Notes PDF
Introductory Problems
1. Find a power series y(x) = ∑n = 0∞anxn that satisfies y’ = y and y(0) = 1.
Answer
1. 1. The series is y(x) = 1 + x + x2/2! + x3/3! + … = ex.
Solution
2. Find a power series y(x) = ∑n = 0∞anxn that satisfies xy” + (1 – x)y’ – y = 0 and y(0) = 2.
Answer
1. The series is y(x) = 2 + 2x + x2 + x3/3 + x4/(4⋅3) + x5/(5⋅4⋅3) +…
Solution
1. Differentiating y(x) = a0 + a1x + a2x2 + a3x3 + a4x4 + … gives: y'(x) = a1 + a22x + a33x2 + a44x3 + a55x4 + …y”(x) = a22 + a33⋅2x + a44⋅3x2 + a55⋅4x3 + a66⋅5x4 + Then: xy'(x) = a1x + a22x2 + a33x3 + a44x4 + a55x5 + …xy”(x) = a22x + a33⋅2x2 + a44⋅3x3 + a55⋅4x4 + a66⋅5x5 + … Equating coefficients of like powers of x gives: xy” y’ – xy’ -y = 0 gives x0 0 a1 – 0 -a0 = 0 a1 = a0 x1 2a2 2a2 – a1 -a1 = 0 4a2 – 2a1 = 0, so a2 = a1/2 x2 3⋅2a3 3a3 – 2a2 -a2 = 0 9a3 – 3a2 = 0, so a3 = a2/3 x3 4⋅3a4 4a4 – 3a3 -a3 = 0 16a4 – 4a3 = 0, so a4 = a3/4
All the coefficients can be expressed in terms of a0, giving an = a0/n! and consequently:
y(x) = a0(1 + x + x2/2 + x3/3! + x4/4! + …) = a0ex.The condition y(0) = 2 implies a0 = 2.
3. Prove that on its domain, (1 – sin(x))-2 = 1 + 2sin(x) + 3sin2(x) + 4sin3(x) + …
Solution
4. By dividing the series, find the coefficients, through the cubic term, of (1 – x)/ex.
Answer
1. The series for (1 – x)/ex begins 1 – 2x + (3/2)x2 – (2/3)x3.
Solution
1. Mimic long division of polynomials.