Question

Solve (1+yz)dx+x(z-x)dy-(1+xy)dz=0

Answer 1

Let $z-x=u$, then $z=x+u$ and $dz=dx+du$.

Rewrite the equation replacing $z$ by $x+u$:

$(1+xy+yu)dx+xudy-(1+xy)dx-(1+xy)du=0$

After simplification we obtain $yudx+xudy-(1+xy)du=0$

Note that $xdy+ydx=d(xy+1)$, so we have $ud(1+xy)-(1+xy)du=0$.

After dividing it by $(1+xy)^2$ we obtain $d\left(\frac{u}{1+xy}\right)=-\frac{ud(1+xy)-(1+xy)du}{(1+xy)^2}=0$, so $\frac{u}{1+xy}=C$

Since $u=z-x$, we have $\frac{z-x}{1+xy}=C$. After expressing $z$ we obtain $z=x+C(1+xy)$

Answer: $z=x+C(1+xy)$