356_pdfsam_math 54 differential equation solutions odd

356_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 5. We can factor the auxiliary equation, r 3 +3 r 2 +28 r + 26 = 0, as follows: r 3 r 2 r +26 = ( r 3 + r 2 )+(2 r 2 +2 r )+(26 r + 26) = r 2 ( r +1)+2 r ( r + 1) + 26( r +1)=( r +1)( r 2 r + 26) = 0 . Thus either r +1=0 r = 1or r 2 r +26=0 r = 1 ± 5 i . Therefore, a general solution is given by y ( x )= c 1 e x + c 2 e x cos 5 x + c 3 e x sin 5 x. 7. Factoring the characteristic polynomial yields 2 r 3 r 2 10 r 7=( 2 r 3 r 2 )+( 3 r 2 3 r 7 r 7) =2 r 2 ( r +1) 3 r ( r 7( r r + 1)(2 r 2 3 r 7) . Thus the roots of the characteristic equation, 2 r 3 r 2 10 r 7 = 0, are r r = 1 , 2 r 2 3 r 7=0 r = 3 ± p 3 2 4(2)( 7) 4 = 3 ± 65 4 , and a general solution is y ( x c 1 e x + c 2 e (3+ 65) x/ 4 + c 3 e (3 65) x/ 4 . 9. In the characteristic equation, r 3 9 r 2 +27 r 27 = 0, we recognize a complete cube, namely, ( r 3) 3 = 0. Thus, it has just one root, r = 3, of multiplicity three. Therefore, a general solution to the given di±erential equation is given by
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