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274_pdfsam_math 54 differential equation solutions odd

274_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 We use the second equation to express z in terms of x and y . z = x + D [ y ] . (5.10) Substituting this expression into the other two equations, we obtain ( D 1)[ x ] 2 y + ( x + D [ y ]) = 0 , 4 x + 4 y + ( D 5)[ x + D [ y ]] = 0 ( D 2)[ x ] + ( D 2)[ y ]) = 0 , ( D 1)[ x ] + ( D 2 5 D + 4) [ y ] = 0 . (5.11) Now we eliminate x by multiplying the first equation by ( D 1), the second equation – by ( D 2), and adding the results. This yields ( D 1)( D 2) + ( D 2)( D 2 5 D + 4) [ y ] = 0 ( D 2)( D 2 4 D + 3) [ y ] = 0 { ( D 2)( D 1)( D 3) } [ y ] = 0 . The roots of the characteristic equation, ( r 2)( r 1)( r 3) = 0, are r = 1 , 2, and 3. Thus, a general solution for y is y = c 1 e t + c 2 e 2 t + c 3 e 3 t . With h := x + y , the first equation in (5.11) can be written in the form
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