What is a solution to the differential equation #dy/dx=2y-1#?

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Answer 1 · 2016-08-23T09:39:14

#y = (C_2 e^{2x}+1)/2#

Substituting #z = 2y-1# in #dy/dx=2y-1#

and considering

#dz/(dx) = 2 dy/(dx)# we have

#1/2 dz/dx = z#. Grouping variables

#dz/z = 2 dx # and integrating

#log_e z = 2 x + C_1# or equivalently

#z = C_2e^{2x} = 2 y -1# and finally

#y = (C_2 e^{2x}+1)/2#