(D^2+4)y=x^2sin2x please completely to do it Solution Let y0 = the complimentary

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Let y0 = the complimentary solution that corresponds to the solution of the homogeneous equation ( D ² + 4 ) y0 = 0 ---> f ( D ) y0 = 0 where the differential operator function f ( D ) is f ( D ) = D ² + 4 …… …… where D = d / d x …… since y0 ? 0 , then f ( D ) = D ² + 4 = 0 …… Consider the auxiliary equation f ( s ) = s ² + 4 = 0 ---> s ² = – 4 …… which gives the two distinct imaginary roots s1 = + 2 i and s2 = – 2 i … The complimentary solution is therefore …… …… y0 = c1 exp ( s1 x ) + c2 exp ( s2 x ) = c1 exp ( + 2 i x ) + c2 exp ( – 2 i x ) Using the definition of complex exponential, we have the equivalent eq. …… y0 = c1 cos ( 2 x ) + c2 sin ( 2 x ) …… c1 , c2 are constants. For the particular solution y1 of the non-homogeneous DE, we use variation of parameters …… y1 = u1 ( x ) cos ( 2 x ) + u2 ( x ) sin ( 2 x ) …… where u1 ( x ) , u2 ( x ) are functions to be determined …… see …… Now , taking the first derivative …… …… y1' = u1' cos ( 2 x ) – 2 u1 sin ( 2 x ) + u2' sin ( 2 x ) + 2 u2 cos ( 2 x ) … where we require that …… u1' cos ( 2 x ) + u2' sin ( 2 x ) = 0 …… so that …… y1' = – 2 u1 sin ( 2 x ) + 2 u2 cos ( 2 x ) …… which is now simpler …… Taking the second derivative …… …… y1'' = – 2 u1' sin ( 2 x ) – 4 u1 cos ( 2 x ) + 2 u2' cos ( 2 x ) – 4 u2 sin ( 2 x ) Substituting in the given DE with y = y1 , we get …… – 2 u1' sin ( 2 x ) + 2 u2' cos ( 2 x ) = x ² sin ( 2 x ) …… The last equation, together with the condition u1' cos ( 2 x ) + u2' sin ( 2 x ) = 0 make up the system of equation that will be solved for u1 ( x ) , u2 ( x ) . Eliminating u2 ( x ) , we get …… u1 ( x ) = – ½ ? x ² sin ² ( 2 x ) d x …… Using sin ² ( 2 x ) = ½ [ 1 – cos ( 4 x ) ] and integrating by parts two times …… u1 ( x ) = – [ x ³ / 12 ] + { [ 8 x ² – 1 ] sin ( 4 x ) / 128 } + x cos ( 4 x ) / 32 …… Eliminating u1 ( x ) , we get …… u2 ( x ) = ½ ? x ² sin ( 2 x ) cos ( 2 x ) d x … Using sin ( 4 x ) = 2 sin ( 2 x ) cos ( 2 x ) and integrating by parts two times …… u2 ( x ) = { [ 1 – 8 x ² ] cos ( 4 x ) + 4 x sin ( 4 x ) } / 128 …… it follows that …… …… y1 = u1 ( x ) cos ( 2 x ) + u2 ( x ) sin ( 2 x ) ………. = – [ x ³ / 12 ] cos ( 2 x ) + { [ 8 x ² – 1 ] sin ( 4 x ) cos ( 2 x ) / 128 } ………… + { x cos ( 2 x ) cos ( 4 x ) / 32 } + [ 1 – 8 x ² ] sin ( 2 x ) cos ( 4 x ) / 128 ………… + 4 x sin ( 2 x ) sin ( 4 x ) / 128 The general solution of the given DE is …… y = y0 + y1 ……

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