# Q5_130 - Math 415 Quiz 5A Assuming the functions y1 and y2...

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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 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.

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