368_pdfsam_math 54 differential equation solutions odd

368_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 Diferentiating y p ( x ) yields y 0 p ( x )=2 C 1 xe 2 x 2 C 1 x 2 e 2 x =2 C 1 ( x x 2 ) e 2 x y 0 p ( x )= 4 C 1 ( x x 2 ) e 2 x +2 C 1 (1 2 x ) e 2 x C 1 (2 x 2 4 x +1) e 2 x y 0 p ( x 4 C 1 (2 x 2 4 x e 2 x C 1 (4 x 4) e 2 x =4 C 1 ( 2 x 2 +6 x 3) e 2 x . By substituting these expressions into the nonhomogeneous equation, we obtain y 0 p ( x )+3 y 0 p ( x ) 4 y p ( x )=4 C 1 ( 2 x 2 x 3) e 2 x +6 C 1 (2 x 2 4 x e 2 x 4 C 1 x 2 e 2 x = e 2 x ⇒− 6 C 1 e 2 x = e 2 x . By equating coefficients, we see that C 1 = 1 / 6. Thus, a general solution to the nonhomoge- neous diferential equation is given by y ( x y h ( x )+ y p ( x c 1 e x + c 2 e 2 x + c 3 xe 2 x 1 6 x 2 e 2 x . 9. Solving the auxiliary equation, r 3 3 r 2 +3 r 1=( r 1) 3 = 0, we Fnd that r = 1 is its root o± multiplicity three. There±ore, a general solution to the associated homogeneous equation is
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