Find the general integral of the equation (y+2) dx=(2x+y-4) dy

Find the general integral of the equation $(y+2)\,dx=(2x+y-4)\,dy$.

Step 1: Eliminate the constants

Here $\dfrac{dy}{dx}=\dfrac{y+2}{2x+y-4}$. Requiring $k+2=0$ and $2h+k-4=0$ gives $k=-2,\ h=3$, so

\[\frac{dY}{dX}=\frac{Y}{2X+Y}.\]

Step 2: Solve the homogeneous equation

With $Y=VX$, $X\dfrac{dV}{dX}=\dfrac{V}{2+V}-V=-\dfrac{V(1+V)}{2+V}$. Using $\dfrac{2+V}{V(1+V)}=\dfrac2V-\dfrac{1}{1+V}$,

\[2\ln|V|-\ln|1+V|=-\ln|X|+C \quad\Longrightarrow\quad \frac{XV^2}{1+V}=K \quad\Longrightarrow\quad Y^2=K(X+Y).\]

Step 3: Undo the translation

With $X=x-3,\ Y=y+2$:

\[\boxed{\,(y+2)^2=C(x+y-1).}\]