Then u01 u02 u03 u04 u2 a2 x u 1 u4 b2 x u3 a1

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Unformatted text preview: g (x) Let u1 = y, u2 = y 0 , u3 = z, u4 = z 0 . Then u01 u02 u03 u04 = u2 = a2 ( x ) u 1 = u4 = b2 ( x ) u3 a1 ( x ) u 2 + f ( x ) b1 ( x ) u4 + g ( x ) , or u0 = Au + F(x) where A= 0 a2 (x) 0 0 1 a1 ( x ) 0 0 0 0 0 b2 ( x ) 0 0 1 b1 ( x ) , F( x ) = 0 f ( x) 0 g ( x) . 2. Consider 3. Consider i.e., x 1e v + 1 y1 + 2 y2 = 0 e x ( u + xv ) = 0 2 For x = 0 we have 1 v + 2 u = 0. But this implies 1 = 2 = 0 since v and u are linearly independent vectors. Thus the only 2x choice of 3x , 2 that works for all x IR is 1 = 2 = 0, 1 (d) On(x ) and W2(y1 , are linearly independent 3x IR. = 0. 2x i.e., On x 0,, W (y) ,y2 ) = det 2 cos(x) 3 cos(x) = 0. y1 x 0 y (x y ) = det on (d) 1 2 2 cos(x) 3 cos(x) x 3x 3. Consider 1 y1 + 2 y2 = 0 ( ). On x < 0, W (y1 , y2 ) = det = 0. x 3x cos(x) 3 cos(x) = 0. On x < 0, W (y1 , y ) = det (a) On x 0 equation2 ( ) becomes x) 3 cos(x) cos( Thus R. 4. Solution:W (y11,,y22) = 0 for all x IIR. Thus W (y y ) = 0 for all x 2x 3x 1 4 + = 0. 4. First find general solut...
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This document was uploaded on 02/09/2014.

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