lec_6 - 6-1 6 Linear Differential Equations of the Second...

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September 15, 2010 6-1 6. Linear Differential Equations of the Second Order A differential equation of the form L ( y ) = g is called linear if L is a linear operator and g = g ( t ) is continuous. The most general second order linear differential equa- tions has the form P ( t ) y 00 + Q ( t ) y 0 + R ( t ) y = G ( t ) where P, Q, R, G are continuous functions defined on an interval I . Assuming that P ( t ) 6 = 0 for t I , we can divide through by P ( t ) and rewrite this d.e. as y 00 + p ( t ) y 0 + q ( t ) y = g ( t ) (1) where p, q, g are all continuous on the interval I . Analogously, we write the IVP y 00 + p ( t ) y 0 + q ( t ) y = g ( t ) , y ( t 0 ) = y 0 , y 0 ( t 0 ) = y 0 0 (2) where p, q, g are all continuous on the interval I , t 0 I , and y 0 , y 0 0 are given constants. The following is an important theorem, usually proved in a more advanced course.
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September 15, 2010 6-2 Theorem(Existence-Uniqueness Theorem for Second Order Linear Differential Equations) . Let p ( t ) , q ( t ) , g ( t ) be continuous functions on the in- terval I , let t 0 I , and let y 0 , y 0 0 be given constants. Then, there is a unique solution y ( t ) to the IVP (2) which is defined on the whole interval I . We are concerned with finding the general solution to (1) , and solving initial value problems. Given equation (1), the associated homogeneous equa- tion is the d.e. y 00 + p ( t ) y 0 + q ( t ) y = 0 (3) A consequence of the next result is that, in order to find the general solution to (1), it suffices to 1. find the general solution y h to (3), (4) and 2. find a particular solution y p to (1). (5) The general solution to (1) is then obtained as y = y h + y p . Theorem. Let y p ( t ) be a particular solution to (1). Then, every solution y ( t ) to (1) can be expressed as
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September 15, 2010 6-3 y ( t ) = y 1 ( t ) + y p ( t ) where y 1 ( t ) is a solution to (3). Conversely, for any solution y 1 ( t ) of (3), the function y ( t ) = y 1 ( t ) + y p ( t ) is a solution to (1). Proof. Let y p ( t ) be a particular solution to (1), and let y ( t ) be any other solution to (1). Consider the function y 1 ( t ) = y ( t ) - y p ( t ) . We clearly have y ( t ) = y 1 ( t )+ y p ( t ). Let us verify that y 1 is a solution to (3) . (6) By linearity, L ( y 1 ) = L ( y ( t ) - y p ( t )) = L ( y ( t )) - L ( y p ( t )) = g ( t ) - g ( t ) = 0 , which verifies (6).
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