5x 0 6 first nd general solution of associated

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Unformatted text preview: First find general solution of associated homogeneous equation: y0 = Ay where A = Characteristic equation is: det 5x 5x . 2 1 + 4 + 3 = 0. 1, 3 1 11 v1 0 1 2 1 Eigenvector for = 1: = ) v2 = v1 . Choose v = 2 1 1 v2 0 1 1 v1 = 0 1 Eigenvector for = 3: 1 1 ) v2 = v1 . Choose v = 2 1 2 11 v2 0 1 General solution of associated homogeneous equation: 1 +c e 2 2 1 = c1 y1 + c2 y2 . 11 22 x x yh (x) = c1 e h 1 1 2 . . 1 1 3x 3x Apply Variation of Parameters to find general solution: y(x) = u1 y1 + u2 y2 where 11 22 x x 4xe 0 0 0 u01 y1 + u02 y2 = 11 22 0 u01 e 1 0 u01 e 1 ) x x x x 0 + u02 e 2 0 u02 e 2 3x 3x 3x 3x = = x x 4xe 0 . Solving gives 0 u01 = 2x 1 0 2x u02 = 2xe2x . 2 Integrating gives 2 u1 = x2 + c1 1 1 2x 2 u2 = xe2x 1 2x 2x 2 e + c2 . 2 General solution: 2 1 y ( x ) = (x 2 + c 1 ) e = c1 e 1 1 1 x x = 1 1 x x 1 2x 3x 1 2x 2 + c 2 ) e 3x 2e 2 2x + (xe2x + c2 e 2 1 1 3x 3x h y h ( x) 3. x (t) = Ax + f , where A = 0 Solution: ✓ 4 2 + + 2 1 3 3 ◆ 1 1 1 x 2 ( x2 + x 1 ) e x 2 21 2 x 1 2 (x x + 2 )e x 2 p y p ( x) and f (t) = ✓ 1/t 4 + 2/t ◆ We first need to find the solution to the corresponding homogeneous equation: 4 2 A I= , 2 1 so det [A I] = ( 4 )( 1 characteristic equation are 0 and ) 4= 5. 2 + 5 , and hence the roots of the To...
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This document was uploaded on 02/09/2014.

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