How do you find the general solution to #dy/dx=2y-1#?

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Answer 1 · 2016-08-01T21:56:26

# y = Ce^(2x) + 1/2#

#dy/dx=2y-1#

separating the variables
#1/(2y-1) *dy/dx=1#

integrating
#int \ 1/(2y-1) dy/dx \ dx =int \ dx#

#int \ 1/(2y-1) \ dy =int \ dx#

#1/2 ln(2y-1) =x + C#

# ln(2y-1) =2x + C#

# 2y-1 = e^ (2x + C) = Ce^(2x)#

# y-1/2 = Ce^(2x)#

# y = Ce^(2x) + 1/2#