Math 217 - November 19, 2010 Series Solutions 1. Find the recursion - PowerPoint PPT Presentation

Math 217 - November 19, 2010 ◮ Series Solutions

1. Find the recursion relation for the equation x 3 y ′ = 2 y 2. Find the recursion relation for the equation y ′′ + y = x 3. Find the recursion relation for the equation y ′′ − 2 y ′ + y = 0

1. Find the recursion relation for the equation x 3 y ′ = 2 y Solution: c 0 = c 1 = c 2 = 0, c n +2 = nc n / 2 and therefore c n = 0 for all n 2. Find the recursion relation for the equation y ′′ + y = x Solution: c 2 = − c 0 / 2, c 3 = − ( c 1 − 1) / 6, c n +2 = − c n / (( n + 1)( n + 2)) for n > 1. 3. Find the recursion relation for the equation y ′′ − 2 y ′ + y = 0 Solution: c n +1 = 2 nc n − c n − 1 n ( n +1)

4. (fun problem) Determine the sum 1 1 − 1 3 + 1 5 − 1 7 + · · ·

4. (fun problem) Determine the sum 1 1 − 1 3 + 1 5 − 1 7 + · · · Solution: π/ 4, this is the series for arctan x with x = 1.

Lecture Problems 5. For the recursive relation, find the series. c n +2 = nc n c 0 = c 1 = c 2 = 0 , 2 6. For the recursive relation, find the series. c 3 = − c 1 − 1 c 2 = − c 0 c n 2 , , c n +2 = − ( n + 1)( n + 2) for n ≥ 2 6 7. Use initial conditions y (0) = 0 and y ′ (0) = 1, use the recursive relation to find the series. c n +1 = 2 nc n − c n − 1 n ( n + 1)

Lecture Problems 5. For the recursive relation, find the series. c n +2 = nc n c 0 = c 1 = c 2 = 0 , 2 Solution: y = 0 6. For the recursive relation, find the series. c 3 = − c 1 − 1 c 2 = − c 0 c n 2 , , c n +2 = − ( n + 1)( n + 2) for n ≥ 2 6 y = x + c 0 cos x + ( c 1 − 1) sin x 7. Use initial conditions y (0) = 0 and y ′ (0) = 1, use the recursive relation to find the series. c n +1 = 2 nc n − c n − 1 n ( n + 1) 1 Solution: c n = ( n − 1)! , y = xe x