475 answers 4 - ANSWERS PAMPHLET 151 2 c1 b y y 2 k1 24.15...

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ANSWERS PAMPHLET 151 2 5 y (0) 5 c 1 ; 1 5 ˙ y (0) 5 2 c 1 1 3 c 2 . c 1 5 2 , c 2 5 1, so y 5 e 2 t (2 cos 3 t 1 sin 3 t ). b ) ¨ y 1 9 y 5 0; r 2 1 9 5 0 , r 5 6 3 i . y 5 k 1 cos 3 t 1 k 2 sin 3 t , ˙ y 5 2 3 k 1 sin 3 t 1 3 k 2 cos 3 t . 2 5 y (0) 5 k 1 , 1 5 ˙ y (0) 5 3 k 2 . k 1 5 2 , k 2 5 1 3 . y 5 2 cos 3 t 1 1 3 sin 3 t . 24.15 y 5 e a t ( c 1 cos b t 1 c 2 sin b t ) , ˙ y 5 e a t [( a c 1 1 b c 2 ) cos b t 1 ( a c 2 2 b c 1 ) sin b t ] , ¨ y 5 e a t [( a 2 2 b 2 ) c 1 1 2 ab c 2 ] cos b t 1 e a t [( a 2 2 b 2 ) c 2 2 2 ab c 1 ] sin b t. a ¨ y 1 b ˙ y 1 cy 5 e a t cos b t [ a ( a 2 2 b 2 ) c 1 1 2 a ab c 2 1 b a c 1 1 b b c 2 1 cc 1 ] 1 e a t sin b t [ a ( a 2 2 b 2 ) c 2 2 2 a ab c 1 1 b a c 2 2 b b c 1 1 cc 2 ] 5 e a t cos b t [ c 1 ( a ( a 2 2 b 2 ) 1 b a 1 c ) 1 c 2 (2 a ab 1 b b )] 1 e a t sin b t [ 2 c 1 (2 a ab 1 b b ) 1 c 2 ( a ( a 2 2 b 2 ) 1 b a 1 c )] . On the other hand, since r 5 a 1 i b is a root of ar 2 1 br 1 c 5 0, 0 5 a ( a 1 i b ) 2 1 b ( a 1 i b ) 1 c 5 [ a ( a 2 2 b 2 ) 1 b a 1 c ] 1 (2 a ab 1 b b ) . So a ( a 2 2 b 2 ) 1 b a 1 c 5 0 and 2 a ab 1 b b 5 0. But these are precisely the coefficients of c 1 and c 2 in the previous expres- sion, and so that expression equals zero. 24.16 a ) 6¨ y 2 ˙ y 2 y 5 0, 6 r 2 2 r 2 1 5 (3 r 1 1)(2 r 2 1) 5 0. y 5 k 1 e 2 t 6 3 1 k 2 e t 6 2 , ˙ y 5 2 1 3 k 1 e 2 t 6 3 1 1 2 k 2 e t 6 2 . 1 5 y (0) 5 k 1 1 k 2 , 0 5 ˙ y (0) 5 2 1 3 k 1 1 1 2 k 2 . So k 1 5 3 5 , k 2 5 2 5 ; and y 5 3 5 e 2 t 6 3 1 2 5 e t 6 2 . b ) ¨ y 1 y 1 2 y 5 0, r 2 1 2 r 1 2 5 0, r 5 2 1 6 i . y 5 e 2 t ( k 1 cos t 1 k 2 sin t ), ˙ y 5 e 2 t [( k 2 2 k 1 ) cos t 2 ( k 1 1 k 2 ) sin t ]. 1 5 y (0) 5 k 1 , 0 5 ˙ y (0) 5 k 2 2 k 1 . So k 1 5 k 2 5 1. y 5 e 2 t (cos t 1 sin t ).
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152 MATHEMATICS FOR ECONOMISTS c ) 4¨ y 2 y 1 y 5 0, 4 r 2 2 4 r 1 1 5 (2 r 2 1) 2 5 0. y 5 k 1 e t 6 2 1 k 2 te t 6 2 , ˙ y 5 1 2 k 1 e t 6 2 1 k 2 e t 6 2 1 1 2 k 2 te t 6 2 . 1 5 y (0) 5 k 1 , 0 5 ˙ y (0) 5 1 2 k 1 1 k 2 = k 1 5 1 , k 2 5 2 1 2 . y ( t ) 5 e t 6 2 2 1 2 te t 6 2 . d ) ¨ y 1 y 1 6 y 5 0, r 2 1 5 r 1 6 5 ( r 1 3)( r 1 2) 5 0. y 5 k 1 e 2 3 t 1 k 2 e 2 2 t , ˙ y 5 2 3 k 1 e 2 3 t 2 2 k 2 e 2 2 t . 1 5 y (0) 5 k 1 1 k 2 , 0 5 ˙ y (0) 5 2 3 k 1 2 2 k 2 = k 1 5 2 2 , k 2 5 3. y 5 2 2 e 2 3 t 1 3 e 2 2 t . e ) ¨ y 2 y 1 9 y 5 0, r 2 2 6 r 1 9 5 ( r 2 3) 2 5 0. y 5 k 1 e 3 t 1 k 2 te 3 t , ˙ y 5 3 k 1 e 3 t 1 k 2 e 3 t 1 3 k 2 te 3 t . 1 5 y (0) 5 k 1 , 0 5 ˙ y (0) 5 3 k 1 1 k 2 = k 1 5 1 , k 2 5 2 3. y 5 e 3 t 2 3 te 3 t . f ) ¨ y 1 ˙ y 1 y 5 0, r 2 1 r 1 1 5 0, r 5 2 1 2 6 i p 3 2 . y 5 e 2 t 6 2 ˆ k 1 cos p 3 2 t 1 k 2 sin p 3 2 t ! ˙ y 5 e 2 t 6 2 ˆ 2 k 1 2 cos p 3 2 t 1 k 2 p 3 2 cos p 3 2 t 2 k 2 2 sin p 3 2 t 2 k 1 p 3 2 sin p 3 2 t ! . 1 5 y (0) 5 k 1 , 0 5 ˙ y (0) 5 2 k 1 2 1 k 2 p 3 2 = k 1 5 1 , k 2 5 1 6 p 3. y 5 e 2 t 6 2 ˆ cos p 3 2 1 1 p 3 sin p 3 2 t ! . 24.17 y [3] 2 y 2 ˙ y 1 2 y 5 0. Look for the solution y 5 e rt . e rt ( r 3 2 2 r 2 2 r 1 2) 5 0 . ( r 2 1)( r 1 1)( r 2 2) 5 0 = r 5 1 , 2 1 , 2 . y 5 k 1 e t 1 k 2 e 2 t 1 k 3 e 2 t .
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