Given: $M(x,y)dx + N(x,y)dy = 0$
Show that there exists some $g(y/x) = \frac{dy}{dx} = \frac{-M(x,y)}{N(x,y)}$
Is my way of doing this problem correct?
FIRST: Solving for $\frac{dy}{dx}$ in $M(x,y)dx + N(x,y)dy = 0$ yields: $\frac{dy}{dx} = \frac{-M(x,y)}{N(x,y)}$
SECOND: Let $v(y/x) = \frac{y}{x}$
-->$y=v(y/x)x$
-->$\frac{dy}{dx} = v^{'}(y/x)*\frac{-y}{x^{2}}*x + v(y/x)$
-->$\frac{dy}{dx} = v(y/x) - \frac{y}{x}*v^{'}(y/x)$
THIRD: Therefore, $\frac{dy}{dx} = g(y/x) = v(y/x) - \frac{y}{x}*v^{'}(y/x)=\frac{-M(x,y)}{N(x,y)}$
-->$g(y/x) = v(y/x) - \frac{y}{x}*v^{'}(y/x)$