Question

Number 5 find the general solution to the differentiabel equation

Answer 1

To solve the differential equation, proceed as follows:

`\begin{gathered} (dy)/(dx)=2y-1 \\ (1)/(2y-1)dy=1dx \end{gathered}`

Integrate each side of the equation:

`\begin{gathered} \int (1)/(2y-1)dy=\int 1dx \\ (1)/(2)\ln |2y-1|=x+C \end{gathered}`

Solve for y:

`\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=(Ce^(2x)+1)/(2) \end{gathered}`

Substitution method:

`\begin{gathered} \int (1)/(2y-1)dy \\ u=2y-1 \\ du=2dy \\ dy=(du)/(2) \\ \int (1)/(2u)du \\ (1)/(2)\int (1)/(u)du \\ (1)/(2)\ln |u| \\ (1)/(2)\ln |2y-1| \end{gathered}`