Answer to Question #190502 in Differential Equations for randz

Question

1. (y − 2)dx − (x − y − 1)dy = 0

Answer 1

$(y - 2)dx = (x - y - 1)dy$

$\frac{{dy}}{{dx}} = \frac{{y - 2}}{{x - y - 1}}$

Let

$x = \widehat x + 3,\,\,y = \widehat y + 2$

Then

$\frac{{d\left( {\widehat y + 2} \right)}}{{d\left( {\widehat x + 3} \right)}} = \frac{{\widehat y + 2 - 2}}{{\widehat x + 3 - \widehat y - 2 - 1}} \Rightarrow \frac{{d\widehat y}}{{d\widehat x}} = \frac{{\widehat y}}{{\widehat x - \widehat y}}$

Let

$\frac{{\widehat y}}{{\widehat x}} = t \Rightarrow \widehat y = t\widehat x \Rightarrow \frac{{d\widehat y}}{{d\widehat x}} = t'\widehat x + t$

Then

$t'\widehat x + t = \frac{{t\widehat x}}{{\widehat x - t\widehat x}} \Rightarrow t'\widehat x = \frac{t}{{1 - t}} - t \Rightarrow t'\widehat x = \frac{{t - t + {t^2}}}{{1 - t}} = \frac{{{t^2}}}{{1 - t}} \Rightarrow \widehat x\frac{{dt}}{{d\widehat x}} = \frac{{{t^2}}}{{1 - t}} \Rightarrow \frac{{1 - t}}{t}dt = \frac{{d\widehat x}}{{\widehat x}} \Rightarrow \left( {\frac{1}{t} - 1} \right)dt = \frac{{d\widehat x}}{{\widehat x}} \Rightarrow \ln t - t = \ln \widehat x + C \Rightarrow \ln \frac{{\widehat y}}{{\widehat x}} - \frac{{\widehat y}}{{\widehat x}} - \ln \widehat x = C$

$x = \widehat x + 3 \Rightarrow \widehat x = x - 3$

$y = \widehat y + 2 \Rightarrow \widehat y = y - 2$

Finally, we have

$\ln \frac{{y - 2}}{{x - 3}} - \frac{{y - 2}}{{x - 3}} - \ln (x - 3) = C$

Answer: $\ln \frac{{y - 2}}{{x - 3}} - \frac{{y - 2}}{{x - 3}} - \ln (x - 3) = C$