336_pdfsam_math 54 differential equation solutions odd

336_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 5 Eliminating x by applying D to the first equation and subtracting the second equation from it yields D D 2 + 1 − D [y ] = 0 Thus on integrating 3 times we get y (t) = C3 + C2 t + C1 t2 . We substitute this solution into the first equation of given system to get x = − (y + y ) = − (2C1 ) + (C3 + C2 t + C1 t2 ) = − (C3 + 2C1 ) + C2 t + C1 t2 . Integrating we obtain x(t) = − (C3 + 2C1 ) + C2 t + C1 t2 dt = C4 − (C3 + 2C1 )t − 1 1 C2 t2 − C1 t3 . 2 3 ⇒ D 3 [y ] = 0. Thus the general solution of the given system is x(t) = C4 − (C3 + 2C1 )t − y (t) = C3 + C2 t + C1 t2 . 3. Writing the system in operator form yields (2D − 3)[x] − (D + 1)[y ] = et , (−4D + 15)[x] + (3D − 1)[y ] = e−t . (5.40) 1 1 C2 t2 − C1 t3 , 2 3 We eliminate y by multiplying the first equation by (3D − 1), the second – by (D + 1), and summing the results. {(2D − 3)(3D − 1) + (−4D + 15)(D + 1)} [x] = (3D − 1)[et ] + (D + 1)[e−t ] ⇒ (D 2 + 9)[x] = et . Since the characterictic equation, r 2 + 9 = 0, has roots r = ±3i, a general solution to the corresponding homogeneous equation is xh (t) = c1 cos 3t + c2 sin 3t. 332 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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