Find the general integral of the equation (2x-4y+6)…

Find the general integral of the equation $(2x-4y+6)\,dx+(x+y-3)\,dy=0$.

Step 1: Eliminate the constants

Here $y'=\dfrac{-2x+4y-6}{x+y-3}$. Requiring $-2h+4k-6=0$ and $h+k-3=0$ gives $h=1,\ k=2$, so

\[\frac{dY}{dX}=\frac{-2X+4Y}{X+Y}.\]

Step 2: Solve the homogeneous equation

With $Y=VX$, $X\dfrac{dV}{dX}=\dfrac{-2+3V-V^2}{1+V}=-\dfrac{(V-1)(V-2)}{1+V}$, so

\[\frac{1+V}{(V-1)(V-2)}\,dV=-\frac{dX}{X}.\]

Partial fractions give $\dfrac{1+V}{(V-1)(V-2)}=\dfrac{-2}{V-1}+\dfrac{3}{V-2}$, so integrating,

\[-2\ln|V-1|+3\ln|V-2|=-\ln|X|+C \quad\Longrightarrow\quad X(V-2)^3=K(V-1)^2.\]

Step 3: Back-substitute

Since $V=Y/X$, this simplifies to $(Y-2X)^3=K(Y-X)^2$. Undoing the translation ($X=x-1,\ Y=y-2$, so $Y-2X=y-2x$ and $Y-X=y-x-1$):

\[\boxed{\,(y-2x)^3=C(y-x-1)^2.}\]