Mehraan7954 Mehraan7954

11-01-2020
Mathematics

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Find the solution of 4x2y′′−4x2y′+y=0,x>04x2y″−4x2y′+y=0,x>0 of the form y1=xr(1+c1x+c2x2+c3x3+⋯)

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femotinga femotinga

13-01-2020

Answer:

r = 1/2

c1 = 3/4

c2 = 27/64

c3 = 405/1408

Step-by-step explanation:

Find the solution of 4x2y′′−4x2y′+y=0,x>04x2y″−4x2y′+y=0,x>0 of the form y1=xr(1+c1x+c2x2+c3x3+⋯)

r = 1/2

c1 = 3/4

c2 = 27/64

c3 = 405/1408

The solution is attached.

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