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1- Explain the the difference between DE with Ordinary points and DE with singular points Consider the following differential equation to be solved using a power series. y"-y=0 Using the substitution y = S find an expression for c, in terms of c for k = 0, 1, 2, .. Fk+2" Find two power series solutions of the given differential equation about the ordinary point x= 0. Compare the series solutions with the solutions of the differential equation obtained using the method of Section 4.3. Try to explain any differences between the two forms of che solution. and O Y1 = 1 + x and y2 O v *x and y2= O y, 1 and yx+ These solutions are recognized as O Y = x and ya = e" O y, =1 and y, -1+ e* O y =1+x and y2 = -1 - x + e* O y, = cosh(x) and y2 = sinh(x) O y, = cos(x) and y2 = sin(x) The general solution obtained from these two solutions is y(x) = which is equivalent to the solution obtained using the method of Section 4.3.

1- Explain the the difference between DE with Ordinary points and DE with singular points Consider the following differential equation to be solved using a power series. y"-y=0 Using the substitution y = S find an expression for c, in terms of c for k = 0, 1, 2, .. Fk+2" Find two power series solutions of the given differential equation about the ordinary point x= 0. Compare the series solutions with the solutions of the differential equation obtained using the method of Section 4.3. Try to explain any differences between the two forms of che solution. and O Y1 = 1 + x and y2 O v *x and y2= O y, 1 and yx+ These solutions are recognized as O Y = x and ya = e" O y, =1 and y, -1+ e* O y =1+x and y2 = -1 - x + e* O y, = cosh(x) and y2 = sinh(x) O y, = cos(x) and y2 = sin(x) The general solution obtained from these two solutions is y(x) = which is equivalent to the solution obtained using the method of Section 4.3.

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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1- Explain the the difference between DE with Ordinary points and DE with singular points Consider the following differential equation to be solved using a power series. y" - y' = 0 Using the substitution y = c x", find an expression for ck + 2 in terms of ck + 1 for k = 0, 1, 2, .... n= 0 Ck + 2 Find two power series solutions of the given differential equation about the ordinary point x = 0. Compare the series solutions with the solutions of the differential equation obtained using the method of Section 4.3. Try to explain any differences between the two forms of the solution. x2 x4 O y, = 1 + and y, = x + 3! x5 7! + 4! 6! 5! O y, = 1 + x and y2 = x2 + x3 x4 x5 2! 3! 4! 5! x2 O y, = x and y, = 1 + x + 2! x3 3! x2 =1 - x4 Y1 and Y2 4! 3! x2 3! x3 x4 O y, = 1 and y, = x + 2! 4! These solutions are recognized as У, 3 х and у, — е* Y, = 1 and y, = -1 + ex Y1 = 1 + x and y, = -1 - x + e* Y1 = cosh(x) and y, = sinh(x) O y, = cos(x) and y, = sin(x) The general solution obtained from these two solutions is y(x) = , which is equivalent to the solution obtained using the method of Section 4.3.
Transcribed Image Text:1- Explain the the difference between DE with Ordinary points and DE with singular points Consider the following differential equation to be solved using a power series. y" - y' = 0 Using the substitution y = c x", find an expression for ck + 2 in terms of ck + 1 for k = 0, 1, 2, .... n= 0 Ck + 2 Find two power series solutions of the given differential equation about the ordinary point x = 0. Compare the series solutions with the solutions of the differential equation obtained using the method of Section 4.3. Try to explain any differences between the two forms of the solution. x2 x4 O y, = 1 + and y, = x + 3! x5 7! + 4! 6! 5! O y, = 1 + x and y2 = x2 + x3 x4 x5 2! 3! 4! 5! x2 O y, = x and y, = 1 + x + 2! x3 3! x2 =1 - x4 Y1 and Y2 4! 3! x2 3! x3 x4 O y, = 1 and y, = x + 2! 4! These solutions are recognized as У, 3 х and у, — е* Y, = 1 and y, = -1 + ex Y1 = 1 + x and y, = -1 - x + e* Y1 = cosh(x) and y, = sinh(x) O y, = cos(x) and y, = sin(x) The general solution obtained from these two solutions is y(x) = , which is equivalent to the solution obtained using the method of Section 4.3.
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