VariationParameters

VariationParameters - Variation of the Parameters This is...

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Variation of the Parameters This is another method we can apply to find a particular solution of a non-homogeneous L.D.E. . We will develop the method in connection with a second order linear O.D.E. with variables coefficients, a 0 (x)y’’ + a 1 (x) y’ + a 2 (x)y = b(x) (1) Suppose that y 1 (x) and y 2 (x) are linearly independent solutions of the corresponding homogeneous equation, a 0 (x)y’’ + a 1 (x) y’ + a 2 (x)y = 0 then the complementary solution of equation (1) is: y c (x) = c 1 y 1 (x) + c 2 y 2 (x) where c 1 and c 2 are arbitrary constants. Replace the arbitrary constants c 1 and c 2 in the complementary solution by two functions v 1 (x) and v 2 (x) to be determined such that y p (x) = v 1 (x)y 1 (x) + v 2 (x)y 2 (x) is a particular solution of equation (1). This is the first condition new impose in order to determine v 1 (x) and v 2 (x). Since there are two functions, we still have a second condition to impose, provided that this second condition does not violate the first condition. Let’s differentiate the function y p : y p ’ = v 1 (x) y 1 ’(x) + v 1 ’(x)y 1 (x) + v 2 (x) y 2 ’(x) + v 2 ’(x)y 2 (x). At this point we impose the second condition. We simplify y p ’ by demanding v 1 ’(x)y 1 (x) + v 2 ’(x)y 2 (x) = 0 So, y p ’ becomes y p ’ = v 1 (x) y 1 ’(x) + v 2 (x) y 2 ’(x) And the second derivative is y p ’’ = v 1 (x) y 1 ’’(x) + v 1 ’(x)y 1 ’(x) + v 2 (x) y 2 ’’(x) + v 2 ’(x)y 2 ’(x). Substituting in equation(1) a 0 (x) [v 1 (x) y 1 ’’(x) + v 1 ’(x)y 1 ’(x) + v 2 (x) y 2 ’’(x) + v 2 ’(x)y 2 ’(x)] + a 1 (x)[ v 1 (x) y 1 ’(x) + v 2 (x) y 2 ’(x)] + a 2 (x) [v 1 (x)y 1 (x) + v 2 (x)y 2 (x)] = b(x) or v 1 (x)[ a 0 (x)y 1 ’’(x) + a 1 (x) y 1 ’(x) + a 2 (x)y 1 (x)] + v 2 (x)[ a 0 (x)y 2 ’’(x) + a 1 (x) y 2 ’(x) + a 2 (x)y 2 (x)] + a 0 (x)[ v 1 ’(x)y 1 ’(x) + v 2 ’(x)y 2 ’(x)] = b(x) Since y 1 and y 2 are solutions of the corresponding homogeneous equation, the first two terms on the left side of the equation are zero. Then, a 0 (x)[ v 1 ’(x)y 1 ’(x) + v 2 ’(x)y 2 ’(x)] = b(x) or v 1 ’(x)y 1 ’(x) + v 2 ’(x)y 2 ’(x) = b(x)/ a 0 (x) So, the two imposed conditions have created a system of two equations that the derivatives of the two unknown functions v 1 and v 2 are satisfying.
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v 1 ’(x)y 1 (x) + v 2 ’(x)y 2 (x) = 0 v 1 ’(x)y 1 ’(x) + v 2 ’(x)y 2 ’(x) = b(x) a 0 (x) ! " # $ # The determinant of the system is the Wronskian of y
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VariationParameters - Variation of the Parameters This is...

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