quiz3sol930

# quiz3sol930 - e-x to put the equation in the right form we...

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TA: R McDougal Name: Math 415 25 October 2007 Quiz III – 9:30 Version Answer each question as completely as you can; remember you must show all work for full credit. You may not consult books, notes, or each other for this quiz. Good luck! Let f ( x ) = e - x . g ( x ) is some other function. Assume the Wronskian of f and g satisﬁes W { f, g } ( x ) = e ix + e - ix 2 , (1) where i 2 = - 1. 1. Use Euler’s Formula to simplify (1). Your ﬁnal answer should not involve i or - 1. (3 points) W { f, g } ( x ) = e ix + e - ix 2 = cos( x ) + i sin( x ) + cos( - ix ) + i sin( - ix ) 2 = cos( x ) + i sin( x ) + cos( ix ) - i sin( ix ) 2 = cos( x ) . 2. Given that g (0) = 1, ﬁnd g ( π/ 2). (7 points) Since W { f, g } ( x ) = cos( x ), it follows that fg 0 - f 0 g = cos x . Thus e - x g 0 + e - x g = cos x . This is a ﬁrst order linear ODE, so after dividing by
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Unformatted text preview: e-x to put the equation in the right form, we can solve it by using an integrating factor of μ = e x . Thus e x g + e x g = e 2 x cos x , so d dx ( e x g ) = e 2 x cos x . Note that Z e 2 x cos xdx = e 2 x sin x-2 Z e 2 x sin xdx = e 2 x sin x-2 ±-e 2 x cos x + 2 Z e 2 x cos xdx ² + c 1 = 1 5 ( e 2 x sin x + 2 e 2 x cos x ) + c 2 . Thus e x g = 1 5 ( e 2 x sin x + 2 e 2 x cos x ) + c 2 , so g = 1 5 ( e x sin x + 2 e x cos x ) + c 2 e-x . Plug-ging in our initial condition, we ﬁnd 1 = 2 / 5 + c 2 , so c 2 = 3 / 5. Thus g ( π/ 2) = 1 5 e π/ 2 + 3 5 e-π/ 2 ....
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