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Unformatted text preview: Nonhomogeneous SOLOD Suppose p,q,g are continuous functions on some interval ( a,b ) , make the differential operator L [ y ]( t ) def = y 00 ( t ) + p ( t ) y ( t ) + q ( t ) y ( t ) , and consider the General SOLODE L [ y ]( t ) = g ( t ) ( NH ) and the Associated HSOLODE L [ y ]( t ) = 0 ( H ) Theorem: Let { y 1 ( t ) ,y 2 ( t ) } be a FS for ( H ) and Y ( t ) one Particular Solution of ( NH ) . Then Y ( t ) + c 1 y 1 ( t ) + c 2 y 2 ( t ) , c 1 ,c 2 R is the general solution of ( NH ) . Proof: Clearly Y + c 1 y 1 + c 2 y 2 solves ( NH ) . Conversely, if y solves ( NH ) , then y Y solves ( H ) and thus is of the form y Y = c 1 y 1 + c 2 y 2 . Hence y = Y + c 1 y 1 + c 2 y 2 . 2 Example: Suppose you are clever or lucky enough to find one solution y 1 of ( H ) and one particular solution Y of ( NH ) . Then use reduction of order to find a sec ond solution y 2 of ( H ) so that { y 1 ,y 2 } is a FS. Then every solution of ( NH ) is of the form Y ( t ) + c 1 y 1 ( t ) + c 2 y 2 ( t ) , c 1 ,c 2 R . Example: Consider the IVP y 00 + 3 x y 3 x 2 y = 5; y (1) = 1 ,y (1) = 0 . Play with it: rewrite it as x 2 y 00 + 3 xy 3 y = 5 x 2 ; y (1) = 1 ,y (1) = 0 and see that y 1 ( x ) def = x is a solution of the 3 associated homogeneous equation y 00 + 3 x y 3 x 2 y = 0 . Reduction of order provides another solu tion y 1 ( x ) Z y 2 1 ( x ) e R 3 x dx dx = x Z x 2 x 3 dx = x  1 4 x 4 = . 25 x 3 Thus { x,x 3 } is a FS. Looking at the orig inal nonhomogeneous equation for a lit tle while you come up with a particular solution Y ( x ) = x 2 . Therefore its general solution is y = x 2 + c 1 x + c 2 x 3 . 4 Now y (1) = 1 + c 1 + c 2 = 1 = c 1 + c 2 = 0 y (1) = 2 + c 1 3 c 2 = 0 = c 1 = 1 / 2 = c 2 = 1 / 2 The solution to the IVP is y ( x ) = x 2 x/ 2 + x 3 / 2 ....
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 Spring '08
 Turner

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