Lesson 11a -Cauchy Euler

# Thenourtwoindependentsolutionsare b a b a 2 4ac 0 r

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Unformatted text preview: al solutions r ”. Then our two independent solutions are: b a (b a ) 2 4ac 0 r (Note that this means that as in the quadratic formula. 2a Why? yx y x ln x r r Then the substituting this in will give us: ax 2 ( x r )' 'bx( x r )' c( x r ) 0 ar (r 1)( x r ) br ( x r ) c( x r ) 0 (ar 2 (b a )r c)( x r ) 0 00 Since we know we can have only 2 independent solutions, our general solution when both are real and distinct is: y c1 x c2 x ln( x) r r Case 2 Continued 0 ax 2 ( x r ln x)' 'bx( x r ln x)' c( x r ln x) 2 r 2 r 1 1 r1 r 1 r1 0 ax (r (r 1) x ln x 2rx x 2 ) bx(rx ln x x ) c( x r ln x) x x x 0 ax 2 (r (r 1) x r 2 ln x 2rx r 2 x r 2 ) bx(rx r 1 ln x x r 1 ) c( x r ln x) 0 [ar 2 ar ]x r ln x 2arx r ax r brx r ln x bx r cx r ln x 0 ([ar 2 ar ] ln x 2ar a br ln x b c ln x) x r 0 ([ar 2 ar ] ln x br ln x c ln x 2ar a b) x r 0 ([ar 2 ar br c] ln x [2ar a b]) x r 00 Since we know we can have only 2 independent solutions, our general solution when both are real and distinct is: y c1 x c2 x ln x r r Case 3 i i Case 3: “r are imaginary solutio...
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## This note was uploaded on 02/07/2014 for the course MATH 1004 taught by Professor Mark during the Summer '00 term at Carleton CA.

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