Q4_1130

Q4_1130 - Can y = sin t 2 be a solution on an interval...

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MATH 415.01 QUIZ 4A Feb 3, 2010 SHOW ALL YOUR WORK! GOOD LUCK! NAME: 1. (6.5 pts.) In the following diﬀerential equation, determine the values of α , if any, for which all solutions tend to zero as t → ∞ . y 00 + (4 - α ) y 0 - 3( α - 1) y = 0 r 2 + (4 - α ) r - 3( α - 1) = 0 .i.e. ( r + 3)( r - ( α - 1)) = 0. Thus, r 1 = - 3 , r 2 = ( α - 1). General Solution is y = c 1 e - 3 t + c 2 e ( α - 1) t Obviously, e - 3 t 0 as t → ∞ . y will tend to zero if we also have e ( α - 1) t 0. This is true if ( α - 1) < 0. i.e. α < 1 2. (6 pts.) Find the Wronskian of 2 x and xe x . W = ± ± ± ± 2 x xe x 2 e x + xe x ± ± ± ± = 2 x ( e x + xe x ) - 2 xe x = 2 x 2 e x

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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.)
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Unformatted text preview: Can y = sin ( t 2 ) be a solution on an interval containing t = 0 of an equation y 00 + p ( t ) y + q ( t ) y = 0 with continuous coeﬃcients? Explain your answer. NO. We have y ( t ) = sin ( t 2 ), then y ( t ) =-2 t cos ( t 2 ). So, y (0) = 0 and 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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This note was uploaded on 07/26/2011 for the course MATH 415 taught by Professor Costin during the Spring '07 term at Ohio State.

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Q4_1130 - Can y = sin t 2 be a solution on an interval...

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