# q5sol - 2 t = sin t cos 2 t E 3 From E 3 we get u 2 = Z ±...

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MA366 Sathaye Quiz 5 Name/ Sec./Row Use the method of variation of parameters to solve the diﬀerential equation y 00 + y = sin( t ) cos( t ) Be sure to show all work. Answer: First, y h is a solution to ( D 2 + 1) y = 0. The roots of D 2 + 1 = 0 are D = ± i , so y h = c 1 cos( t ) + c 2 sin( t ). We guess that y = u 1 cos( t ) + u 2 sin( t ) where u 1 ,u 2 are functions. By the usual method of variation of parameters we get: D ( y ) = u 1 ( - sin( t )) + u 2 (cos( t )) + ( u 0 1 cos( t ) + u 0 2 sin( t )) . We set u 0 1 cos( t ) + u 0 2 sin( t ) = 0 ( E 1) Then D 2 ( y ) = - u 1 cos( t ) - u 2 sin( t ) + ( - u 0 1 sin( t ) + u 0 2 cos( t )) . So, our equation becomes: ( D 2 +1) y = - u 1 cos( t ) - u 2 sin( t )+ u 1 cos( t )+ u 2 sin( t )+( - u 0 1 sin( t ) + u 0 2 cos( t )) = sin( t ) cos( t ) . This simpliﬁes to - u 0 1 sin( t ) + u 0 2 cos( t ) = sin( t ) cos( t ) ( E 2) . If we calculate sin( t ) E 1 + cos( t ) E 2 then we have: u 0 2 (sin 2 ( t ) + cos
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Unformatted text preview: 2 ( t )) = sin( t ) cos 2 ( t ) ( E 3) . From E 3 we get u 2 = Z ± sin( t ) cos 2 ( t ) dt ² =-cos 3 ( t ) / 3 + a 2 ( E 4) . Similarly cos( t ) E 1-sin( t ) E 2 gives: u 1 (cos 2 ( t ) + sin 2 ( t ) =-sin 2 ( t ) cos( t ) ( E 5) . From E 5 we get u 1 = Z-± sin 2 ( t ) cos( t ) dt ² =-sin 3 ( t ) / 3 + a 1 ( E 6) . • Finally, from E 4 ,E 6 we deduce the solution: y = u 1 cos( t ) + u 2 sin( t ) = cos( t ) ±-sin 3 ( t ) / 3 + a 1 ² + sin( t ) ±-cos 3 ( t ) / 3 + a 2 ² . 1...
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## This note was uploaded on 01/18/2012 for the course MATH 366 taught by Professor Edraygoins during the Fall '09 term at Purdue.

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