375_pdfsam_math 54 differential equation solutions odd

# 375_pdfsam_math 54 differential equation solutions odd - 2...

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Exercises 6.3 33. Let us write given equation in operator form. ( D 3 2 D 2 3 D +10 ) [ y ]=(34 x 16) e 2 x 10 x 2 +6 x +34 . By inspection, r = 2 is a root of the characteristic equation, r 3 2 r 2 3 r + 10 = 0. Using, say, long division we Fnd that r 3 2 r 2 3 r +10=( r +2) ( r 2 4 r +5 ) =( r +2) ± ( r 2) 2 +1 ² and so the other two roots of the auxiliary equation are r =2 ± i . This gives a general solution to the corresponding homogeneous equation y h ( x )= c 1 e 2 x +( c 2 cos x + c 3 sin x ) e 2 x . According to the nonhomogeneous term, we look for a particular solution to the original equation of the form y p ( x )= x ( C 0 x + C 1 ) e 2 x + C 2 x 2 + C 3 x + C 4 , where the factor x in the exponential term appears due to the fact that r = 2 is a root of the characteristic equation. Substituting y p ( x ) into the given equation and simplifying yield ( D 3 2 D
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Unformatted text preview: 2 − 3 D + 10 ) [ y p ( x )] = (34 x − 16) e − 2 x − 10 x 2 + 6 x + 34 ⇒ (34 C x + 17 C 1 − 16 C ) e − 2 x + 10 C 2 x 2 + (10 C 3 − 6 C 2 ) x +10 C 4 − 3 C 3 − 4 C 2 = (34 x − 16) e − 2 x − 10 x 2 + 6 x + 34 . Equating corresponding coeﬃcients, we obtain the system 34 C = 34 , 17 C 1 − 16 C = − 16 , 10 C 2 = − 10 , 10 C 3 − 6 C 2 = 6 , 10 C 4 − 3 C 3 − 4 C 2 = 34 ⇒ C = 1 , C 1 = 0 , C 2 = − 1 , C 3 = 0 , C 4 = 3 . Thus, y p ( x ) = x 2 e − 2 x − x 2 + 3 and y ( x ) = y h ( x ) + y p ( x ) = c 1 e − 2 x + ( c 2 cos x + c 3 sin x ) e 2 x + x 2 e − 2 x − x 2 + 3 371...
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