Solve: (x-4y-4).dx-(x-6y-5).dy=0
dy/dx=(x-4(y+1)-1)/(x-6(y+1)+1)
dy/dx=((x-1)-4(y+1))/((x-1)-6(y+1) )
Let u=x+1, v=y+1.
du=dx, dv=dy.
Then, dy/dx=dv/du=(u-4v)/(u-6v)=(1-4w)/(1-6w).
w=v/u
v=u∙w
dv/du=w+u∙dw/dv
w+u∙dw/dv=(1-4w)/(1-6w)
So, u∙dw/du=(1-4w-w+6w^2)/(1-6w)=(6w^2-5w+1)/(1-6w)
u∙dw/du=((3w-1)(2w-1))/(1-6w)
Integrating this expression,
∫(1-6w)/((3w-1)(2w-1)) dw=∫du/u
Converting to partial fractions,
∫3/((3w-1) )-4/((2w-1)) dw=∫du/u
ln(3w-1)-2 ln(2w-1)+lnA=lnu
ln(A(3w-1)/(2w-1)^2 )=lnu
A(3w-1)/(2w-1)^2 =u
A(3w-1)=u(4w^2-4w+1)
3Aw-A=4uw^2-4uw+u
4uw^2-(4u+3A)w+(u+A)=0
Using the quadratic formula,
w={(4u+3A)±√((4u+3A)^2-4∙(4u)∙(u+A) )}/(8u)
w={(4u+3A)±√(16u^2+24uA+9A^2-16u^2-16uA)}/(8u)
w={(4u+3A)±√(-8uA+9A^2 )}/(8u)
v/u={(4(x+1)+3A)±√(-8(x+1)A+9A^2 )}/(8u)
y+1={(4(x+1)+3A)±A√(-(8/A)(x+1)+9)}/(8)
Answer: y(x)=-1+1/8k {(4k(x+1)+3)±√(-8k(x+1)+9)}, A=1/k