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In this exercise we consider finding the general solution of the first order linear equation y' - 6xy = 0. This equation has an ordinary point at x = 0 and therefore has a power series solution in the form an = 0 a2 = 3 ao We learned how to easily solve problems like this in several different ways but here we want to consider the power series method. (1) Insert the formal power series into the differential equation and derive the recurrence relation an-2 for n = = 1, 2, ... NOTE that an is an arbitrary constant and a₁ = 0. The recurrence relation then implies that a2k+1 0 for k = 0, 1, 2, .... Thus we see that the solution series can be written using only the - even terms as y = y = ao, a4= 18/4 y = ao (2) Use the recurrence relation to find the first few coefficients in terms of a ∞ Σ n=0 ∞ Anxn Σ k=0 k=0 n Azkx 2k (3) Write out a few more terms if needed but try to guess the pattern for the general term a2k9n/(n+4) ao for all k = 1, 2, … … (4) Therefore the general solution, with arbitrary constant a can be written as ao, a6 18/4 ao, ag = ₂,2k X 27/8 Notice The general solution of this first order linear equation is y = aºе³ 3x²

In this exercise we consider finding the general solution of the first order linear equation y' - 6xy = 0. This equation has an ordinary point at x = 0 and therefore has a power series solution in the form an = 0 a2 = 3 ao We learned how to easily solve problems like this in several different ways but here we want to consider the power series method. (1) Insert the formal power series into the differential equation and derive the recurrence relation an-2 for n = = 1, 2, ... NOTE that an is an arbitrary constant and a₁ = 0. The recurrence relation then implies that a2k+1 0 for k = 0, 1, 2, .... Thus we see that the solution series can be written using only the - even terms as y = y = ao, a4= 18/4 y = ao (2) Use the recurrence relation to find the first few coefficients in terms of a ∞ Σ n=0 ∞ Anxn Σ k=0 k=0 n Azkx 2k (3) Write out a few more terms if needed but try to guess the pattern for the general term a2k9n/(n+4) ao for all k = 1, 2, … … (4) Therefore the general solution, with arbitrary constant a can be written as ao, a6 18/4 ao, ag = ₂,2k X 27/8 Notice The general solution of this first order linear equation is y = aºе³ 3x²

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 general solution of the first order linear equation y' - 6xy = 0. This equation has an ordinary point at x = 0 and therefore has a power series solution in the form an = 0 NOTE that ao is an arbitrary constant and a a2k+1 even terms as We learned how to easily solve problems like this in several different ways but here we want to consider the power series method. (1) Insert the formal power series into the differential equation and derive the recurrence relation an-2 for n = 1, 2, .. a2 ao 0. The recurrence relation then implies that = 0 for k = 0, 1, 2, . Thus we see that the solution series can be written using only the = Y 3 a2k 9n/(n+4) = Y ao, a4 = 18/4 ao (2) Use the recurrence relation to find the first few coefficients in terms of y = ao ∞ n=0 k=0 = n Anxn ∞ k=0 a zk x ² k ao, a6 (3) Write out a few more terms if needed but try to guess the pattern for the general term ao for all k = 18/4 1, 2, . (4) Therefore the general solution, with arbitrary constant a can be written as αχ, ag x 2k Notice The general solution of this first order linear equation is y = = a0 € ³x² 27/8
Transcribed Image Text:In this exercise we consider finding the general solution of the first order linear equation y' - 6xy = 0. This equation has an ordinary point at x = 0 and therefore has a power series solution in the form an = 0 NOTE that ao is an arbitrary constant and a a2k+1 even terms as We learned how to easily solve problems like this in several different ways but here we want to consider the power series method. (1) Insert the formal power series into the differential equation and derive the recurrence relation an-2 for n = 1, 2, .. a2 ao 0. The recurrence relation then implies that = 0 for k = 0, 1, 2, . Thus we see that the solution series can be written using only the = Y 3 a2k 9n/(n+4) = Y ao, a4 = 18/4 ao (2) Use the recurrence relation to find the first few coefficients in terms of y = ao ∞ n=0 k=0 = n Anxn ∞ k=0 a zk x ² k ao, a6 (3) Write out a few more terms if needed but try to guess the pattern for the general term ao for all k = 18/4 1, 2, . (4) Therefore the general solution, with arbitrary constant a can be written as αχ, ag x 2k Notice The general solution of this first order linear equation is y = = a0 € ³x² 27/8
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