What is the solution for y^'=((xy+3x-y-3))/((xy-2x+4y-8)) ?

The solution for y^'=((xy+3x-y-3))/((xy-2x+4y-8)) is y=-5\W(-(x+4}{5e^{\frac{x+c_{1}+3)/5}})-3

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y^'=((xy+3x-y-3))/((xy-2x+4y-8))

y^{^{'}} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

y = - 5 W (- \frac{x + 4}{5 e ^{\frac{x + c _{1} + 3}{5}}}) - 3

Solution steps

y^{'} (x) = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

Solve separable ODE:

y = - 5 W (- \frac{x + 4}{5 e ^{\frac{x + c _{1} + 3}{5}}}) - 3

y = - 5 W (- \frac{x + 4}{5 e ^{\frac{x + c _{1} + 3}{5}}}) - 3

\frac{4}{( 3 s - 1 ) ^{5}}

\int \frac{1}{5 + 4 sec ( x )} d x

y^{^{'}} + \frac{y}{x} = 2 y^{2}

\frac{d}{d x} (\frac{4 x + 2}{( 3 - 2 x ) ^{3}})

\frac{\partial}{\partial x} (x e^{14 x y})

What is the solution for y^'=((xy+3x-y-3))/((xy-2x+4y-8)) ?
The solution for y^'=((xy+3x-y-3))/((xy-2x+4y-8)) is y=-5\W(-(x+4}{5e^{\frac{x+c_{1}+3)/5}})-3