Is there a simpler method to this solve this equation
$$(x-2y+1)\text dx+(4x-3y-6)\text dy=0$$
$$\frac{\text dy}{\text dx}=\frac{2y-x-1}{4x-3y-6}$$
$$\frac{\text d Y}{ \text d X}=\frac{2(Y+k)-(X+h)-1}{4(X+h)-3(Y+k)-6}$$
$$\frac{\text d Y}{ \text d X}=\frac{2Y-X+(2k-h-1)}{4X-3Y+(4h-3k-6)}$$
$$2k-h-1=0\qquad4h-3k-6=0$$ $$h=2k-1\qquad 4(2k-1)-3k-6=0$$ $$5k-10=0\qquad k=2\qquad h=2(2)-1\qquad h=3$$ $$y=Y+2\qquad\qquad x=X+3$$ $$Y=y-2\qquad\qquad X=x-3$$
$$\frac{\text d Y}{\text d X}=\frac{2Y-X}{4X-3Y}=\frac{2\left(\frac{Y}{X}\right)-1}{4-3\left(\frac{Y}{X}\right)}$$
$$\left(\frac{Y}{X}\right)=V\qquad Y=VX$$
$$\frac{\text d Y}{\text d X}=X\frac{\text d V}{\text d X}+V$$
$$X\frac{\text d V}{\text d X}+V=\frac{2V-1}{4-3V}$$
$$X\frac{\text d V}{\text d X}=\frac{2V-1}{4-3V}-V$$
$$X\frac{\text d V}{\text d X}=\frac{2V-1}{4-3V}-V\left(\frac{4-3V}{4-3V}\right)$$
$$X\frac{\text d V}{\text d X}=\frac{3V^2-2V-1}{4-3V}$$
$$\frac{\text dX}{X}=\frac{4-3V}{3V^2-2V-1}\text dV$$
$$\int\frac{\text dX}{X}=\int\frac{4-3V}{3V^2-2V-1}\text dV$$
$$\ln X=\int\frac{4-3V}{3V^2-2V-1}\text dV$$
$$3V^2-2V-1=(V-1) \overline {\bigg)3V^2-2V-1}$$
$$\qquad3V+1$$$$=(V-1) \overline {\bigg)3V^2-2V-1}$$$$\quad\qquad3V^2-3V$$$$\qquad\qquad\qquad\qquad V-1$$$$\qquad\qquad\qquad\qquad \underline{V-1}$$$$\qquad\qquad\qquad\qquad 0$$
$$3V^2-2V-1=(V-1)(3V+1)$$
$$\frac{4-3V}{3V^2-2V-1}=\frac{4-3V}{(V-1)(3V+1)}=\frac{\alpha}{(V-1)}+\frac{\beta }{(3V+1)}=\frac{(3V+1)\alpha+(V-1)\beta}{(V-1)(3V+1)}$$
$$4-3V=(3V+1)\alpha+(V-1)\beta$$ $$4-3V=3V\alpha+\alpha +V\beta-\beta$$ $$4-3V=\alpha-\beta+(3\alpha +\beta)V$$ $$4=\alpha-\beta$$ $$-3=3\alpha+\beta$$ $$4+\beta=\alpha$$ $$-3=3(4+\beta)+\beta$$ $$-15=4\beta$$ $$\beta=\frac{-15}{4}$$ $$4=\alpha-\left(\frac{-15}{4}\right)$$ $$\alpha=4-\frac{15}{4}=\frac{16}{4}-\frac{15}{4}=\frac14$$
$$\ln X=\int\frac{1/4}{V-1}-\frac{15/4}{3V+1}\text d V$$
$$\ln X=\frac{1}{4}\int\frac{1}{V-1}-\frac{15}{3V+1}\text d V$$
$$4\ln X=\ln(V-1)-5\ln(3V+1)+c$$
$$\ln \left(X^4\right)=\ln\left({\frac{(V-1)}{(3V+1)^5}}\right)+c$$
$$X^4=\frac{V-1}{(3V+1)^5}\times e^c$$
$$(3V+1)^5X^4=e^c(V-1)$$
$$(3V+1)^5X^5=AX(V-1)$$
$$(3XV+X)^5=A(XV-X)$$
$$(3X\tfrac{Y}{X}+X)^5=A(X\tfrac{Y}{X}-X)$$
$$(3Y+X)^5=A(Y-X)$$
$$(3(y-2)+(x-3))^5=A((y-2)-(x-3))$$
$$(3y+x-9)^5=A(y-x+1)$$