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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 definition 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. Factoring 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 ± ( 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 + y 6 y + 4 y = 0 is given by 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 differential equation is r 3 6 r 2
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