p2spring10_SOLUTIONS_ONLY - Since 5 < 3, those...

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some answers: 1b) y ( t ) = c 1 cos( t ) + c 2 sin( t ) + cos( at ) / (1 - a 2 ) + 2 sin( bt ) / (1 - b 2 ) 1c) 1 / (1 - x 2 3 ) when - 3 < x < 3. 1d) Either the alternating series test or the Taylor remainder gives that the error is less than the maximum value of the next term x 3 6 0 . 001 6 . 2a) Use some identity such as cos( nt ) = ( e int + e - int ) / 2 to simplify. 2b) You must find numbers c 1 , c 2 , c 3 not all 0, so that c 1 e it + c 2 e - it + c 3 sin( t ) = 0 for all t . We know that e it = cos( t ) + i sin( t ) so e - it = cos( t ) - i sin( t ). Adding, find e it + e - it = - 2 i sin( t ). Rearrange it: e it + e - it + 2 i sin( t ) = 0. 2c) - 2 3 π 3a) Try y ( t ) = c 1 e 3 t + c 2 e - 3 t + Ae - t . You need (1 - 9) A = 5, so A = - 5 / 8. Then initial conditions give y ( t ) = 13 24 e 3 t - 22 24 e - 3 t - 5 8 e - t . 3b) As before y ( t ) = c 1 e 3 t + c 2 e - 3 t . Then y ( - 1) = c 1 e - 3 + c 2 e 3 = 0 and y (1) = c 1 e 3 + c 2 e - 3 = 0. So c 1 = - e 6 c 2 = e 12 c 1 forces c 1 = 0. Then c 2 = 0. 4a) You need A = 3 / ( c 4 ω 4 - 3 c 2 ω 2 + 1). 4b) You didn’t divide by 0 in 4a) unless c 4 ω 4 - 3 c 2 ω 2 + 1 = 0. That is a quadratic equation for ( ) 2 , so the special cases are when ( ) 2 = 3 ± 9 - 4 2 .
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Unformatted text preview: Since 5 &lt; 3, those are both positive, and the resonances are c = q 3 5 2 . 5) y (0) = 0 gives c 1 = 0. The DE gives ( EI/ ) k 4 = 2 . Then y ( L ) = 0 gives sin( kL ) = 0 so k = /L , 2 /L , 3 /L etc. When k = n/L the frequency must be = p EI/ ( n/L ) 2 for nonzero solution. No restriction on c 2 . 6a) With p ( x ) = c 3 s 3 + c 1 s you need 0 = (1-s 2 )(6 c 3 s )-2 s ( c 1 + 3 c 3 s 2 ) + 12( c 3 s 3 + c 1 s ) = (6 c 3 + 10 c 1 ) s + (-12 c 3 + 12 c 3 ) s 3 . So c 3 is arbitrary, c 1 = 6 c 3 / 10, so p ( s ) = 3 s-5 s 3 or any multiple. 6b) u = 3 x 2 z +3 y 2 z-2 z 3 (or a multiple), so u xx + u yy + u zz = 6 z +6 z-12 z = 0. 7...
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