Tutorial8

Tutorial8 - and if follows that(as b(s2 1(cs d(s2 1 = s2 for all s By setting s = 1 and s = 1 respectively in the last equation we obtain the pair

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( as + b )( s 2 + 1) + ( cs + d )( s 2 - 1) = s 2 for all s . By setting s = 1 and s = - 1, respectively in the last equation, we obtain the pair of equations 2( a + b ) = 1 , 2( - a + b ) = 1 , and therefore a = 0 and b = 1 / 2 . If we set s = 0, then b - d = 0 so d = 1 / 2 . Finally, equating the coefficients of the cubic terms on each side, we find that a + c = 0, so c = 0 . Thus Y ( s ) = 1 / 2 s 2 - 1 + 1 / 2 s 2 + 1 , and from line 4 and 6 of Table 5.1, the solution of the initial value problem is y = sinh t + sin t 2 . 2 Let u c ( t ) be the unit step function u c ( t ) = 0 , t < c 1 , t c. Then we have the following useful relations L{ u c ( t ) f ( t - c ) } = e - cs F ( s ) , L{ e ct f ( t ) } = F ( s - c ) . In particular, L{ u c ( t ) } = e - cs s . Let δ ( t ) be the Dirac function, then Z -∞ δ ( t - c ) f ( t ) dt = f ( c ) . If c > 0, then from above it follows L{ δ ( t - c ) f ( t ) } = e - cs f ( c ) . In particular,
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This note was uploaded on 01/28/2011 for the course MATH 150 taught by Professor T.qian during the Spring '09 term at HKUST.

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Tutorial8 - and if follows that(as b(s2 1(cs d(s2 1 = s2 for all s By setting s = 1 and s = 1 respectively in the last equation we obtain the pair

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