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eng math 501e_2011 2012 fall_week2_applications

eng math 501e_2011 2012 fall_week2_applications -...

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1 Application 1 (Reduction of order – How to obtain a second solution) ( ) 2 1 1 '' 2 ' 2 0 has y as a first solution. Find another independent solution. x y xy y x - - + = = Solution ( ) 2 2 2 2 2 2 2 2 ( 1) 2 2 1 2 2 ' side x 1 , 2 2 '' ' 0 1 1 '' ( ) ' ( ) ( ) 2 2 ( ) 1 1 1 , 2 ( 1) 1 1 . . 1 .( 1) pdx In x Let s divide each by x y y y x x y P x y q x y r x x x P x pdx dx x x t x dt xdx dt In t In x t U e U e y x U x x - - - - + = - - + + = = - → - = - - = - = = = - = = = - 2 2 2 2 1 2 1 1 1 1 1- 1 . 1 U x u Udx dx x x x y u y x x y x x = - = = = + = = + = + 1 1 2 2 1 2 2 1 1 2 2 1 1 2 1 2 Note: ( ) ( ) 0 ( 0 and 0) or ( ) k y x k y x k k k k y y y y k k y and y are proportional linearly dependent + = = - = - Application 2 (Homogeneous equations with constant coefficients – Two distinct real roots) Solve the initial value problem. '' ' 2 0 , (0) 4 , '(0) 5 y y y y y + - = = = - Solution 2 1 2 2 1 2 1 2 2 1 2 1 2 1 2 1 2 1 2 2 2 0 ( 2)( 1) 0 1 , 2 . . (0) 4 4 ' . 2 . '(0) 5 5 2 4 2 5 1 , 3 3 x x x x x x r r r r r r y c e c e y c c y c e c e y c c c c c c c c y e e - - - + - = + - = = = - = + = = + = - = - → - = - + = - = - = = = + 1 2 1 1 2 1 1 2 1 2 1 1 2 1,2 1 1 2 : hom ( ) * ( ) . .
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