Question

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?

Answer 1

\[\int\limits{1 \over y^2-4y+3} \quad dy = \int\limits 1 \quad dx\]
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
\[y = {3-Ae^{2x} \over 1-Ae^{2x}}\]