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Tutorial8 - and if follows that(as b(s2 1(cs d(s2 1 = s2...

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and if follows that ( 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, L{ δ ( t - c ) } = e - cs . Ex 2. Find the solution of the initial value problem y 00 + 4 y = g ( t ) , y (0) = 0 , y 0 (0) = 0 , 2
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where g ( t ) = 0 , 0 t < 5 , ( t - 5) / 5 , 5 t < 10 , 1 , t 10 . The graph of the function g ( t ) is as follows 2 4 6 8 10 12 14 0.2 0.4 0.6 0.8 1
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