Unformatted text preview: 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 = 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 x2 Z e 2 x sin xdx = e 2 x sin x2 ±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 ex . Plugging 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 eπ/ 2 ....
View
Full Document
 Fall '07
 COSTIN
 Math, Differential Equations, Equations, Cos, TA, Leonhard Euler, Euler's formula, 1 3 g

Click to edit the document details