Quiz6Bsolutions

Quiz6Bsolutions - x ) + c 2 sin( x ). Then, y =-c 1 sin( x...

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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: x ) + c 2 sin( x ). Then, y =-c 1 sin( x ) + c 2 cos( x ). So, y (0) = c 2 = 0 i.e. c 2 = 0. Also, y ( ) = c 1 sin( ) = 0. So, = n . i.e. = n . Thus, eigenvelues are n = n 2 with corresponding eigenfunctions y n = cos( nx ) for n = 1 , 2 , 3 ,.... Case3:: =- 2 < ,r = : y = c 1 cosh( x )+ c 2 sinh( x ). Then, y = c 1 sinh( x )+ c 2 cosh( x ). So, y (0) = c 2 = 0 i.e. c 2 = 0. Also, y ( ) = c 1 sinh( ) = 0. This implies that sinh( ) = 0.i.e. = 0. Not possible. Thus c 1 = 0. Then the solution is y = 0 . There is no nontrivial solution....
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