198_pdfsam_math 54 differential equation solutions odd

198_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 So, y 0 p 7 y 0 p =(6 A 2 θ +2 A 1 ) 7 ( 3 A 2 θ 2 +2 A 1 θ + A 0 ) = 21 A 2 θ 2 +(6 A 2 14 A 1 ) θ +2 A 1 7 A 0 = θ 2 . Comparing the corresponding coefficients, we fnd A 2 , A 1 ,and A 0 . 21 A 2 =1 , 6 A 2 14 A 1 =0 , 2 A 1 7 A 0 =0 A 2 = 1 / 21 , A 1 =3 A 2 / 7= 1 / 49 , A 0 =2 A 1 / 7= 2 / 343 . Hence y p ( θ )= 1 21 θ 3 1 49 θ 2 2 343 θ. 25. We look For a particular solution oF the Form y p ( t )= t s ( A cos 3 t + B sin 3 t ) e 2 t .S ince r =2+3 i is not a root oF the auxiliary equation, which is r 2 +2 r +4=0,wecantake s =0 . Thus , y p ( t )=( A cos 3 t + B sin 3 t ) e 2 t y 0 p ( t )=[(2 A +3 B )cos3 t +( 3 A +2 B )sin3 t )] e 2 t y 0 p ( t )=[( 5 A +12 B )cos3 t +( 12 A 5 B )sin3 t )] e 2 t . Next, we substitute y p , y 0 p ,and y 0
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