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OpenStudy (anonymous):

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?

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OpenStudy (anonymous):

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

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