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336_pdfsam_math 54 differential equation solutions odd

# 336_pdfsam_math 54 differential equation solutions odd -...

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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 D 3 [ y ] = 0 . Thus on integrating 3 times we get y ( t ) = C 3 + C 2 t + C 1 t 2 . We substitute this solution into the first equation of given system to get x = ( y + y ) = (2 C 1 ) + ( C 3 + C 2 t + C 1 t 2 ) = ( C 3 + 2 C 1 ) + C 2 t + C 1 t 2 . Integrating we obtain x ( t ) = ( C 3 + 2 C 1 ) + C 2 t + C 1 t 2 dt = C 4 ( C 3 + 2 C 1 ) t 1 2 C 2 t 2 1 3 C 1 t 3 . Thus the general solution of the given system is x ( t ) = C 4 ( C 3 + 2 C 1 ) t 1 2 C 2 t 2 1 3 C 1 t 3 , y ( t ) = C 3 + C 2 t + C 1 t 2 . 3. Writing the system in operator form yields (2 D 3)[ x ] ( D + 1)[ y ] = e t , ( 4 D + 15)[ x ] + (3 D 1)[ y ] = e t . (5.40) We eliminate
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