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Quiz6Bsolutions - μx c 2 sin μx Then y =-μc 1 sin μx...

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MATH 415.01 QUIZ 6B Feb 17, 2010 SHOW ALL YOUR WORK! GOOD LUCK! NAME: 1. (6.5 pts.) A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 10 cm/sec, and if there is no damping, determine the position u of the mass at any time t . SOLUTION: m = 100, k = 980 / 5 and γ = 0. Also, u 0 (0) = 0 , u 0 (0) = 10. Then the equation is: 100 u 00 + 980 / 5 u = 0. i.e. 100 r 2 + 980 / 5 = 0 So, r = 14 i . Then, u = c 1 cos(14 t ) + c 2 sin(14 t ) Use initial conditions, u (0) = c 1 = 0 and u 0 (0) = 14 c 2 = 10.Thus u ( t ) = 5 7 sin(14 t ) 2. (6 pts.) Find the eigenvalues and eigenfunctions of the given boundary value problem. y 00 + λy = 0 , y 0 (0) = 0 , y 0 ( π ) = 0 SOLUTION: r 2 + λ = 0 , r 2 = - λ Case1: λ = 0 , r = 0 : y = c 1 + c 2 x . Then y 0 = c 2 , y 0 (0) = y 0 ( π ) = c 2 = 0. Thus, y = c 1 . Thus, λ = 0 is an eigenvalue, with eigenvector y 0 = 1. Case2: λ = μ 2 > 0 , r = ± μi : y = c 1 cos(
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Unformatted text preview: μx ) + c 2 sin( μx ). Then, y =-μc 1 sin( μx ) + μc 2 cos( μx ). So, y (0) = μc 2 = 0 i.e. c 2 = 0. Also, y ( π ) = μc 1 sin( μπ ) = 0. So, μπ = nπ . i.e. μ = n . Thus, eigenvelues are λ n = n 2 with corresponding eigenfunctions y n = cos( nx ) for n = 1 , 2 , 3 ,.... Case3:: λ =-μ 2 < ,r = ± μ : y = c 1 cosh( μx )+ c 2 sinh( μx ). Then, y = μc 1 sinh( μx )+ μc 2 cosh( μx ). So, y (0) = μc 2 = 0 i.e. c 2 = 0. Also, y ( π ) = μc 1 sinh( μπ ) = 0. This implies that sinh( μπ ) = 0.i.e. μ = 0. Not possible. Thus c 1 = 0. Then the solution is y = 0 . There is no nontrivial solution....
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