To solve the differential equation, proceed as follows:
[tex]\begin{gathered} \frac{dy}{dx}=2y-1 \\ \frac{1}{2y-1}dy=1dx \end{gathered}[/tex]
Integrate each side of the equation:
[tex]\begin{gathered} \int \frac{1}{2y-1}dy=\int 1dx \\ \frac{1}{2}\ln |2y-1|=x+C \end{gathered}[/tex]
Solve for y:
[tex]\begin{gathered} \ln |2y-1|=2x+C \\ e^{\ln |2y-1|}=e^{2x+C} \\ 2y-1=e^{2x}e^C \\ 2y=Ce^{2x}+1 \\ y=\frac{Ce^{2x}+1}{2} \end{gathered}[/tex]
Substitution method:
[tex]\begin{gathered} \int \frac{1}{2y-1}dy \\ u=2y-1 \\ du=2dy \\ dy=\frac{du}{2} \\ \int \frac{1}{2u}du \\ \frac{1}{2}\int \frac{1}{u}du \\ \frac{1}{2}\ln |u| \\ \frac{1}{2}\ln |2y-1| \end{gathered}[/tex]
