Math 217 - November 29, 2010 Series Solutions 1. What is an - - PowerPoint PPT Presentation

Math 217 - November 29, 2010

◮ Series Solutions

• 1. What is an ordinary point?
• 2. What is a singular point?
• 3. Why do we care about ordinary points and singular points?
• 4. If you find a series solution at an ordinary point, what can you say

about the radius of convergence for the series solution? 5. (1 − t2)y′′ − 6ty′ − 4y = 0 Start with the series y = ∞

n=0 cntn.

(a) Find a recurrance relation for the coefficients. (b) Use the recurrance relation to find c0, c1, c2, c3, c4, c5, and c6. (c) Try to find the solution to the equation.

6. (2x − x2)y′′ − 6(x − 1)y′ − 4y = 0 Start with the series y = ∞

n=0 cn(x − 1)n.

(a) Find a recurrance relation for the coefficients. (b) Use the recurrance relation to find c0, c1, c2, c3, c4, c5, and c6. Hint: 2x − x2 = 1 − (x − 1)2 (c) Try to find the solution to the equation.

• 1. What is an ordinary point?

Solution: See page 518.

• 2. What is a singular point?
• 3. Why do we care about ordinary points and singular points?
• 4. If you find a series solution at an ordinary point, what can you say

about the radius of convergence for the series solution? 5. (1 − t2)y′′ − 6ty′ − 4y = 0 Start with the series y = ∞

n=0 cntn.

(a) Find a recurrance relation for the coefficients. (b) Use the recurrance relation to find c0, c1, c2, c3, c4, c5, and c6. Solution: cn+2 = n + 4 n + 2cn (c) Try to find the solution to the equation.

6. (2x − x2)y′′ − 6(x − 1)y′ − 4y = 0 Start with the series y = ∞

n=0 cn(x − 1)n.

(a) Find a recurrance relation for the coefficients. (b) Use the recurrance relation to find c , c , c , c , c , c , and c .

Lecture Problems

• 7. Use a substitution to transform the differential equations

(a) (x + 1)2y ′′ − 2(x + 1)y ′ − y = x, y(−1) = 5, y ′(−1) = 2 (b) (x + 1)2y ′′ − 2(x + 1)y ′ − y = x, y(3) = 5, y ′(3) = 2

Lecture Problems

• 7. Use a substitution to transform the differential equations

(a) Solution: Use t = x + 1. (x + 1)2y ′′ − 2(x + 1)y ′ − y = x, y(−1) = 5, y ′(−1) = 2 t2y ′′ − 2ty ′ − y = t − 1 (b) Solution: Use t = x − 3. (x + 1)2y ′′ − 2(x + 1)y ′ − y = x, y(3) = 5, y ′(3) = 2 (t + 4)2y ′′ − 2(t + 4)y ′ − y = t + 3

• 8. Given the recurrance relation, complete the solution.

cn+2 = −n + 4 n + 2cn

• 9. Given the recurrance relation, complete the solution.

cn+2 = (n − 3)(n − 4) (n + 1)(n + 2)cn

• 8. Given the recurrance relation, complete the solution.

cn+2 = −n + 4 n + 2cn Solution: c2n =(−1)n(n + 1)c0 c2n+1 =(−1)n 1 3(2n + 3)c1

• 9. Given the recurrance relation, complete the solution.

cn+2 = (n − 3)(n − 4) (n + 1)(n + 2)cn Solution: y = c0(1 + 6x2 + x4) + c1(x + x3)

• 10. Given the recurrance relation, find the solution to the DE

c2 = 1 2c0, cn+2 = cn − 2cn−1

• 11. Given the recurrance relation, find the solution to the DE

c2 = c3 = c4 = c5 = 0, cn+2 = cn−4 + n2cn+2 (n + 1)(n + 2)