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Example 2 Eliminate the arbitrary constants of y – Ae²r + Bxe™*. Solution: Let y = Aeir + Bxer be Equation (1). Substituting Equation (1) to Equations (2) & (3). we get Differentiating the equation twice, we got y' - 2y + Bet y' - 2Aer + 28xe" + Be" or and y" - 4y + 48e From y' - 2y + Be", we got yo' = 2(Ae²x + Bxeż*) + Bež* as Equation (2) and y" - 4Ae + 48xei« + 28e²« + 28e Be - y' – 2y y"- 4Ae+ 4Bxe + 4Be²* Substitute Be" - y' – 2y to y" = 4y + 48e²" and we got y" = 4y + 4(y' – 2y) or y" - 4(Ae + Bxe) + 4Be" as Equation (3). Simplify and finaly we get y" - 4y + 4y = 0

Example 2 Eliminate the arbitrary constants of y – Ae²r + Bxe™*. Solution: Let y = Aeir + Bxer be Equation (1). Substituting Equation (1) to Equations (2) & (3). we get Differentiating the equation twice, we got y' - 2y + Bet y' - 2Aer + 28xe" + Be" or and y" - 4y + 48e From y' - 2y + Be", we got yo' = 2(Ae²x + Bxeż*) + Bež* as Equation (2) and y" - 4Ae + 48xei« + 28e²« + 28e Be - y' – 2y y"- 4Ae+ 4Bxe + 4Be²* Substitute Be" - y' – 2y to y" = 4y + 48e²" and we got y" = 4y + 4(y' – 2y) or y" - 4(Ae + Bxe) + 4Be" as Equation (3). Simplify and finaly we get y" - 4y + 4y = 0

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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Example 2 Eliminate the arbitrary constants of y - Ae* + Bxe*. Solution: Let y = Ae* + Bxe* be Equation (1). Substituting Equation (1) to Equations (2) & (3). we get Differentiating the equation twice, we get y' - 2y + Be y' = 2Ae2x + 2Bxe + Be or and y' = 2(Ae2x + Bxe2*) + Bez* as Equation (2) y" - 4y + 4Be and From y' = 2y + Be", we got y" = 4Ae2x + 4Bxe* + 2Bex + 2Bex y" = 4Ae* + 4Bxe + 4Be* Be - y' - 2y Substitute Be* = y' - 2y to y" = 4y + 4Be* and we get y" = 4y + 4(y' – 2y) or y" = 4(Ae* + Bxe) + 1Be as Equation (3). Simplify and finally we get y" - 4y + ty - 0
Transcribed Image Text:Example 2 Eliminate the arbitrary constants of y - Ae* + Bxe*. Solution: Let y = Ae* + Bxe* be Equation (1). Substituting Equation (1) to Equations (2) & (3). we get Differentiating the equation twice, we get y' - 2y + Be y' = 2Ae2x + 2Bxe + Be or and y' = 2(Ae2x + Bxe2*) + Bez* as Equation (2) y" - 4y + 4Be and From y' = 2y + Be", we got y" = 4Ae2x + 4Bxe* + 2Bex + 2Bex y" = 4Ae* + 4Bxe + 4Be* Be - y' - 2y Substitute Be* = y' - 2y to y" = 4y + 4Be* and we get y" = 4y + 4(y' – 2y) or y" = 4(Ae* + Bxe) + 1Be as Equation (3). Simplify and finally we get y" - 4y + ty - 0
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