<s>[INST] Can someone help me finish the solution to this differential equation? Find the general solution of $(x-4y-3)dx-(x-6y-5)dy=0.$
The rest of my solution:
$x-4y=3$
$x-6y=5$
The point of intersection is $(-1, -1)$.
Let
$x=u-1$; $dx=du$
$y=v-1$; $dy=dv$
$(u-4v)du-(u-6v)dv=0$
Let $u=vz; du=vdz+zdv$
$(vz-4v)(vdz+zdv)-(vz-6v)dv=0$
$(z-4)(vdz+zdv)-(z-6)dv=0$
$(z^2-5z+6)dv-v(z-4)dz=0$
$\frac{dv}{v}-\frac{z-4}{(z^2-5z+6)}dz=0$
$\int\frac{dv}{v}-\int \frac{z-4}{(z^2-5z+6)}dz=0$
$\int\frac{dv}{v}-2\int \frac{dz}{(z-2)}+\int\frac{dz}{z-3}=0$
$\ln \lvert v\rvert -2\ln \lvert z-2\rvert +\ln \lvert z-3 \rvert + \ln \lvert c \rvert=0$
$\ln\lvert cv(z-3) \rvert =2\ln \lvert z-2 \rvert$
$\ln \lvert (z-2)^2 \rvert = \ln\lvert cv(z-3) \rvert $
I do not know how the final answer became $(x-2y-1)^2=c(x-3y-2).$
 [/INST] You find :
$$\ln((z-2)^2)=\ln(cv(z-3))$$
$$(z-2)^2=cv(z-3)$$
$z=\frac{u}{v}$
$$(\frac{u}{v}-2)^2=cv(\frac{u}{v}-3)$$
$$(u-2v)^2=cv^2(u-3v)$$
$u=x+1$ and $v=y+1$
$$((x+1)-2(y+1))^2=c(y+1)^2((x+1)-3(y+1))$$
$$(x-2y-1)^2=c(y+1)^2(x-3y-2)$$
</s>