Question

Solve the given differential equation by separation of variables.

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{x}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{3}}{\mathbf{x}\mathbf{y}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{-}\mathbf{8}}$

Short answer

$y=x+5\ln \left|\frac{y+3}{x+4}\right|+C$

Step-by-step solution

Definition of separable equation.

A first-order differential equation of the form

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{h}\mathbf{\left(}\mathit{y}\mathbf{\right)}$

is said to be separable or to have separable variables.

Step 2: Grouping factor.

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

Factor By Grouping

$$\begin{gathered} \frac{dy}{dx}=\frac{x(y+3)-(y+3)}{x(y-2)+4(y-2)} \\ \frac{dy}{dx}=\frac{(y+3)(x-1)}{(y-2)(x+4)} \\ \frac{y-2}{y+3}dy=\frac{x-1}{x+4}dx \\ \left(1-\frac{5}{y+3}\right)dy=\left(1-\frac{5}{x+4}\right)dx\frac{(y-2+5)-5}{y+3}dy=\frac{(x-1+5)-5}{x+4}dx \end{gathered}$$

Integrate both sides

$$\begin{gathered} y-5\ln |y+3|=x-5\ln |x+4|+C \\ y=x+5\ln \left|\frac{y+3}{x+4}\right|+C \end{gathered}$$