find solutions of the pffafian differential equation 
Find the solution of:

$$y(x+4)(y+z)\,dx - x(y+3z)\,dy + 2xy\,dz=0$$
We know it is integrable, but for the solution, we tried some methods and got stuck. We are studying from the book of Ian Sneddon.
 A: Hint:
$y(x+4)(y+z)~dx-x(y+3z)~dy+2xy~dz=0$
$\dfrac{(x+4)(y+z)~dx}{x}-\dfrac{(y+3z)~dy}{y}+2~dz=0$
which can find the clear solving route that $z\to y\to x$
A: Suppose your Pfaffin DE is solvable, that is, $X\cdot curl X= 0$ is satisfied for $X= (P, Q, R)$ with $P=y(x+4)(y+z)$, $Q=-x(y+3z)$, and $R=2xy$.
Let $x$ treat as constant, so $dx=0$. Thus we have $-x(y+3z)dy+2xydz=0$, or $-(y+3z)dy+2ydz=0$, which is a linear ODE in $y$ and $z$:
$$\frac{dz}{dy}-\frac{3}{2y}z=\frac{1}{2}$$ with the solution $c_1 
= U(x, y, z)=zy^{-3/2}+y^{-1/2}$. Due to the method of solution of the Pfaffin DE there is $\mu$ such that $\mu Q=\frac{\partial U}{\partial y}$ and $\mu R=\frac{\partial U}{\partial z}$. Here it is not difficult to see that $\mu=\displaystyle \frac{1}{2xy^{5/2}}$. Then we compute $K=\mu P-\frac{\partial U}{\partial x}=\frac{x+4}{2x}U=(1/2+2/x)U$. This implies a linear ODE in $U$: $\frac{dU}{dx}+\frac{x+4}{2x}U=0$. Solving this we get $$ x^2e^{x/2}U(x, y, z)=x^2e^{x/2}(zy^{-3/2}+y^{-1/2}) =c_2.$$
