Method of Variation of Params SLP 9_06_11

# Method of Variation of Params SLP 9_06_11 - Prof S.L...

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1 Prof. S.L. Phoenix 9/06/11 Math 3100 Method of Variation of Parameters for Higher Order Linear ODE’s Suppose we have the equation     1 1 1 1 n n n n L y y p x y p x y p x y g x  (1) which is eqn. (2) in E&P, except we use g x instead of f x . From the homogeneous equation 0 L y we can get n linearly independent complementary solutions 1 2 , ,..., n y x y x y x . To get a particular solution, ( ) y x Y x , we develop a general useful formula based on the method of variation of parameters. This method begins with the form 1 1 2 2 n n Y x u x y x u x y x u x y x   (2) where 1 2 , , , n u x u x u x  arbitrary, but so far unknown functions. At this stage, this form is general enough to cover all possibilities for the particular solution. In fact it looks like overkill , but as we shall see, we will eventually apply 1 n constraints. To be a particular solution we must get the form (2) to satisfy the differential equation (1). This means we will need to take n derivatives of Y x , which through repeated applications of the chain rule will cause a big mess unless we make some clever simplifications that we can get to work out in the end. For the first derivative we use the chain rule of differentiation to give us 1 1 2 2 1 1 2 2 n n n n Y x u x y x u x y x u x y x u x y x u x y x u x y x     (3) but we simplify this by setting the last half of the sum equal to zero, i.e., we apply the constraint 1 1 2 2 0 n n u x y x u x y x u x y x   (4) Thus we now only have 1 1 2 2 n n Y x u x y x u x y x u x y x   (5) Taking a derivative of (5) we get

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