Exam 7: Laplace Transform

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Let f(t)={1 if 0≤t≤3t−2 if t>3}f ( t ) = \left\{ \begin{array} { l l } 1 & \text { if } 0 \leq t \leq 3 \\t - 2 & \text { if } t > 3\end{array} \right\} . Then L{f(t)}\mathcal { L } \{ f ( t ) \} is

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When the Laplace transform is applied to the system =3x-y =x+y x(0)=2,y(0)=1 the resulting transformed system is

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The Laplace transform of cos⁡t\cos t is

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The solution of the system in the previous problem is

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The Laplace transform of ∫0tτ2cos⁡(t−τ)dτ\int _ { 0 } ^ { t } \tau ^ { 2 } \cos ( t - \tau ) d \tau is

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The inverse Laplace transform of F(s)=(5s+2)/(s2⋅(s2−1))F ( s ) = ( 5 s + 2 ) / \left( s ^ { 2 } \cdot \left( s ^ { 2 } - 1 \right) \right) is

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When the Laplace transform is applied to the system =4x-y =2x+y x(0)=1,y(0)=0 the resulting transformed system is

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The Laplace transform of sin⁡(2t)\sin ( 2 t ) is

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The Laplace transform of tette ^ { t } is

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A uniform beam of length L has a concentrated load, w0w _ { 0 } , at x=L/2x = L / 2 . It is embedded at the left end and free at the right end. The correct initial value problem for the vertical deflection, y(x)y ( x ) , at a distance x from the embedded end is

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The Laplace transform of t2sin⁡(3t)t ^ { 2 } \sin ( 3 t ) is

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The Laplace transform of e−te ^ { - t } is

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