Q5A_130

# Q5A_130 - y ( t ) = C 1 e 2 t cos(2 t )+ C 2 e 2 t sin(2 t...

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Math 415 Quiz 5A Assuming the functions y 1 and y 2 are linearly independent solutions of y 00 + p ( t ) y 0 + q ( t ) y = 0 and that y 3 = y 1 + y 2 and y 4 = y 1 - y 2 , show that y 3 and y 4 also form a linearly independent set of solutions. We know that the Wronskian of y 1 and y 2 , W ( y 1 ,y 2 )( t ) is nonzero on the interval on which the solution is guaranteed to exist. So let’s consider the Wronskian of y 3 and y 4 . W ( y 3 ,y 4 ) = ± ± ± ± ± y 1 + y 2 y 1 - y 2 y 0 1 + y 0 2 y 0 1 - y 0 2 ± ± ± ± ± = ( y 1 + y 2 )( y 0 1 - y 0 2 ) - ( y 1 - y 2 )( y 0 1 + y 0 2 ) = ( y 1 y 0 1 - y 1 y 0 2 + y 2 y 0 1 - y 2 y 0 2 ) - ( y 1 y 0 1 + y 1 y 0 2 - y 2 y 0 1 - y 2 y 0 2 ) = - 2 y 1 y 0 2 + 2 y 2 y 0 1 = - 2 W ( y 1 ,y 2 ) 6 = 0 Hence, y 3 and y 4 are linearly independent. Find the solution of the initial value problem: y 00 - 4 y 0 + 8 y = 0 , y ( π/ 2) = 0 , y 0 ( π/ 2) = - 2 . Describe its behavior as t increases. The characteristic equation is r 2 - 4 r + 8, which has roots r = 2 ± 2 i . So the general solution is
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Unformatted text preview: y ( t ) = C 1 e 2 t cos(2 t )+ C 2 e 2 t sin(2 t ). Plugging in the initial values tells us that-C 1 e π = 0 and-2 C 1 e π-2 C 2 e π =-2. So C 1 = 0 and C 2 = e-π , and our solution is y ( t ) = e 2 t-π sin(2 t ). As for the behavior as t → ∞ , e 2 t-π grows without bound, and sin(2 t ) oscillates between positive and negative one, so the function oscillates without bound, getting closer and closer to both positive and negative inﬁnity by turns. 1...
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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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