Higher Order Linear Differential Equations Familiar stuff Example - PowerPoint PPT Presentation

Higher Order Linear Differential Equations Math 240 Linear DE Linear differential operators Higher Order Linear Differential Equations Familiar stuff Example Homogeneous equations Math 240 — Calculus III Summer 2015, Session II Tuesday, July 28, 2015

Higher Order Agenda Linear Differential Equations Math 240 Linear DE Linear differential operators Familiar stuff 1. Linear differential equations of order n Example Linear differential operators Homogeneous equations Familiar stuff An example 2. Homogeneous constant-coefficient linear differential equations

Higher Order Introduction Linear Differential Equations Math 240 Linear DE Linear We now turn our attention to solving linear differential differential operators equations of order n . The general form of such an equation is Familiar stuff Example a 0 ( x ) y ( n ) + a 1 ( x ) y ( n − 1) + · · · + a n − 1 ( x ) y ′ + a n ( x ) y = F ( x ) , Homogeneous equations where a 0 , a 1 , . . . , a n , and F are functions defined on an interval I . The general strategy is to reformulate the above equation as Ly = F, where L is an appropriate linear transformation. In fact, L will be a linear differential operator .

Higher Order Linear differential operators Linear Differential Equations Recall that the mapping D : C k ( I ) → C k − 1 ( I ) defined by Math 240 D ( f ) = f ′ is a linear transformation. This D is called the Linear DE derivative operator. Higher order derivative operators Linear differential D k : C k ( I ) → C 0 ( I ) are defined by composition: operators Familiar stuff D k = D ◦ D k − 1 , Example Homogeneous equations so that D k ( f ) = d k f dx k . A linear differential operator of order n is a linear combination of derivative operators of order up to n , L = D n + a 1 D n − 1 + · · · + a n − 1 D + a n , defined by Ly = y ( n ) + a 1 y ( n − 1) + · · · + a n − 1 y ′ + a n y, where the a i are continous functions of x . L is then a linear transformation L : C n ( I ) → C 0 ( I ) . (Why?)

Higher Order Examples Linear Differential Equations Math 240 Example Linear DE Linear If L = D 2 + 4 xD − 3 x , then differential operators Familiar stuff Ly = y ′′ + 4 xy ′ − 3 xy. Example Homogeneous We have equations L (sin x ) = − sin x + 4 x cos x − 3 x sin x, � x 2 � = 2 + 8 x 2 − 3 x 3 . L Example If L = D 2 − e 3 x D, determine � 2 x − 3 e 2 x � = − 12 e 2 x − 2 e 3 x + 6 e 5 x 1. L � � 3 sin 2 x = − 3 e 3 x sin 2 x − 6 cos 2 x 2. L

Higher Order Homogeneous and nonhomogeneous equations Linear Differential Equations Math 240 Consider the general n -th order linear differential equation Linear DE a 0 ( x ) y ( n ) + a 1 ( x ) y ( n − 1) + · · · + a n − 1 ( x ) y ′ + a n ( x ) y = F ( x ) , Linear differential operators Familiar stuff where a 0 � = 0 and a 0 , a 1 , . . . , a n , and F are functions on an Example interval I . Homogeneous equations If a 0 ( x ) is nonzero on I , then we may divide by it and relabel, obtaining y ( n ) + a 1 ( x ) y ( n − 1) + · · · + a n − 1 ( x ) y ′ + a n ( x ) y = F ( x ) , which we rewrite as Ly = F ( x ) , where L = D n + a 1 D n − 1 + · · · + a n − 1 D + a n . If F ( x ) is identically zero on I , then the equation is homogeneous , otherwise it is nonhomogeneous .

Higher Order The general solution Linear Differential Equations Math 240 If we have a homogeneous linear differential equation Linear DE Ly = 0 , Linear differential operators its solution set will coincide with Ker( L ) . In particular, the Familiar stuff kernel of a linear transformation is a subspace of its domain. Example Homogeneous equations Theorem The set of solutions to a linear differential equation of order n is a subspace of C n ( I ) . It is called the solution space . The dimension of the solutions space is n . Being a vector space, the solution space has a basis { y 1 ( x ) , y 2 ( x ) , . . . , y n ( x ) } consisting of n solutions. Any element of the vector space can be written as a linear combination of basis vectors y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) + · · · + c n y n ( x ) . This expression is called the general solution .

