385_pdfsam_math 54 differential equation solutions odd

# 385_pdfsam_math 54 differential equation solutions odd - 3...

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Exercises 6.4 Therefore, according to formula (12) on page 340 of the text, a particular solution, y p ( x ), will be given by y p ( x )= e x Z 3 e x g ( x ) 6 e 2 x dx + e x Z e 3 x g ( x ) 6 e 2 x dx + e 2 x Z 2 g ( x ) 6 e 2 x dx y p ( x )= 1 2 e x Z e x g ( x ) dx + 1 6 e x Z e x g ( x ) dx + 1 3 e 2 x Z e 2 x g ( x ) dx. 11. First, we ±nd a fundamental solution set to the corresponding homogeneous equation, x 3 y 0 3 xy 0 +3 y =0 . (6.25) Here we involve the procedure of solving Cauchy-Euler equations discussed in Problem 38, Section 4.3. Thus, let x = e t .Then dx/dt = e t = x and so the chain rule yields dy dt = dy dx dx dt = x dy dx , d 2 y dt 2 = d dt ± dy dt ² = d dx ± x dy dx ² dx dt = ³ dy dx + x d 2 y dx 2 ´ x == x dy dx + x 2 d 2 y dx 2 = dy dt + x 2 d 2 y dx 2 , x 2 d 2 y dx 2 = d 2 y dt 2 dy dt , d 3 y dt 3 = d dt ± d 2 y dt 2 ² = d dx ± x dy dx + x 2 d 2 y dx 2 ² dx dt = ³ dy dx +3 x d 2 y dx 2 + x 2 d 3 y dx 3 ´ x = x dy dx +3 x 2 d 2 y dx 2 + x 3 d 3 y dx 3 = dy dt +3 ± d 2 y dt 2 dy dt ² + x 3 d 3 y dx 3 =3 d 2 y dt 2 2 dy dt + x
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Unformatted text preview: 3 d 3 y dx 3 ⇒ x 3 d 3 y dx 3 = d 3 y dt 3 − 3 d 2 y dt 2 + 2 dy dt . Substituting these expressions into (6.25), we obtain ³ d 3 y dt 3 − 3 d 2 y dt 2 + 2 dy dt ´ − 3 ³ dy dt ´ + 3 y = 0 ⇒ d 3 y dt 3 − 3 d 2 y dt 2 − dy dt + 3 y = 0 . The auxiliary equation corresponding to this linear homogeneous equation with constant co-eﬃcients is r 3 − 3 r 2 − r + 3 = 0 ⇒ r 2 ( r − 3) − ( r − 3) = 0 ⇒ ( r − 3)( r + 1)( r − 1) = 0 , 381...
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