with respect to y: f y( x, y) x cos y g ( y) Comparing them,we see 1 x cos y x cos y g ( y) that g ( y ) 1 then g ( y) y We do not need the arbitrary constant here Thus So f ( x, y) x sin y y x sin y y C is the solution . 机动 目录 上页 下页 返回 结束 Integrating factors The differential equation 2 x y 2 xyy 0 is not exact. If the differential equation P ( x, y ) Q ( x, y ) dy 0 dx is not exact, we look for an integrating factor I(x,y) Such that ,after multiplication by I(x,y), the resulting equation dy I ( x, y ) P( x, y ) I ( x, y )Q( x, y ) 0 (10 ) dx is exact. 机动 目录 上页 下页 返回 结束 Equation (10) is exact if ( IP) ( IQ) Y x That is I y P IPy I xQ IQx or PI y QI x I (Qx Py) (11) In general,it is harder to solve this partial differential equation than the original differential equation. But it is sometimes possible to find I that is a function of x or y alone. 机动 目录 上页 下页 返回 结束 Suppose I is a function of x alone. Then I y 0 So equation 11 becomes if Py Qx Q dI Py Qx I dx Q (12) is a function of x alone , Then Equation 12 is a first –order linear (and separable ) ordinary differential equation. It can be solved for I(x) Then Equation 10 is exact and can be solved In Example 1 or 2. 机动 目录 上页 下页 返回 结束 Example 3 Solve the differential equation 2 x y xyy 0 2 Solution : Here P( x, y) 2 x y 2 P 2y y since But P y Py Qx Q Q x Q ( x, y ) xy Q y x the given equation is not exact . 2y y 1 xy x is a function of x alone. So by Equation 12 there is an integrating factor I that satisfies dI 1 I dx x we get I ( x ) x. dI I dx x ( just a particcular one ) 机动 目录 上页 下页 返回 结束 Multiplying the original equation by x ,we get x(2 x y 2 xyy) 0 (13) 2 x 2 xy 2 x 2 yy 0 If we let p( x, y) 2 x 2 xy 2 , q( x, y) x 2 y p q Then 2 xy y x So Equation 13 is now exact.thus there is a function f such that f x ( x, y ) 2 x 2 xy 2 , f y ( x, y) x2 y Integrating the first of these equations,we get 2 3 1 2 2 f ( x, y ) x x y q ( y ) 3 2 so f y ( x, y) x2 y g ( y) Comparison then gives g ( y ) 0 ,so g is a constant (which we can take to be 0) 机动 目录 上页 下页 返回 结束 therefore 2 3 1 2 2 f ( x, y ) x x y 3 2 2 3 1 2 2 The solution is x x y C 3 2 机动 目录 上页 下页 返回 结束 15.3homework 15.4 Strategy for solving firstorder equations In solving first-order differential equations we used the technique for separable equations in Sections 8.1 and the method for linear equations in Section 15.2.(on the text). We also develops methods for solving homogeneous equations in Section 15.1 (on the text) and exact equations in Section 15.3 (on the text)。 In this section we present a miscellaneous collection of first-order differential equations and part of the problem is to recognize which technique should be used on each equation. 机动 目录 上页 下页 返回 结束 As with the strategy of integration (Section7.6) and the strategy of testing series (Section 10.7),the main idea is to classify the equation according to its form. Here , however , the important thing is not so much the form of the functions involved e as it is the form of the equation itself . 机动 目录 上页 下页 返回 结束 Recall that a separable equation can be written in the form (1) dy g ( x) f ( y ) dx that is ,the expression for dy/dx can be factored and a product of a function of x and a function of y.A linear equation can be put into the form (2) dy P( x) y Q(x) dx 机动 目录 上页 下页 返回 结束 A homogeneous equation can be expressed in the form dy y (3) g( ) dx x An exact equation has the form (4) dy P ( x, y ) Q ( x, y ) 0 dx P where y Q x If an equation has none of these forms ,we can ,as a last resort , attempt to find and integrating factor and thus equation exact . 