365_pdfsam_math 54 differential equation solutions odd

# 365_pdfsam_math 54 differential equation solutions odd - r...

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Exercises 6.3 Using the identities e ax =cosh ax +sinh ax, e ax =cosh ax sinh ax, we can express the solution in terms of hyperbolic and trigonometric functions as follows. y ( x )= C 1 e k/ ( EI ) x + C 2 e k/ ( EI ) x + C 3 sin 4 r k EI x + C 4 cos 4 r k EI x = C 1 cosh 4 r k EI x +sinh 4 r k EI x  + C 2 cosh 4 r k EI x sinh 4 r k EI x  + C 3 sin 4 r k EI x + C 4 cos 4 r k EI x = c 1 cosh 4 r k EI x + c 2 sinh 4 r k EI x + c 3 sin 4 r k EI x + c 4 cos 4 r k EI x , where c 1 := C 1 + C 2 , c 2 := C 1 C 2 , c 3 := C 3 ,and c 4 := C 4 are arbitrary constants. EXERCISES 6.3: Undetermined Coeﬃcients and the Annihilator Method, page 337 1. The corresponding homogeneous equation for this problem is y 0 2 y 0 5 y 0 +6 y =0wh ich has the associated auxiliary equation given by r 3 2 r 2 5 r + 6 = 0. By inspection we see that r
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Unformatted text preview: r 3 − 2 r 2 − 5 r + 6 = ( r − 1)( r 2 − r − 6) = ( r − 1)( r − 3)( r + 2) = 0 . Thus, the roots to the auxiliary equation are given by r = 1, 3, and − 2, and a general solution to the homogeneous equation is y h ( x ) = c 1 e x + c 2 e 3 x + c 3 e − 2 x . The nonhomogeneous term, g ( x ) = e x + x 2 , is the sum of an exponential term and a polynomial term. Therefore, according to Section 4.5, this equation has a particular solution of the form y p ( x ) = x s 1 C 1 e x + x s 2 ( C 2 + C 3 x + C 4 x 2 ) . 361...
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