# \frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8} Solve it using variable separation

Question
Differential equations
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}{y}+{3}{x}-{y}-{3}}}{{{x}{y}-{2}{x}+{4}{y}-{8}}}}$$
Solve it using variable separation

2021-02-11
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}{y}+{3}{x}-{y}-{3}}}{{{x}{y}-{2}{x}+{4}{y}-{8}}}}={\frac{{{x}{\left({y}+{3}\right)}-{1}{\left({y}+{3}\right)}}}{{{x}{\left({y}-{2}\right)}+{4}{\left({y}-{2}\right)}}}}$$
$$\displaystyle={\frac{{{\left({y}+{3}\right)}{\left({x}-{1}\right)}}}{{{\left({y}-{2}\right)}{\left({x}+{4}\right)}}}}$$
$$\displaystyle{\frac{{{y}-{2}}}{{{y}+{3}}}}{\left.{d}{y}\right.}={\frac{{{x}-{1}}}{{{x}+{4}}}}{\left.{d}{x}\right.}$$

### Relevant Questions

Solve the equation:
$$\displaystyle{\left({x}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\left({y}^{{2}}+{1}\right)}$$
Solve the equation:
$$\displaystyle{\left({a}-{x}\right)}{\left.{d}{y}\right.}+{\left({a}+{y}\right)}{\left.{d}{x}\right.}={0}$$
Find $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}$$ using the rules for finding derivatives.
$$\displaystyle{y}={\left({2}{x}^{{{2}}}-{6}\right)}{\left({8}{x}^{{{2}}}-{9}{x}+{9}\right)}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=$$
Find the indicated derivatives.
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\ {\quad\text{if}\quad}\ {y}={2}{x}^{{{3}}}$$
find the indicated derivatives. $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}{\quad\text{if}\quad}{y}={2}{x}^{{{3}}}$$
Given $$\displaystyle{f{{\left({x},{y}\right)}}}={2}{x}^{{{2}}}-{x}{y}^{{{3}}}+{4}{y}^{{{6}}}$$, find
$$\displaystyle{{f}_{{\times}}{\left({x},{y}\right)}}=$$
$$\displaystyle{{f}_{{{x}{y}}}{\left({x},{y}\right)}}=$$
$$\displaystyle{y}={\frac{{{7}{x}^{{{5}}}}}{{{5}}}}-{2}{x}+{9}{e}^{{{x}}}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{e}^{{-{x}}}}}{{{y}+{e}^{{{y}}}}}}$$
Give the correct answer and solve the given equation $$[x-y \arctan(\frac{y}{x})]dx+x \arctan (\frac{y}{x})dy=0$$
Given that, $$\displaystyle{y}^{{{\sin{{x}}}}}={x}^{{{{\cos}^{{{2}}}{x}}}},{f}\in{d}:{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}$$