L97V - Fall 2003 Math 308/501502 9 Matrix Methods for...

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Fall 2003 Math 308/501–502 9 Matrix Methods for Linear Systems 9.7V Nonhomogeneous Lin Sys: Variation of Parameters Wed, 26/Nov c ± 2003, Art Belmonte Summary Definitions Let x 1 ,..., x n be a fundamental set of solutions to x 0 = Ax .The matrix X = [ x 1 x n ] whose columns are the x k is called a fundamental matrix. The system x 0 = Ax + f is a nonhomogeneous linear system . Here A = A ( t ) is a coefficient matrix that may (in general) depend on the independent variable t and f = f ( t ) is a vector-valued function of t . In the following c = [ c 1 ; c 2 ; ... ; c n ] is a constant column vector. Solutions of a nonhomogeneous linear system Theorem If x p is a particular solution of the nonhomogeneous system x 0 = Ax + f and X is a fundamental matrix for the corresponding homogeneous system x 0 = Ax ,thena general solution of the nonhomogeneous solution is given by x = x p + x h = x p + Xc = x p + n X k = 1 c k x k . Variation of parameters for systems A general solution of x 0 = Ax + f (a nonhomogeneous first-order linear system in normal form! ) is obtained as follows. 1. Find a fundamental set of solutions, X ,of x 0 = Ax . 2. Solve the linear system Xv p = f to obtain v p = v 0 ,the derivative of v , the variable parameter. 3. Integrate to obtain the variable parameter, v = R v p dt. 4. Construct a particular solution, x p = Xv . 5. Form a general solution, x = x p + x h = x p + Xc , where c = [ c 1 ; c 2 ; ; c n ] , a constant column vector. Some formulas x p = X ( t ) v ( t ) = X ( t ) Z X ( t ) - 1 f ( t ) dt x p = X ( t ) Z t t 0 X (w) - 1 f (w) d w (solution with x p ( t 0 ) = 0 ) x ( t ) = X ( t ) ± X ( t 0 ) - 1 x 0 + Z t t 0 X (w) - 1 f (w) d w ² (solution of the IVP x 0 = Ax + f , x ( t 0 ) = x 0 ) “Hand” Examples Example A Find a general solution to x 0 = Ax + f ,where A = ³ - 31 0 - 38 ´ and f = ³ 3 4 ´ . Solution 1. Find a fundamental set of solutions. Eigenvalues: r k = 2 , 3. Eigenvectors: v k = [2 ; 1] , [5 ; 3]. Fundamental set of solutions: x 1 = e 2 t [2 ; 1] , x 2 = e 3 t [5 ; 3].
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This note was uploaded on 03/29/2008 for the course MATH 308 taught by Professor Comech during the Spring '08 term at Texas A&M.

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L97V - Fall 2003 Math 308/501502 9 Matrix Methods for...

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