机动 目录 上页 下页 返回 结束 In each of these cases ,some preliminary algebra may be required in order to put a given equation into one of the preceding forms . (This step is analogous to Step l of the strategy for integration: algebraic implification. ) 机动 目录 上页 下页 返回 结束 It may happen that a given equation is of more than one type . For instance, the equation xy y is separable because dy/dx can be written as dy 1 ( )y dx x (compare with Equation 1) It is also linear ,since we can write the equation as dy 1 ( ) y 0 dx x dy y dx x (compare with Equation 2) 机动 目录 上页 下页 返回 结束 Furthermore ,it is also homogeneous because we can write it as dy y dx x (compare with Equation 2). In such a case we could solve the equation using any one of the corresponding methods ,although one of the methods might be easier than the others . 机动 目录 上页 下页 返回 结束 In the following examples we identify the type of each equation without working our the details of the solutions . Example1 y 1 x y xy Initially ,this equation may not appear to be in any of the forms of Equations 1,2,3,or 4,but observe that we can factor the right side and therefore write the equation as y (1 x)(1 y ) We now recognize the equation ad being separable and we can solve it using the methods of Section 15.1. 机动 目录 上页 下页 返回 结束 Example 2 x 2 y 2 2 xyy 0 The equation is clearly mot separable , nor separable, nor is it linear . p y 2 y Since it is not exact . qy 2 y But if we solve for y’, we get y2 x2 1 y 1 y 2 xy 2x y x Which shows that y’ is a function of y/x and the equation is homogeneous (see Equation 3). (We could have anticipated this because the expressions x2,y2,and 2xy are all of degree 2.) The change of variable v=y\x converts the equation into a separable equation . 机动 目录 上页 下页 返回 结束 3x 2 2 xy 3 y 3x 2 y 2 Example3 This equation is not separable homogeneous . ,linear ,or We suspect is might be exact, so we write it in the form (3x 2 2 xy 3 ) (3x 2 y 2 ) y 0 If then p( x, y) 3x2 2 xy3 , q( x, y) 3x 2 y 2 py 6 xy 2 qx Therefore ,the equation is indeed exact and can be dolled by the methods of Section 15.3. 机动 目录 上页 下页 返回 结束 15.4homework Example4 ( x 1) y xy x3 x 2 If we put the equation in the form x 1 y x y x x2 We recognize it as having the form of Equation 2. It is therefore linear and can be solved using the integrating factor e (1 1 )dx x 15.6 Nonhomogeneous Linear Equations In this section we deal with second-order nonhomogeneous linear differential equations with constant coefficient : (1) ay by cy G (x) where a, b, 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 ) x ( cy ) x ( p y ) x ( y where y p is a particular solution of Equation 1 and y c is the general solution of the complementary Equation 2. Proof We want to verify that if y is any solution of Equation 1, then y y p is a solution of the complementary Equation 2. Indeed ) p y y (c ) p y y (b ) p y y ( a p yc yc pyb yb pya ya ) p yc pyb pya( ) yc yb ya( 0 )x ( G )x ( G We know from Section 15.5 that the solution of the complementary equation is 2y 2c 1y 1c cy , where 1y and 2y 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. 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 is 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) Case 1 G(x) is a polynomial It is reasonable to guess that the particular solution y p 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 y p (x) a polynomial (of the same degree as G ) into the differential equation and determine the coefficients. Example 1 Solve the equation 2 y y 2y x . Case 2 G(x) is of the form Ce kx , where C and k are constants. We take as a trial solution a function of the same form kx y Ae ` p , because the derivatives of e kx are constant kx e multiples of . Example 2 Case 3 3x Solve the equation y 4 y e . G(x) is either Ccoskx or Csinkx Because of the rules for differentiating the sine and