WrittenHW06-Solutions

WrittenHW06-Solutions - Math 216 (Section 50) Written...

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Unformatted text preview: Math 216 (Section 50) Written Homework #6 Solutions Fall 2007 1 ( 3.51 ) Let us assume y p ( x ) = Ae 3 x . Plugging this into the equation, we have 9 Ae 3 x + 16 Ae 3 x = e 3 x = 25 Ae 3 x = e 3 x = 25 A = 1 = A = 1 25 . Therefore y p ( x ) = 1 25 e 3 x . 2 ( 3.53 ) Let us assume y p ( x ) = A cos(3 x ) + B sin(3 x ) . Plugging this into the equation, we have [- 9 A cos(3 x )- 9 B sin(3 x )]- [- 3 A sin(3 x ) + 3 B cos(3 x )]- 6 [ A cos(3 x ) + B sin(3 x )] = 2 sin(3 x ) = (- 15 A- 3 B ) cos(3 x ) + (3 A- 15 B ) sin(3 x ) = 2 sin(3 x ) = - 15 A- 3 B = 0 and 3 A- 15 B = 2 Solving them gives A = 1 39 , B =- 5 39 . Therefore y p ( x ) = 1 39 cos(3 x )- 5 39 sin(3 x ) = 1 39 [cos(3 x )- 5 sin(3 x )] 3 ( 3.55 ) Since sin 2 x = 1- cos 2 x 2 = 1 2- 1 2 cos 2 x. We can assume y p ( x ) = A cos 2 x + B sin2 x + C. Plugging this into the equation, we have- 4 A cos 2 x- 4 B sin2 x- 2 A sin2 x + 2 B cos 2 x + A cos 2 x + B sin2 x + C = 1 2- 1 2 cos 2 x = (- 3 A + 2 B ) cos 2 x + (- 2 A- 3 B ) sin2 x + C = 1 2- 1 2 cos 2 x = - 3 A + 2 B =- 1 2 ,- 2 A- 3 B = 0 , C = 1 2 . Solving the equations for A and B gives A = 3 26 , B =- 1 13 . Therefore y p ( x ) = 3 26 cos 2 x- 1 13 sin2 x + 1 2 = 1 26 (3 cos 2 x- 2 sin2 x + 13) . 1 4 ( 3.536 ) Let us first find the complementary solution. The characteristic equation is r 4- 4 r 2 = 0 = r 2 ( r 2- 4) = 0 = r 2 ( r + 2)( r- 2) = 0 = r = 0 , 2 , where r = 0 has multiplicity 2. So the complementary solution can be written as follows: y c ( x ) = c 1 + c 2 x + c 3 e 2 x + c 4 e- 2 x . Let us next find the particular solution y p ( x ). Assume y p ( x ) = x 2 ( A + Bx + Cx 2 ) = Ax 2 + Bx 3 + Cx 4 ....
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This homework help was uploaded on 04/02/2008 for the course ENGR 101 taught by Professor Ringenberg during the Fall '07 term at University of Michigan.

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WrittenHW06-Solutions - Math 216 (Section 50) Written...

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