## anonymous 5 years ago dy/dx=y^2-4y+3 find the exact solution and plot a slope field so i got 2 roots y=-4 and y=1 and my critical points are x=1 and x=3 where to go from here?

$\int\limits{1 \over y^2-4y+3} \quad dy = \int\limits 1 \quad dx$ using partial fractions ${1 \over 2} \int\limits {1 \over y-3} - {1 \over y-1} \quad dy = \int\limits 1 \quad dx$ and integrating gives ${1 \over 2} (\log(y-3) - \log(y-1) =x+c$ using laws of logs $\log \quad {y-3 \over y-1} = 2x+c_2$ raising to the powers of e ${y-3 \over y-1} = e^{2x+c_2} = Ae^{2x}$ by the simple algebraic rearangement gives $y = {3-Ae^{2x} \over 1-Ae^{2x}}$