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Unformatted text preview: Answer Guide 6 Math 427K: Unique Number 55160 Wednesday, October 12, 2011 1. Suppose that x ( t ) = cos( ! 1 t ) & cos( ! 2 t ) . Make the substitutions & = ! 1 + ! 2 2 and ¡ = ! 1 & ! 2 2 and verify that x ( t ) = & 2 sin( &t ) sin( ¡t ) : Solution: Following the hint, let A = &t = ! 1 + ! 2 2 t and B = ¡t = ! 1 & ! 2 2 t . Then A + B = ! 1 t and A & B = ! 2 t . Using the angle addition formulas, we easily get x ( t ) = cos( A + B ) & cos( A & B ) (1a) = f cos A cos B & sin A sin B g & f cos A cos B + sin A sin B g (1b) = & 2 sin A sin B (1c) = & 2 sin( &t ) sin( ¡t ) : (1d) For the next four problems, &nd the general solution: 2. x 000 & x 00 & 12 x = 0 : Solution: The characteristic equation is 0 = r 3 & r 2 & 12 r = r ( r + 3)( r & 4) : So the general solution is x ( t ) = C 1 e & 3 t + C 2 + C 2 e 4 t : ( F ) 3. x 0000 & 4 x 000 + 7 x 00 & 16 x + 12 x = 0 : Solution: The characteristic equation is 0 = r 4 & 4 r 3 + 7 r 2 & 16 r + 12 = ( r & 1)( r & 3)( r & 2 i )( r + 2 i ) : So the general solution is...
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This note was uploaded on 10/24/2011 for the course MATH 427K taught by Professor Delallave during the Spring '11 term at University of Texas at Austin.
 Spring '11
 DELALLAVE
 Math, Differential Equations, Equations

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