q5sol - q x is continuous for x 6 = 0 • g x is continuous...

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MAP 2302 — SOLUTION TO QUIZ 5 — FALL 2002 [6 points ] Use Theorem 2 (The Existence/Uniqueness Theorem for 2nd order linear DE’s) to discuss the existence and uniqueness of a solution to the Initial Value Problem: x 2 d 2 y dx 2 + (1 - x ) dy dx - 3 y = tan x, y (1) = 0 , y 0 (1) = π 2 . Solution : The DE is equivalent to d 2 y dx 2 + (1 - x ) x 2 dy dx - 3 x 2 y = tan x x 2 , provided x 6 = 0. Let p ( x ) = (1 - x ) x 2 , q ( x ) = - 3 x 2 , g ( x ) = tan x x 2 . Then p ( x ) is continuous for x 6 = 0.
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Unformatted text preview: q ( x ) is continuous for x 6 = 0. • g ( x ) is continuous for x 6 = 0, ± π 2 , ± 3 π 2 ... . The three functions p ( x ), q ( x ) and g ( x ) are all continous on the open interval (0 , π 2 ), which contains x = 1. The Theorem implies that the IVP has a unique solution on the open interval (0 , π 2 ). 1...
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