# L09 - Nonhomogeneous SOLOD Suppose p,q,g are continuous...

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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 non–homogeneous 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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L09 - Nonhomogeneous SOLOD Suppose p,q,g are continuous...

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