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WrittenHW03-Solutions-1

# WrittenHW03-Solutions-1 - Math 216(Section 50 Written...

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Math 216 (Section 50) Written Homework #3 – Solutions Fall 2007 1 ( 2.4–2 ) The IVP (Initial Value Problem) to solve is y = f ( x, y ) , y (0) = 1 2 , where f ( x, y ) = 2 y. With ( x 0 , y 0 ) = (0 , 1 / 2), and a given value of h , Euler’s method looks like the following: y 1 = y 0 + h · f ( x 0 , y 0 ) = y 0 + h · 2 y 0 , y 2 = y 1 + h · f ( x 1 , y 1 ) = y 1 + h · 2 y 1 , y 3 = y 2 + h · f ( x 2 , y 2 ) = y 2 + h · 2 y 2 , y 4 = y 3 + h · f ( x 3 , y 3 ) = y 3 + h · 2 y 3 , . . . h = 0 . 25 Using Euler’s method shown above, one obtains y 1 = y 0 + h · 2 y 0 = 1 / 2 + 0 . 25 · (2 · 1 / 2) = 0 . 75 , y 2 = y 1 + h · 2 y 1 = 0 . 75 + 0 . 25 · (2 · 0 . 75) = 1 . 125 . h = 0 . 1 Similarly, y 1 = y 0 + h · 2 y 0 = 1 / 2 + 0 . 1 · (2 · 1 / 2) = 0 . 6 , y 2 = y 1 + h · 2 y 1 = 0 . 6 + 0 . 1 · (2 · 0 . 6) = 0 . 72 , y 3 = y 2 + h · 2 y 2 = 0 . 72 + 0 . 1 · (2 · 0 . 72) = 0 . 864 , y 4 = y 3 + h · 2 y 3 = 0 . 864 + 0 . 1 · (2 · 0 . 864) = 1 . 0368 , y 5 = y 4 + h · 2 y 4 = 1 . 0368 + 0 . 1 · (2 · 1 . 0368) = 1 . 24416 1 . 244 . Exact solution Using the given solution y ( x ) = 1 2 e 2 x , y (1 / 2) 1 . 359 . 2 ( 2.4–6 ) The IVP to solve is y = f ( x, y ) , y (0) = 2 , where f ( x, y ) = - 2 xy. With ( x 0 , y 0 ) = (0 , 2), and a given value of h , Euler’s method looks like the following: y 1 = y 0 + h · f ( x 0 , y 0 ) = y 0 - 2 hx 0 y 0 , y 2 = y 1 + h · f ( x 1 , y 1 ) = y 1 - 2 hx 1 y 1 , y 3

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