19. Solve dy/dx={xy+3x-y-3}/{xy-2x+4y-8} by separable method.

$\begin{equation*} \frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8}, \end{equation*}$

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

$\begin{equation*} \frac{dy}{dx}=\frac{x(y+3)- (y+3)}{x(y-2)+4(y-2)}. \end{equation*}$

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

$\begin{align*} \frac{dy}{dx}&=\frac{x(y+3)- (y+3)}{x(y-2)+4(y-2)}\end{align*}$

$\begin{align*} &= \quad \frac{x(y+3)- (y+3)}{(y-2)(x+4)}, \end{align*}$

$\begin{align*} \frac{dy}{dx}&=\frac{(x-1)(y+3)}{(x+4)(y-2)} \end{align*}$

By rearranging with respect to variables we get:

$\begin{align*} \frac{y-2}{y+3}dy&=\frac{x-1}{x+4}dx\end{align*}$

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

$\begin{align*} \int \frac{y-2}{y+3}dy&=\int \frac{x-1}{x+4}dx \end{align*}$

$\begin{align*} \int \frac{y+3}{y+3}-\frac{5}{y+3}dy&=\int \frac{x+4}{x+4}-\frac{5}{x+4}dx \end{align*}$

$\begin{align*} \int (1-\frac{5}{y+3})dy&=\int (1-\frac{5}{x+4})dx \end{align*}$

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

$\begin{align*} y-5\ln(y+3)+C_1=x-5\ln(x+4)+C_2 \end{align*}$

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

$\begin{align*} y-{5\ln(y+3)}&=x-{5\ln(x+4)}+lnC \end{align*}$

where $C$ is an arbitrary constant.

Now, we can simplify this expression

$\begin{align*} \frac{(y+3)^5}{e^y}&=\frac{c(x+4)^5}{e^x} \end{align*}$

$\begin{align*} (y+3)^5e^x&=c(x+4)^5e^y \end{align*}$