Ma2aPracLectures19-21

Ma2aPracLectures19-21 - Lecture 19 Linear ODEs of higher...

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Unformatted text preview: Lecture 19 Linear ODEs of higher order Solutions can be obtained by solving linear system of first order ODEs. Idea : Replace one equation by many equations, thereby reducing the order of the ODE to just first order. n 1: A n-th 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 coecients a j could depend on x . We say that ( ) has constant coecients 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 n-space 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 coecients, a solution matrix is given by ( 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 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 ( x ) = e Ax . If ( 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 coecients, 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 1 , 2 ,. .. n with corresponding eigenvectors v 1 , v 2 ,. .. v n , then a basis of solution for ( ) is given by y (1) = v 1 e t , y (2) = v 2 e 2 t , .. . , y ( n ) = v n e t . 2 If v ( j ) = v ( j ) 1 . . . v ( j ) n , then a basis of solutions of the n-th order ODE ( ) is given by v (1) 1 e 1 t , v (2) 1 e 2 t , . . . , v ( n ) 1 e n t Useful fact 2 : Recall that The eigenvalues of A are determined by solving the characteristic equation: | A | = 0. n = 2: A = ( 1 a 2 a 2 ) , A = ( t a 2 + a 1 ) | A | = 2 + a 1 + a 2 General formula : | A | = 2 = n + a 1 n 1 + + a n 1 + 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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Ma2aPracLectures19-21 - Lecture 19 Linear ODEs of higher...

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