p2spring09_SOLUTIONS_ONLY - f , if you are careful to point...

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some answers: 1b) x ( t ) = c 1 e t + e - t c 2 cos(2 t ) + c 3 sin(2 t ) · - 29 8 e - t - 6 2a) a = - 1 4 2b) 2 cos( t ) cos(10 t ) 3a) c 1 + c 2 e - c m t 3b) P = - ma m 2 ω 2 + c 2 , Q = - c P 3c) Using parts a) and b) and linearity it is c 1 + c 2 e - 2 t + - 3 5 2 +2 2 ( cos(5 t ) - 2 5 sin(5 t ) ) + - 7 9 2 +2 2 ( cos(9 t ) - 2 9 sin(9 t ) ) 4a) 9 is near 3 π so f (9) = 0 4b) f ( t ) = 4 π sin( t ) + 4 π sin(2 t ) + 4 3 π sin(3 t ) + 4 5 π sin(5 t ) + . . . 4c) g is a continuous even function whose graph is π π ‘V’ shaped on the interval [ - π 2 , π 2 ] and g ( t ) = π when π 2 ≤ | t | ≤ π . 4d) This g is periodic, although periodicity is not an automatic thing when integrating periodic functions. The R π - π f ( t ) dt = 0. So the integral of f over any interval of length 2 π is 0. So for any number a you have g ( a + 2 π ) - g ( a ) = R a +2 π 0 f ( t ) dt - R a 0 f ( t ) dt = R a - a f ( t ) dt = 0 and so g has period 2 π . Or, you might argue using your graph of
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Unformatted text preview: f , if you are careful to point out that the difference between g ( a + 2 π ) and g ( a ) is always the integral of f over that length, which is always 0. 5a) This is a geometric series which converges to 1 / (1-e x ) when e x < 1. So any negative x is alright. 5b) y ( x ) = 1 3 x 3-1 15 x 5 + ··· . 6) The functions are any multiples of sin( cx ) where c cos( c ) + 3 sin( c ) = 0, and λ = c 2 . So you need c =-3 tan( c ). A sketch of the graph of-3 tan( c ) as a function of c will show what is needed. 7...
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This note was uploaded on 01/21/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).

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