269_pdfsam_math 54 differential equation solutions odd

269_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.2 This homogeneous linear equation with constant coefficients has the characteristic equation r 2 2 r + 5 = 0 with roots r =1 ± 2 i . Thus a general solution is y = c 1 e t cos 2 t + c 2 e t sin 2 t. Therefore, x = ( c 1 e t cos 2 t + c 2 e t sin 2 t ) 0 ( c 1 e t cos 2 t + c 2 e t sin 2 t ) = ( c 1 e t cos 2 t 2 c 1 e t sin 2 t + c 2 e t sin 2 t +2 c 2 e t cos 2 t ) ( c 1 e t cos 2 t + c 2 e t sin 2 t ) =2 c 2 e t cos 2 t 2 c 1 e t sin 2 t. 15. In operator form, the system becomes 2 z +( D 5)[ w ]=5 t, ( D 4)[ z ] 3 w 7 t. We multiply the Frst equation by 3, the second equation by ( D 5), and add the resulting equations. {− 6+( D 5)( D 4) } [ z ]=3(5 t )+( D 5)[17 t ]= 70 t +17 ( D 2 9 D +14 ) [ z 70 t . Solving the characteristic equation, r 2 9 r + 14 = 0, we obtain r , 7. Hence, a general solution to the corresponding homogeneous equation is
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