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quiz3_ODE_solutions

# quiz3_ODE_solutions - Name UF ID number Quiz 3 MAP 2302...

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Name : UF ID number : Quiz 3, MAP 2302, Fall’11 Show your work! Write your name on every piece of paper you turn in! 1 . Find a general solution of the equation y primeprimeprime - 3 y primeprime + 3 y prime - y = e x 2 . Find the Laplace transform of f ( t ) = t sin 2 t 3 . Find the function whose Laplace transform is F ( s ) = 7 s 2 + 23 s + 30 ( s - 2)( s 2 + 2 s + 5) 4 . Use the method of Laplace transforms to solve the initial value problem: y primeprime + 3 ty prime - 6 y = 1 , y (0) = y prime (0) = 0 Hint: Recall dy dx + p ( x ) y = q ( x ) y ( x ) = 1 μ ( x ) integraldisplay μ ( x ) q ( x ) dx, μ ( x ) = exp parenleftbiggintegraldisplay p ( x ) dx parenrightbigg 5 . Find the Laplace transform of the solution of the initial value problem: y primeprime + 3 y prime + 2 y = e - 3 t u ( t - 2) , y (0) = 2 , y prime (0) = - 3 where u ( t ) is the step function ( u ( t ) = 1 if t 0 and u ( t ) = 0 elsewhere). Extra credit : Find the solution. 6 ( Extra credit ). Consider the initial value problem y primeprime + 2 y prime + 5 y = g ( t ), y (0) = y prime (0) = 0, where g ( t ) = 1 a = const > 0 if 0 < t < a and g ( t ) = 0 elsewhere. Find the solution of this problem in the limit when a 0.

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Some properties of the Laplace transform F ( s ) = L ( f ) = integraldisplay 0 e - st f ( t ) dt, s > 0 L ( t n ) = n ! s n +1 , n = 0 , 1 , 2 , ..., 0! = 1 , s > 0 L ( e at t n ) = n ! ( s - a ) n +1 , n = 0 , 1 , 2 , ..., s > 0 L ( e at sin( bt )) = b ( s - a ) 2 + b 2 , s > a L ( e at cos( bt )) = s - a ( s - a ) 2 + b 2 , s > a L ( f ( n ) ) = s n F ( s ) - s n - 1 f (0) - s n - 2 f prime (0) - · · · - f ( n - 1) (0) , n = 1 , 2 , ...
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quiz3_ODE_solutions - Name UF ID number Quiz 3 MAP 2302...

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