Quiz6Asolutions

Quiz6Asolutions - cos(8 t ) Now, use initial conditions: u...

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MATH 415.01 QUIZ 6A Feb 17, 2010 SHOW ALL YOUR WORK! GOOD LUCK! NAME: 1. (6.5 pts.) A mass that weighs 8 lb stretches a spring 6 in. The system is acted on by an external force of 8 sin(8 t ) lb. If the mass is pulled down 3 in. and then released, determine the position of the mass at any time. SOLUTION: m = 8 / 32 = 1 / 4, γ = 0, k = 8 / (6 / 12) = 16. Also, we have u (0) = 3 / 12 = 1 / 4 , u 0 (0) = 0 i.e. The equation is: 1 / 4 u 00 + 16 u = 8 sin(8 t ). Characteristic equation 1 / 4 r 2 + 16 = 0. So, r 2 = - 64. Then u c = c 1 cos(8 t ) + c 2 sin(8 t ). Find particular solution: U ( t ) = ( A cos(8 t ) + B sin(8 t )) t , U 0 ( t ) = A cos(8 t ) + B sin(8 t ) + t ( - 8 A sin(8 t ) + 8 B cos(8 t )) and U 00 ( t ) = - 16 A sin(8 t ) + 16 B cos(8 t ) + t ( - 64 A cos(8 t ) - 64 B sin(8 t )) Then, solve for A and B. 1 / 4 U 00 + 16 U = 8 sin(8 t ) i.e. - 4 A sin(8 t ) + 4 B cos(8 t ) + t ( - 16 A cos(8 t ) - 16 B sin(8 t )) + 16( A cos(8 t ) + B sin(8 t )) t = 8 sin(8 t ) i.e. - 4 A sin(8 t ) + 4 B cos(8 t ) = 8 sin(8 t ) . So, we get , A = - 2, B = 0. Thus the solution is u = c 1 cos(8 t ) + c 2 sin(8 t ) - 2 t
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Unformatted text preview: cos(8 t ) Now, use initial conditions: u (0) = c 1 = 1 / 4 and u (0) = 8 c 2-2 = 0 . i.e. c 2 = 1 / 4 . Final answer becomes, u = 1 / 4 cos(8 t ) + 1 / 4 sin(8 t )-2 t cos(8 t ) 2. (6 pts.) Find the solution of the given boundary value problem below or show that it has no solution. y 00 + y = x, y (0) = 0 , y ( ) = 0 SOLUTION: Homogeneous solution: r 2 +1 = 0 and r = i . y c = c 1 cos( x )+ c 2 sin( x ). Particular solution: Y ( x ) = Ax + B . Y = A and Y 00 = 0. A = 1 and B = 0. General solution: y = c 1 cos( x ) + c 2 sin( x ) + x . Use initial conditions: y (0) = c 1 = 0. Thus, y ( ) = c 2 sin( ) + = 0. This gives us = 0. Impossible. Thus there is no solution....
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