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initial value problem ly" — xy' + 3y = 0 subject to the initial condition y(0) = 3, y' (0) = 2.

initial value problem ly" — xy' + 3y = 0 subject to the initial condition y(0) = 3, y' (0) = 2.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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In this exercise we consider finding the first five coefficients in the series solution of the second order linear initial value problem 1y" — xy' + 3y = 0 subject to the initial condition y(0) = 3, y′ (0) = 2. Since the equation has an ordinary point at x = 0 and it has a power series solution in the form Cn = has been factored out. (1) Insert the formal power series into the differential equation and derive the recurrence relation Cn-2 for n = 2, 3, … . . The solution to this initial value problem can be written in the form y(x) = coy₁(x) + C₁Y2(x) where co and c₁ are determined from the initial conditions. The function y₁ (x) is an even function and y₂(x) is an odd function. For this example, from the initial conditions, we have c = A2 = The function y₁ (x) is an infinite series y₁ (x) = 1 + " Y a 4 ∞ n=0 = = Cn xn (2) Use the recurrence relation to find the first few coefficients of the infinite series ∞ and C1 = 2k a2kx² NOTE note that the constant co a6 Finally the polynomial y₂(x) NOTE The function y₂(x) is an odd degree polynomial with first term ä. In other words, note that the constant c₁ has been factored out.
Transcribed Image Text:In this exercise we consider finding the first five coefficients in the series solution of the second order linear initial value problem 1y" — xy' + 3y = 0 subject to the initial condition y(0) = 3, y′ (0) = 2. Since the equation has an ordinary point at x = 0 and it has a power series solution in the form Cn = has been factored out. (1) Insert the formal power series into the differential equation and derive the recurrence relation Cn-2 for n = 2, 3, … . . The solution to this initial value problem can be written in the form y(x) = coy₁(x) + C₁Y2(x) where co and c₁ are determined from the initial conditions. The function y₁ (x) is an even function and y₂(x) is an odd function. For this example, from the initial conditions, we have c = A2 = The function y₁ (x) is an infinite series y₁ (x) = 1 + " Y a 4 ∞ n=0 = = Cn xn (2) Use the recurrence relation to find the first few coefficients of the infinite series ∞ and C1 = 2k a2kx² NOTE note that the constant co a6 Finally the polynomial y₂(x) NOTE The function y₂(x) is an odd degree polynomial with first term ä. In other words, note that the constant c₁ has been factored out.
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