Question

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​

Answer 1

It looks like the differential equation is

$\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$

Factorize the right side by grouping.

$xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)$

$xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)$

Now we can separate variables as

$\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx$

Integrate both sides.

$\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx$

$\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx$

$\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}$

You could go on to solve for $y$ explicitly as a function of $x$, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.