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springps08_soln

springps08_soln - 18.03 PS8 Solutions-Spring 09 1a L(eat...

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18.03 PS8 Solutions—Spring 09 1a. L ( e at f ( t )) = integraldisplay 0 e st e at f ( t ) dt = integraldisplay 0 e ( s a ) t f ( t ) dt = F ( s - a ) 1b. L ( u ( t - a ) f ( t - a )) = integraldisplay 0 e st u ( t - a ) f ( t - a ) dt = integraldisplay a e s ( τ + a ) u ( τ ) f ( τ ) = e as integraldisplay 0 e f ( τ ) = e as F ( s ) 1c. Let g ( t ) = t + 1, so L ( g ( t )) = L ( t ) + L (1) = 1 s 2 + 1 s = s + 1 s 2 . Define G ( s ) = s + 1 s 2 . By (a), F ( s ) = L ( f ( t )) = L (( t + 1) e t +1 ) = e L ( e t ( t + 1)) = eG ( s - 1) = es ( s - 1) 2 . By (b), L ( u ( t - 1) te t ) = L ( u ( t - 1) f ( t - 1)) = e s F ( s ) = e 1 s s ( s - 1) 2 . 1d. L ( f ( at )) = integraldisplay 0 e st f ( at ) dt = integraldisplay 0 e s ( τ/a ) f ( τ ) 1 a = 1 a integraldisplay 0 e ( s/a ) τ f ( τ ) = 1 a F parenleftBig s a parenrightBig 2. Let Y ( s ) = L ( y ( t )). Then L ( y ) = sY ( s ) - y (0) = sY ( s ) - 2 and L ( y ′′ ) = L (( y ) ) = s ( sY ( s ) - 2) - y (0) = s 2 Y ( s ) - 2 s . So, y ′′ + y = 1 = L ( y ′′ ) + L ( y ) = L (1) = [ s 2 Y ( s ) - 2 s ] + [ sY ( s ) - 2] = 1 /s . Thus (using partial fractions), Y ( s ) = 2 s + 2 + 1 /s s 2 + s = 2 s 2 + 2 s + 1 s 2 ( s + 1) = 1 s 2 + 1 s + 1 s + 1 = L ( t ) + L (1) + L ( e t ) .
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