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Unformatted text preview: Lecture 19 Linear ODE’s of higher order Solutions can be obtained by solving linear system of first order ODE’s. Idea : Replace one equation by many equations, thereby reducing the order of the ODE to just first order. n ≥ 1: A nth order linear ODE is of the form d n u dx n + a 1 d n − 1 u dx n − 1 + a 2 d n − 2 u dx n − 2 + ··· + a n − 1 du dx + a n u = f ( x ) (*) Here the coeﬃcients a j could depend on x . We say that ( ∗ ) has constant coeﬃcients iff each a j is independent of x . ( ∗ ) is homogeneous if f ( x ) = 0. First, look at the homogeneous case: Define a vector y = y 1 y 2 . . . y n in nspace where y 1 = u, y 2 = du dx , y 3 = d 2 u dx 2 , . . ., y n = d n − 1 u dx n − 1 . Then ( ∗ ) can be rewritten suggestively as dy 1 dx = y 2 dy 2 dx 2 = y 3 . . . dy n − 1 dx n − 1 = y n dy n dx n = − a n y 1 − a n − 1 y 2 .. . . . . − a 1 y n This can be viewed as a matrix of linear equations, and so gives rise to the 1 linear system: d dx y = A y , where A = 1 0 1 · · · · 1 − a n · · · − a 2 − a 1 (**) Whether or not A has constant coeﬃcients, a solution matrix is given by Φ( x ) = e ∫ x A ( t ) dt , as long as A is integrable as a function of x , which is the case, for example, when A is continuous. (If A is not defined at 0, then we can take some other lower limit where A is defined; the effect of changing the lower limit is to multiply Φ by a constant.) Of course, when A is a constant matrix , ∫ x A ( t ) dt is just Ax , and we get the more familiar expression Φ( x ) = e Ax . If Φ( x ) = ( y (1) y (2) .. . y ( n ) ) , then the y ( j ) are the different solution vec tors: y ( j ) = y ( j ) 1 y ( j ) 1 . . . y ( j ) n What we seek is a set of independent solutions to the original ODE ( ∗ ). If A has constant coeﬃcients, then for every j ≤ n , y ( j ) 2 = dy ( j ) 1 dx ,.. . ,y ( j ) n = dy ( j ) n − 1 dx . Useful fact 1 : If A is a constant matrix with n distinct eigenvalues λ 1 ,λ 2 ,. .. λ n with corresponding eigenvectors v 1 , v 2 ,. .. v n , then a basis of solution for ( ∗∗ ) is given by y (1) = v 1 e λt , y (2) = v 2 e λ 2 t , .. . , y ( n ) = v n e λt . 2 If v ( j ) = v ( j ) 1 . . . v ( j ) n , then a basis of solutions of the nth order ODE ( ∗ ) is given by v (1) 1 e λ 1 t , v (2) 1 e λ 2 t , . . . , v ( n ) 1 e λ n t Useful fact 2 : Recall that The eigenvalues of A are determined by solving the characteristic equation:  λ − A  = 0. n = 2: A = ( 1 − a 2 − a 2 ) , λ − A = ( λ − t a 2 λ + a 1 ) ⇒  λ − A  = λ 2 + a 1 λ + a 2 General formula :  λ − A  = λ 2 = λ n + a 1 λ n − 1 + ··· + a n − 1 λ + a n As we started with the n th order linear, homogeneous ODE: d 2 y dx + a 1 d n − 1 u dx n − 1 + .. .a n − 1 du dx + a n u = 0 , we can read off the characteristic equation directly from the ODE.we can read off the characteristic equation directly from the ODE....
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 Fall '08
 Makarov,N
 Equations, Ode, Trigraph, Complex number

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