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268_pdfsam_math 54 differential equation solutions odd

268_pdfsam_math 54 differential equation solutions odd - t...

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Chapter 5 11. From the second equation, we obtain u = ( D 2 + 2) [ v ] / 2. Substitution into the first equation eliminates u and gives ( D 2 1 ) 1 2 ( D 2 + 2 ) [ v ] + 5 v = e t ( D 2 1 ) ( D 2 + 2 ) 10 [ v ] = 2 e t ( D 4 + D 2 12 ) [ v ] = 2 e t . (5.4) Solving the characteristic equation, r 4 + r 2 12 = 0, r 4 + r 2 12 = 0 ( r 2 + 4 ) ( r 2 3 ) = 0 r = ± 2 i, ± 3 , we conclude that a general solution to the corresponding homogeneous equation is v h ( t ) = c 1 cos 2 t + c 2 sin 2 t + c 3 e 3 t + c 4 e 3 t . A particular solution to (5.4) has the form v p ( t ) = ce t . Substitution yields ( D 4 + D 2 12 ) ce t = ce t + ce t 12 ce t = 10 ce t = 2 e t c = 1 5 . Therefore, v = v h + v p = c 1 cos 2 t + c 2 sin 2 t + c 3 e 3 t + c 4 e 3 t + e t / 5 and u = 1 2 ( D 2 + 2 ) [ v ] = 1 2 c 1 cos 2 t + c 2 sin 2 t + c 3 e 3 t + c 4 e 3 t + 1 5 e t c 1 cos 2 t + c 2 sin 2
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Unformatted text preview: t + c 3 e √ 3 t + c 4 e − √ 3 t + 1 5 e t ´ = c 1 cos 2 t + c 2 sin 2 t − 5 2 c 3 e √ 3 t − 5 2 c 4 e − √ 3 t − 3 10 e t . By replacing ( − 5 / 2) c 3 by c 3 and ( − 5 / 2) c 4 by c 4 we obtain the same answer as given in the text. 13. Expressing x ±rom the second equation and substituting the result into the frst equation, we obtain x = y − y ⇒ d ( y − y ) dt = ( y − y ) − 4 y ⇒ y − 2 y + 5 y = 0 . 264...
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