$$\frac{dy}{dx}=\frac{2x+3}{y+x-2}$$
Using the substitution: $y=vx$ I turned this expression into $$\frac{dv}{dx}\cdot x +v=\frac{2x+3}{vx+x-2}$$
Usually, at this stage, you can cancel a few things out and then proceed to use the separable equation, take integrals and get the general solution but I don't think my expression is separable!
Any help?
$x=x_1-\frac32 \Rightarrow dx=dx_1$ and $2x+3=2x_1$
$y=y_1+\frac72 \Rightarrow dy=dy_1$ and $x+y-2=x_1+y_1$
$ \frac{dy}{dx}=\frac{dy_1}{dx_1}=\frac{2x_1}{x_1+y_1} \Rightarrow(x_1+y_1)dy_1=2x_1dx_1$
$y_1=ux_1\Rightarrow dy_1=x_1du+udx_1 $
$dy_1=x_1du+udx_1 \Rightarrow (x_1+y_1)dy_1=2x_1dx_1$
$=(x_1du+udx_1)(x_1+ux_1)=x_1^2du+ux_1dx_1+ux_1^2du+u^2x_1dx_1$
$x_1^2du+ux_1dx_1+ux_1^2du+u^2x_1dx_1-2x_1dx_1=(u+1)x_1^2du+x_1(u-u^2-2)dx_1=0$
$(u+1)x_1^2du+x_1(u-u^2-2)dx_1=0 \Rightarrow \frac{dx_1}{x_1}+\frac{du}{2-u}=0$
$\frac{dx_1}{x_1}+\frac{du}{2-u}=0\Rightarrow ln(x_1)+ln(2-u)=c$
$\Rightarrow ln(x_1)+ln(2-u)=c \Rightarrow ln(x+\frac32)+ln(2-\frac{y-\frac72}{x-\frac32})=c$