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Chem Differential Eq HW Solutions Fall 2011 193

Chem Differential Eq HW Solutions Fall 2011 193 - a = 0 and...

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Section A.5 The Power Series Method, Part II A193 So y + 2 xy + y = 0 implies that ( m + 2)( m + 1) a m +2 + (2 m + 1) a m = 0 a m +2 = - (2 m + 1) ( m + 2)( m + 1) a m for all m 0 . So a 0 and a 1 are arbitrary; a 2 = - 1 2 a 0 , a 4 = - 5 4 · 3 a 2 = 5 4! a 0 , a 6 = - 9 6 · 5 5 4! a 0 = - 9 · 5 6! a 0 , . . . a 3 = - 3 3 · 2 a 1 a 5 = - 7 5 · 4 a 3 = 7 · 3 5! a 1 a 7 = - 11 · 7 · 3 7! a 1 . . . So y = a 0 1 - 1 2 x 2 + 5 4! x 4 - 9 · 5 6! x 6 + · · · + a 1 x - 3 3! x 3 + 7 · 3 5! x 5 - 11 · 7 · 3 7! x 7 + · · · . 13. To solve (1 - x 2 ) y - 2 xy + 2 y = 0, y (0) = 0, y (0) = 3, follow the steps in Example 5 and you will arrive at the recurrence relation a m +2 = m ( m + 1) - 2 ( m + 2)( m + 1) a m = ( m + 2)( m - 1) ( m + 2)( m + 1) a m = m - 1 m + 1 a m , m
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Unformatted text preview: a = 0 and a 1 = 3. So a 2 = a 4 = ··· = 0 and, from the recurrence relation with m = 1, a 3 = (1-1 1 + 1 a = 0 . So a 5 = a 7 = ··· = 0 and hence y = 3 x is the solution. 17. Put the equation (1-x 2 ) y ±±-2 xy ± + 2 y = 0 in the form y ±±-2 x 1-x 2 + 2 1-x 2 y = 0 ....
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