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Answer 1 · 2017-12-04T04:13:24

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

.

#d/dx(y)=2y-4#

First Order Separable Ordinary Differential Equation has the form of:

#N(y)dy=M(x)dx# or

#N(y)y'=M(x)#

#N(y)=1/(2y-4)#

#M(x)=1#

#int1/(2y-4)dy=int1dx#

#1/2ln(2y-4)+C_2=x+C_1#

Now, we combine constants:

#1/2ln(2y-4)=x+C#

#ln(2y-4)=2x+C#

Using the definition of Log, we isolate #y#:

#e^(2x+C)=2y-4#

#2y=e^(2x+C)+4#

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