s251ex2(fa00)

s251ex2(fa00) - u ( t ) is 2 u 00 + 4 u + 20 u = 0, u (0) =...

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Math 251 Fall 2000 Midterm Exam II 1. y 2 ( t ) = t - 2 (or any of its nonzero constant multiples). 2. a) The homogeneous solution is y c ( t ) = C 1 e 3 t + C 2 te 3 t . The right-hand side of the equa- tion is 3 e 2 t . The form of the particular solution is, therefore, Y ( t ) = Ae 2 t . Substitute Y ( t ) into the equation and solve: Y 00 - 6 Y 0 + 9 Y = 3 e 2 t . The result is A = 3, so Y ( t ) = 3 e 2 t . Finally, the solution is y ( t ) = y c ( t ) + Y ( t ) = C 1 e 3 t + C 2 te 3 t + 3 e 2 t . b) Y ( t ) = ( At + B ) cos 2 t + ( Ct + D ) sin 2 t + Et 3 + Ft 2 + Gt + H + It 2 e 3 t . 3. a) Since the Wronskian is W ( y 1 (1) , y 2 (1)) = ¯ ¯ ¯ ¯ y 1 y 2 y 0 1 y 0 2 ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ t te t 1 te t + e t ¯ ¯ ¯ ¯ = t 2 e t 6 = 0 when t > 0, the functions y 1 ( t ) = t and y 2 ( t ) = te t do form a set of fundamental solutions. b) Since y 1 and y 2 form a set of fundamental solutions, the general solution is y ( t ) = C 1 y 1 + C 2 y 2 = C 1 t + C 2 te t . 4. Constants: m = 2, g = 10, L = 1; mg = 20 = kL so k = 1; the resistive force is F r = γ | V | = 8 when | V | = 2, so the damping constant is γ = 4. The initial displacement is 1, and the initial velocity is 3 3 - 1. a) The required initial value problem for
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Unformatted text preview: u ( t ) is 2 u 00 + 4 u + 20 u = 0, u (0) = 1, u (0) = 3 3-1. Its solution is u ( t ) = e-t cos 3 t + 3 e-t sin 3 t . b) lim t u ( t ) = 0 c) The system is underdamped, therefore, the mass will oscillate (i.e., pass through the equilibrium position) innitely many times. 5. a) y ( t ) = e 2 t cos 3 t + 2 e 2 t sin 3 t . b) f ( t ) = (1-u 2 ( t ))(3 t 2 ) + u 2 ( t )(3 t 2 + sin t ) = 3 t 2 + u 2 ( t ) sin t . F ( s ) = L{ 3 t 2 } + e-2 s L{ sin( t + 2 ) } = 6 s 3 + e-2 s L{ sin t } = 6 s 3 + e-2 s 1 s 2 + 1 . 6. y ( t ) = 1 4 e 2 t-1 4 e-2 t + u 2 ( t )[ 1 4 e 2( t-2 )-1 4 e-2( t-2 ) ] 1...
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