179_pdfsam_math 54 differential equation solutions odd

179_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.2 (d) By the defnition oF cosh t , y 3 ( t )=cosh t = e t + e t 2 = 1 2 e t + 1 2 e t = 1 2 y 1 ( t )+ 1 2 y 2 ( t ) , and given Functions are linearly dependent on ( −∞ , ). 41. The auxiliary equation For this problem is r 3 + r 2 6 r + 4 = 0. ±actoring yields r 3 + r 2 6 r +4 = ( r 3 r 2 ) + ( 2 r 2 2 r ) +( 4 r +4) = r 2 ( r 1) + 2 r ( r 1) 4( r 1) = ( r 1)( r 2 +2 r 4) . Thus the roots oF the auxiliary equation are r =1 and r = 2 ± p ( 2) 2 4(1)( 4) 2 = 1 ± 5 . ThereFore, the Functions e t , e ( 1 5) t ,and e ( 1+ 5) t are solutions to the given equation, and they are linearly independent on ( −∞ , ) (see Problem 40). Hence, a general solution to y 0 + y 0 6 y 0 +4 y =0isg ivenby y ( t )= c 1 e t + c 2 e ( 1 5) t + c 3 e ( 1+ 5) t . 42. The auxiliary equation associated with this di²erential equation is r 3 6
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