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Quiz4Bsolutions

# Quiz4Bsolutions - y(0 = 0 Now assume that y t = sin t 2 is...

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MATH 415.01 QUIZ 4B Feb 3, 2010 SHOW ALL YOUR WORK! GOOD LUCK! NAME: 1. (6.5 pts.) Find the solution of the initial value problem y 00 - y = 0 , y (0) = 5 2 , y 0 (0) = 3 2 r 2 - 1 = 0 i.e. ( r - 1)( r + 1) = 0. So, r 1 = 1 , r 2 = - 1. Thus, general solution is y = c 1 e t + c 2 e - t Then, y 0 = c 1 e t - c 2 e - t . Use initial conditions, c 1 + c 2 = 5 / 2 c 1 - c 2 = 3 / 2 Thus, c 1 = 2 and c 2 = 1 / 2 Finally solution becomes y = 2 e t + 1 2 e - t 2. (6 pts.) Can y = sin ( t 2 ) be a solution on an interval containing t = 0 of an equation y 00 + p ( t ) y 0 + q ( t ) y = 0 with continuous coefficients? Explain your answer. NO. We have y ( t ) = sin ( t 2 ), then y 0 ( t ) = - 2 t cos ( t 2 ). So, y (0) = 0 and
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Unformatted text preview: y (0) = 0. Now, assume that, y ( t ) = sin ( t 2 ) is a solution of an equation y 00 + p ( t ) y + q ( t ) y = 0. Then, it should also satisfy initial conditions y (0) = 0 and y (0) = 0. But, this initial value problem has a UNIQUE solution y = 0. (check your notes). There is only one solution y = 0 and obviously sin ( t 2 ) 6 = 0 . Hence, y ( t ) = sin ( t 2 ) is NOT a solution of this diﬀerential equation....
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