357_pdfsam_math 54 differential equation solutions odd

# 357_pdfsam_math 54 differential equation solutions odd -...

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Exercises 6.2 13. The auxiliary equation in this problem is r 4 +4 r 2 +4 = 0. This can be factored as ( r 2 +2) 2 =0. Therefore, this equation has roots r = 2 i , 2 i , 2 i , 2 i , which we see are repeated and complex. Therefore, a general solution to this problem is given by y ( x )= c 1 cos ± 2 x ² + c 2 x cos ± 2 x ² + c 3 sin ± 2 x ² + c 4 x sin ± 2 x ² . 15. The roots to this auxiliary equation, ( r 1) 2 ( r +3)( r 2 +2 r +5) 2 =0 ,are r =1 , 1 , 3 , 1 ± 2 i, 1 ± 2 i, whe reweno tetha t1and 1 ± 2 i are repeated roots. Therefore, a general solution to the diFerential equation with the given auxiliary equation is y ( x )= c 1 e x + c 2 xe x + c 3 e 3 x +( c 4 + c 5 x ) e x cos 2 x +( c 6 + c 7 x ) e x sin 2 x. 17. ±rom the diFerential operator, replacing D by r , we obtain the characteristic equation ( r +4)( r 3)( r +2) 3 ( r 2 +4 r +5) 2 r 5 =0 , whose roots r +4=0 r = 4
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Unformatted text preview: ⇒ r = 3 , ( r + 2) 3 = 0 ⇒ r = − 2 of multiplicity 3 , ( r 2 + 4 r + 5) 2 = 0 ⇒ r = − 2 ± i of multiplicity 2 , r 5 = 0 ⇒ r = 0 of multiplicity 5 . Therefore, a general solution is given by y ( x ) = c 1 e − 4 x + c 2 e 3 x + ( c 3 + c 4 x + c 5 x 2 ) e − 2 x + ( c 6 + c 7 x ) e − 2 x cos x + ( c 8 + c 9 x ) e − 2 x sin x + c 10 + c 11 x + c 12 x 2 + c 13 x 3 + c 14 x 4 . 19. ±irst, we ²nd a general solution to the given equation. Solving the auxiliary equation, r 3 − r 2 − 4 r + 4 = ( r 3 − r 2 ) − (4 r − 4) = ( r − 1)( r 2 − 4) = ( r − 1)( r + 2)( r − 2) = 0 , 353...
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