Solving by variation of parameters y" – 3y' – 4y = 5x we find that: u_1=xe^x-x and u_2=(-x)/4 e^(-4x)+1/16 e^(-4x) u_1=xe^x+x and u_2=(-x)/4 e^(-4x)-1/16
Solving by variation of parameters y" – 3y' – 4y = 5x we find that: u_1=xe^x-x and u_2=(-x)/4 e^(-4x)+1/16 e^(-4x) u_1=xe^x+x and u_2=(-x)/4 e^(-4x)-1/16
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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Transcribed Image Text:Solving by variation of parameters y" – 3y' – 4y :
= 5x we find that:
u_1=xe^x-x and u_2=(-x)/4 e^(-4x)+1/16
e^(-4x)
u_1=xe^x+x and u_2=(-x)/4 e^(-4x)-1/16
e^(-4x)
u_1=xe^x-x and u_2=(-x)/4 e^(-4x)-(1/16
Je^(-4x)
O None of them
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