cosine functions, we take as a trial particular solution a function of the form y p ( x) A cos kx B sin kx Example 3 Solve the equation y y 2 y sin x. Case 4 G(x) is a product of functions of the preceding types We take the trial solution to be a product of functions of the same type. For instance, in solving the differential equation y 2 y 4 y x cos 3 x we would try y p ( x) ( Ax B) cos 3x (Cx D) sin 3x Case 5 G(x) is a sum of functions of these types We use the easily verified principle of superposition, which says if y p1 and y p 2 are solutions of ay by cy G1 ( x) ay by cy G2 ( x) respectively, then y p1 y p 2 is a solution of ay by cy G1 ( x) G2 ( x) Example 4 Solve y 4 y xex cos 2 x. Finally we note that the recommended trial solution y p sometimes turns out to be a solution of the complementary equation and therefore cannot be a solution of the nonhomogeneous equation. In such cases we multiply the recommended trial solution by x (or x 2 if necessary) so that no term in y p (x) is a solution of the complementary equation. Example 5 Solve y y sin x. Summary of undetermined coefficient G(x) = First try yp Cn x n C1 x C0 An x n A1 x A0 Ce kx Aekx C cos kx D sin kx A cos kx B sin kx Modification: If any term of y p is a solution of the complementary equation, multiply y p by x (or x 2 if necessary) . •The method of variation of parameters Suppose we have already solved the homogeneous equation ay by cy 0 and written the solution as (4) )x ( 2y 2c )x ( 1y 1c )x (y where y1 and y 2 are linearly independent solutions. Let us replace the constants c1 and c2 in Equation 4 by arbitrary functions u1 ( x) and u2 ( x) . We look for a particular solution of the nonhomogeneous equation of the form ay by cy G (x) (5) )x ( 2y )x ( 2u )x ( 1y )x ( 1u )x ( p y Differentiating Equation 5, we get (6) ) 2y 2u 1y 1u( ) 2y 2u 1y 1u( py Since u1 and u2 are arbitrary functions, we can impose two conditions on them. One condition is that y p 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 us impose the condition that (7) 0 2y 2u 1y 1u Then 2y 2u 1y 1u 2y 2u 1y 1u py Substituting in the differential equation, we get G ) 2y 2u 1y 1u(c ) 2y 2u 1y 1u(b )2y 2u 1y 1u 2y 2u 1y 1u(a or (8) G ) 2y 2u 1y 1u(a ) 2yc 2yb 2ya( 2u ) 1yc 1yb 1ya( 1u But y1 and equation, so y 2 are solutions of the complementary ay1 by1 cy1 0 and and Equation 8 simplifies to (9) G ) 2y 2u 1y 1u(a ay2 by2 cy2 0 Equation 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 then the particular solution is given by Equation 5. Example 6 Solve the equation y y tan x, 0 x 2. 15.8 Series Solutions So far, the only second-order differential equations that we have been able to solve are linear equations with constant coefficients. Even a simple-looking equation like (1) y 2 xy y 0 is not easy to solve. But it is important to be able to solve equations such as Equation 1 since they arise from physical problems and, in particular, in connection with the Schrödinger equation(薛定谔方程) in quantum mechanics. In such a case, we use the method of power series; that is, we look for a solution of the form y f ( x) cn x n c0 c1 x c2 x 2 c3 x 3 n 0 The method is to substitute this expression into the differential equation and determine the values of the coefficients c0 , c1 , c2 Example 1 Use power series to solve the equation y y 0. Solution We assume there is a solution of the form (2) y f ( x) c0 c1 x c2 x 2 c3 x 3 cn x n n 0 By Theorem 10.9.2 we can differentiate power series term by term. So (3) y c1 2c2 x 3c3 x 2 ncn x n 1 n 1 y 2c2 2 3c3 x n(n 1)cn x n 2 n2 Rewrite y as y (n 2)( n 1)cn 2 x n n 0 In order to compare the expression for y and y more easily, we substituting these expressions into the