20. Solve dy/dx={xy+2y-x-2}/{xy-3y+x-3} by separable method.

$\begin{equation*} \frac{dy}{dx}=\frac{xy+2y-x-2}{xy-3y+x-3}, \end{equation*}$

we can rearrange the terms in the numerator and denominator to get:

$\begin{equation*} \frac{dy}{dx}=\frac{x(y-1)+2 (y-1)}{x(y+1)-3(y+1)}. \end{equation*}$

Now we can use the method of separation of variables by multiplying both sides by the denominator and then separating the variables as follows:

$\begin{align*} \frac{dy}{dx}&=\frac{x(y-1)+2 (y-1)}{x(y+1)-3(y+1)}\end{align*}$

$\begin{align*} \frac{dy}{dx}&=\frac{(x+2)(y-1)}{(x-3)(y+1)} \end{align*}$

By rearranging with respect to variables we get:

$\begin{align*} \frac{y+1}{y-1}dy&=\frac{x+2}{x+3}dx\end{align*}$

Integrating both sides with respect to their respective variables, we obtain:

$\begin{align*} \int \frac{y+1}{y-1}dy&=\int \frac{x+2}{x+3}dx \end{align*}$

$\begin{align*} \int \frac{y-1}{y-1}+\frac{2}{y-1}dy&=\int \frac{x+3}{x+3}+\frac{5}{x+3}dx \end{align*}$

$\begin{align*} \int (1+\frac{2}{y-1})dy&=\int (1+\frac{5}{x+3})dx \end{align*}$

Now we can simplify this expression, therefore, we have:

$\begin{align*} y+2\ln(y-1)+lnC_1=x+5\ln(x+3)+lnC_2 \end{align*}$

Therefore, the general solution to the differential equation is given by:

$\begin{align*} y+2\ln(y-1)=x+5\ln(x+3)+lnC \end{align*}$

where $C$ is an arbitrary constant.

Now, we can simplify this expression

$\begin{align*} e^y(y-1)^2&=e^xC(x+3)^5 \end{align*}$