(2) - Assignment Solutions

(2) - Assignment Solutions - Written Assignment # 2 and the...

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Written Assignment # 2 and the solutions (Math-315, 2010 ) (1) (*) Given the following inhomogeneous equation: P ( D )=( D + 2)( D 2 + 4) y = h ( x ) = 2e - 2 x + x sin 2 x. (a) Find the annihilator for the above inhomogeneous term. (b) Determine the form of a particular solution and derive its ±nal expression. (c) Find the general solution for the equation. Solution: (a) Let y P = y p 1 + y p 2 , and P ( D ) y p 1 = h 1 ( x ) = 2e - 2 x , P ( D ) y p 2 = h 2 ( x )= x sin 2 x. Then, the annihilator for h 1 ( x )is Q 1 ( D D + 2); the annihilator for h 2 ( x Q 2 ( D D 2 + 4) 2 ; the annihilator for h ( x Q ( D D + 2)( D 2 + 4) 2 . (b) Particular solution y p 1 is subject to the homogeneous equation: Q 1 ( D ) P ( D )[ y p 1 ]=( D + 2) 2 ( D 2 + 4)[ y p 1 ]=0 . Hence we have y p 1 = C 1 e - 2 x + C 2 x e - 2 x + C 3 sin 2 x + C 4 cos 2 x One may set C 1 = C 3 = C 4 = 0, since P ( D )[ C 1 e - 2 x + C 3 sin 2 x + C 4 cos 2 x ] = 0. To ±nd the constant C 2 , we substitute y p 1 = C 2 x e - 2 x into the inhomogeneous equation. It follows that P ( D ) y p 1 =( D + 2)( D 2 + 4) y p 1 D + 2)[8 C 2 x e - 2 x - 4 C 2 e - 2 x ]=8 C 2 e - 2 x = 2e - 2 x . We derive C 2 = 1 4 . So that, y p 1 ( x 1 4 x e - 2 x . (c) Furthermore, the particular solution y p 2 is subject to the homogeneous equation: Q 2 ( D ) P ( D )[ y p 2 D + 2)( D 2 + 4) 3 [ y p 2 . Hence we have y p 2 = d 1 e - 2 x + d 2 sin 2 x + d 3 cos 2 x + d 4 x sin 2 x + d 5 x cos 2 x + d 6 x 2 sin 2 x + d 7 x 2 cos 2 x One may set d 1 = d 2 = d 3 = 0, since P ( D )[ d 1 e - 2 x + d 2 sin 2 x + d 3 cos 2 x ] = 0. To ±nd the other constants, we substitute y p 2 = d 4 x sin 2 x + d 5 x cos 2 x + d 6 x 2 sin 2 x + d 7 x 2 cos 2 x 1
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2 into the relative inhomogeneous equation: P ( D ) y p 2 =( D + 2)( D 2 + 4) y p 2 = h ( x ) . It is calculated that ( D + 2)( D 2 + 4) ± d 4 x sin 2 x ² D + 2) ± 4 d 4 cos 2 x ² =8 d 4 cos 2 x - 8 d 4 sin 2 x ; ( D + 2)( D 2 + 4) ± d 5 x cos 2 x ² D + 2) ± - 4 d 5 sin 2 x ² = - 8 d 5 cos 2 x - 8 d 5 sin 2 x ; ( D + 2)( D 2 + 4) ± d 6 x 2 sin 2 x ² D + 2) ± 2 d 6 sin 2 x +8 d 6 x cos 2 x ² = 12 d 6 cos 2 x +4 d 6 sin 2 x + 16 d 6 x cos 2 x - 16 d 6 x sin 2 x ; ( D + 2)( D 2 + 4) ± d 7 x 2 cos 2 x ² D + 2) ± 2 d 7 cos 2 x - 8 d 7 x sin 2 x ² = - 12 d 7 sin 2 x d 7 cos 2 x - 16 d 7 x cos 2 x - 16 d 6 x sin 2 x ; It follows that - 16( d 6 + d 7 ) x sin 2 x + 16( d 6 - d 7 ) x cos 2 x + ± 8 d 4 - 8 d 5 + 12 d 6 d 7 ² cos 2 x + ± - 8 d 4 - 8 d 5 d 6 - 12 d 7 ² sin 2 x = h 1 ( x )= x sin 2 x. We Fnd that d 4 = 4 64 ,d 5 = - 1 64 6 = d 7 = - 1 32 . We Fnally obtain y P = y p 1 + y p 2 = 1 4 x e - 2 x + 1 63 x (3 sin 2 x - cos 2 x ) - 1 32 x 2 (sin 2 x + cos 2 x ) . (d) ±ind the general solution for the equation. The associated homogeneous equation P ( D )[ y H ] = 0, has general solution y H = A 1 e - 2 x + B 1 cos 2 x + C 1 sin 2 x.
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(2) - Assignment Solutions - Written Assignment # 2 and the...

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