Higher Order The Wronskian Linear Differential Equations Math 240 We can use the Wronskian � � y 1 ( x ) y 2 ( x ) · · · y n ( x ) Linear DE � � Linear � � y ′ y ′ y ′ 1 ( x ) 2 ( x ) · · · n ( x ) differential � � operators � � W [ y 1 , y 2 , . . . , y n ]( x ) = . . . ... Familiar stuff . . . � � . . . Example � � � � Homogeneous y ( n − 1) y ( n − 1) y ( n − 1) ( x ) ( x ) · · · ( x ) � � equations n 1 2 to determine whether a set of solutions is linearly independent. Theorem Let y 1 , y 2 , . . . , y n be solutions to the n -th order differential equation Ly = 0 whose coefficients are continuous on I . If W [ y 1 , y 2 , . . . , y n ]( x ) = 0 at any single point x ∈ I , then { y 1 , y 2 , . . . , y n } is linearly dependent. To summarize, the vanishing or nonvanishing of the Wronskian on an interval completely characterizes the linear dependence or independence of a set of solutions to Ly = 0 .

Higher Order The Wronskian Linear Differential Equations Math 240 Linear DE Linear differential Example operators Familiar stuff Verify that y 1 ( x ) = cos 2 x and y 2 ( x ) = 3 − 6 sin 2 x are Example solutions to the differential equation y ′′ + 4 y = 0 on ( −∞ , ∞ ) . Homogeneous equations Determine whether they are linearly independent on this interval. � � 3 − 6 sin 2 x cos 2 x � � W [ y 1 , y 2 ]( x ) = � � − 2 sin 2 x − 12 sin x cos x � � = − 6 sin 2 x cos 2 x + 6 sin 2 x cos 2 x = 0 They are linearly dependent. In fact, 3 y 1 − y 2 = 0 .

Higher Order Nonhomogeneous equations Linear Differential Equations Consider the nonhomogeneous linear differential equation Math 240 Ly = F . The associated homogeneous equation is Ly = 0 . Linear DE Linear Theorem differential operators Suppose { y 1 , y 2 , . . . , y n } are n linearly independent solutions to Familiar stuff Example the n -th order equation Ly = 0 on an interval I , and y = y p is Homogeneous equations any particular solution to Ly = F on I . Then every solution to Ly = F on I is of the form y = � c 1 y 1 + c 2 y 2 + · · · + c n y n + y p , �� � = + y p y c for appropriate constants c 1 , c 2 , . . . , c n . This expression is the general solution to Ly = F . The components of the general solution are ◮ the complementary function , y c , which is the general solution to the associated homogeneous equation, ◮ the particular solution , y p .

Higher Order Something slightly new Linear Differential Equations Math 240 Linear DE Linear differential operators Familiar stuff Example Theorem Homogeneous If y = u p and y = v p are particular solutions to Ly = f ( x ) and equations Ly = g ( x ) , respectively, then y = u p + v p is a solution to Ly = f ( x ) + g ( x ) . Proof. We have L ( u p + v p ) = L ( u p ) + L ( v p ) = f ( x ) + g ( x ) . Q . E . D .

Higher Order An example Linear Differential Equations Math 240 Example Linear DE Determine all solutions to the differential equation Linear differential y ′′ + y ′ − 6 y = 0 of the form y ( x ) = e rx , where r is a constant. operators Familiar stuff Example Homogeneous Substituting y ( x ) = e rx into the equation yields equations e rx ( r 2 + r − 6) = r 2 e rx + re rx − 6 e rx = 0 . Since e rx � = 0 , we just need ( r + 3)( r − 2) = 0 . Hence, the two solutions of this form are y 1 ( x ) = e 2 x y 2 ( x ) = e − 3 x . and Could this be a basis for the solution space? Check linear independence. Yes! The general solution is y ( x ) = c 1 e 2 x + c 2 e − 3 x .

Higher Order An example Linear Differential Equations Math 240 Example Linear DE Determine the general solution to the differential equation Linear differential operators y ′′ + y ′ − 6 y = 8 e 5 x . Familiar stuff Example Homogeneous equations We know the complementary function, y c ( x ) = c 1 e 2 x + c 2 e − 3 x . For the particular solution, we might guess something of the form y p ( x ) = ce 5 x . What should c be? We want 8 e 5 x = y ′′ p + y ′ p − 6 y p = (25 c + 5 c − 6 c ) e 5 x . Cancel e 5 x and then solve 8 = 24 c to find c = 1 3 . The general solution is y ( x ) = c 1 e 2 x + c 2 e − 3 x + 1 3 e 5 x .

Higher Order Introduction Linear Differential Equations Math 240 Linear DE Linear We just found solutions to the linear differential equation differential operators y ′′ + y ′ − 6 y = 0 Familiar stuff Example Homogeneous of the form y ( x ) = e rx . In fact, we found all solutions. equations This technique will often work. If y ( x ) = e rx then y ′ ( x ) = re rx , y ′′ ( x ) = r 2 e rx , y ( n ) ( x ) = r n e rx . . . . , So if r n + a 1 r n − 1 + · · · + a n − 1 r + a n = 0 then y ( x ) = e rx is a solution to the linear differential equation y ( n ) + a 1 y ( n − 1) + · · · + a n − 1 y ′ + a n y = 0 . Let’s develop this approach more rigorously.