Unformatted text preview: Since W { f, g } ( x ) = e2 x cos( x ), it follows that fgf g = e2 x cos x . Thus ex g + ex g = e2 x cos x . This is a ﬁrst order linear ODE, so after dividing by ex 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 = cos x , so d dx ( e x g ) = cos x . Thus g ( x ) = ex (sin x + c ). Since g (0) = 1, c = 1. That is, g ( x ) = ex (1 + sin x ) . Thus g ( π/ 2) = 2 eπ/ 2 ....
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 Fall '07
 COSTIN
 Math, Differential Equations, Equations, TA, Leonhard Euler, Euler's formula, e2x cos

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