differential equation, we obtain n ( n 2 )( n 1 ) c x c x n 0 n2 n n 0 n 0 or (5) n [( n 2 )( n 1 ) c c ] x 0 n2 n n 0 If two power series are equal, then the corresponding coefficients must be equal (from Section 10.10). Therefore, the coefficients of x n in Equation 5 must be 0: (6) (n 2)(n 1)cn 2 cn 0 cn cn 2 (n 1)( n 2) Equation 6 is called a recursion relation. If c0 and c1 are known, this equation allows us to determine the remaining coefficients recursively by putting n = 0, 1, 2, 3, …in succession. Put n 0: Put n 1: Put n 2: Put n 3: Put n 4: Put n 5: c0 c2 1 2 c1 c3 23 c0 c0 c2 c4 3 4 1 2 3 4 4! c3 c1 c1 c5 4 5 2 3 4 5 5! c0 c0 c4 c6 56 4! 5 6 6! c5 c1 c1 c7 67 5! 6 7 7! By now we see the pattern: For the even coefficients, For the odd coefficients, c0 (2n)! c1 n c2 n 1 (1) (2n 1)! c2 n (1) n Putting these values back into Equation 2, we write the solution as y c0 c1 x c2 x 2 c3 x 3 c4 x 4 c5 x 5 2n x2 x4 x6 x c0 (1 (1) n ) 2! 4! 6! (2n)! 2 n 1 x3 x5 x7 x c1 ( x (1) n ) 3! 5! 7! (2n 1)! 2n 2 n 1 x x c0 (1) n c1 (1) n (2n)! (2n 1)! n 0 n 0 Notice that there are two arbitrary constants c0 and c1 , which is to be expected. Note 1: We recognize the series obtained in Example 1 as being the Maclaurin series for cosx and sinx. (See Equation 10.10.16 and 10.10.15.) Therefore, we could write the solution as y( x) c0 cos x c1 sin x But we are not usually able to express power series solutions of differential equations in terms of known functions. Example 2 y 2 xy y 0. Solve Solution We assume there is a solution of the form y f ( x ) cn x n n 0 Then y ncn x n 1 n 1 and y n(n 1)cn x n2 n2 (n 2)( n 1)cn 2 x n n 0 Substituting in the differential equation, we get (n 2)(n 1)cn2 x 2 x ncn x n n 0 n 1 n 1 cn x n 0 n 0 n ( n 2 )( n 1 ) c x 2 nc x c x n n 0 n2 n n 0 n n 1 n 0 n [( n 2 )( n 1 ) c ( 2 n 1 ) c ] x 0 n2 n n 0 This is true if the coefficient of xn is 0: (n 2)( n 1)cn 2 (2n 1)cn 0 (7) cn 2 2n 1 cn (n 1)( n 2) n 0, 1, 2, 3, We solve this recursion relation by putting n = 0, 1, 2, 3, … successively in Equation 7: Put n 0: Put n 1: Put n 2: Put n 3: 1 c2 c0 1 2 1 c3 c1 23 3 3 3 c4 c2 c0 c0 3 4 1 2 3 4 4! 5 1 5 1 5 c5 c3 c1 c1 45 2 3 4 5 5! Put Put Put Put 7 3 7 3 7 n 4 : c6 c4 c0 c0 56 4! 5 6 6! 9 1 5 9 1 5 9 n 5 : c7 c5 c1 c1 67 5! 6 7 7! 11 3 7 11 n 6 : c8 c6 c0 7 8 8! 13 1 5 9 13 n 7 : c9 c7 c1 89 9! In general, the even coefficients are given by c2 n 3 7 11 (4n 5) c0 (2n)! and the odd coefficients are given by 1 5 9 (4n 3) c2 n 1 c1 (2n 1)! The solution is y c0 c1 x c2 x 2 c3 x 3 c4 x 4 1 2 3 4 3 7 6 3 7 11 8 c0 (1 x x x x ) 2! 4! 6! 8! 1 3 1 5 5 1 5 9 7 1 5 9 13 9 c1 ( x x x x x ) 3! 5! 7! 9! Or (8) 1 2 3 7 (4n 5) 2 n y c0 (1 x x ) 2! (2n)! n2 1 5 9 (4n 3) 2 n 1 c1 ( x x ) (2n 1)! n 1 Note 2: In Example 2 we had to assume that the differential equation had a series solution. But now we could verify directly that the function given by Equation 8 is indeed a solution. Note 3: Unlike the situation of Example 1, the power series that arise in the solution of Example 2 do not define elementary functions. The functions and 1 2 3 7 (4n 5) 2 n y1 ( x) 1 x x 2! (2n)! n2 1 5 9 (4n 3) 2 n 1 y2 ( x) x x (2n 1)! n 1 are perfectly good functions but they cannot be expressed in terms of familiar functions. We can use these power series expressions for y1 and y2 to compute approximate values of the functions and even to graph them. Note 4: If we are asked to solve the initial-value problem y 2 xy y 0 y (0) 0 y(0) 1 we would observe from Theorem 10.10.5 that c0 y(0) 0 c1 y(0) 1 This would simplify the calculations in Example 2, since all of the even coefficients would be 0. The solution to the initial-value problem is 1 5 9 (4n 3) 2 n 1 y ( x) x x (2n 1)! n 1