6.4we - y ( t ) to (1) is given by (5) y ( t ) = ( y I ( t...

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DIFFERENTIAL EQUATIONS WITH DISCONTINUOUS FORCING FUNCTIONS KIAM HEONG KWA Consider the initial value problem (1) y 00 ( t ) + αy 0 ( t ) + βy ( t ) = u a ( t ) , y (0) = y 0 , y 0 (0) = y 0 0 , where α,β R and u a is the Heaviside function (2) u a ( t ) = ( 0 if t < a, 1 if t a for some a 0. Note that the non-homogeneous term u a in (1) is discontinuous at t = a . It is an instance of the class of discontinuous forcing functions . In general, the discontinuity of u a at t = a is man- ifested in a corresponding discontinuity of the solution y ( t ) to (1) at the same point. Without the aid of the Laplace transform, this initial value problem (1) has to be solved in two steps as follows. First of all, we note that for 0 t < a , the solution y ( t ) satisfies the initial value problem (3) y 00 ( t ) + αy 0 ( t ) + βy ( t ) = 0 , y (0) = y 0 , y 0 (0) = y 0 0 . Let y I ( t ) be the solution to (3). Next, we note that for t a , the solution y ( t ) to (1) fulfills the other initial value problem, i.e., (4) y 00 ( t ) + αy 0 ( t ) + βy ( t ) = 1 , y ( a ) = y I ( a ) , y 0 ( a ) = y 0 I ( a ) . Let y II ( t ) be the solution to (4). Then the solution
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Unformatted text preview: y ( t ) to (1) is given by (5) y ( t ) = ( y I ( t ) if 0 t &lt; a, y II ( t ) if t a. On the other hand, the application of the Laplace transform to the initial value problem (1) has the advantage of solving the problem in Date : February 22, 2011. 1 2 KIAM HEONG KWA a more direct manner. Let Y ( s ) = L y ( s ). Then taking the Laplace transform of (1) yields ( s 2 + s + ) Y ( s )-( s + ) y-y = e-as s , from which it follows that (6) Y ( s ) = ( s + ) y + y s 2 + s + + e-as s ( s 2 + s + ) . The solution y ( t ) to (1) is then obtained by inverting the transform: (7) y ( t ) = L-1 [ Y ( s )]( t ) = L-1 ( s + ) y + y s 2 + s + + e-as s ( s 2 + s + ) ( t ) ....
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6.4we - y ( t ) to (1) is given by (5) y ( t ) = ( y I ( t...

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