Solving PDE using Lagrange method of characteristics The problem in this question is found in section 1.3.2 "Determining the parametric structure of models" by D.J. Cole et al.
I have a linear, 6-dimensional PDE:
$$
-\frac{\partial f}{\partial \phi_3}\frac{\phi_3}{p_4} + \frac{\partial f}{\partial p_4} = 0
$$
where 
$$
f \mathrm{\quad is\ shorthand\ for\quad} f(\phi_1,\phi_2,\phi_3, p_2, p_3, p_4)
$$
But the remaining variables simply have zero-valued coefficients. I need to find the solution to be 
$$ 
\quad f(\phi_1,\phi_2,\phi_3, p_2, p_3, p_4) = F(\phi_1,\phi_2, p_2, p_3, \phi_3 p_4)
$$
So far I have tried this approach
$$
\frac{\partial f}{\partial p_4} = \frac{\partial f}{\partial \phi_3}\frac{\phi_3}{p_4}
$$
$$
\frac{1}{\partial p_4} = \frac{1}{\partial \phi_3}\frac{\phi_3}{p_4}
$$
$$
\frac{p_4}{\partial p_4} = \frac{\phi_3}{\partial \phi_3}
$$
$$
\int\frac{1}{p_4} \partial p_4= \int\frac{1}{\phi_3}\partial \phi_3
$$
$$
\mathrm{ln}|p_4|= \mathrm{ln}|\phi_3| + C_1
$$
And to try the characteristic method
$$
\frac{\partial\phi_3}{-\phi_3 / p_4} = \partial p_4 = \frac{\partial f}{0}
$$
$$
-\int \frac{1}{\phi_3} \partial \phi_3 = \int 1\ \partial p_4
$$
$$
-\mathrm{ln}|\phi_3| = p_4 + C_2
$$
Though neither of these approaches seems to lead in the right direction and I do not have sufficient information to know where to be going.
 A: $$
-\frac{\partial f}{\partial \phi_3}\frac{\phi_3}{p_4} + \frac{\partial f}{\partial p_4} = 0 \tag 1$$
Explanation of the first part :
Consider the total derivative of a function $f(\phi_3\:,\:p_4)$ :
$$df=\frac{\partial f}{\partial \phi_3}d\phi_3+\frac{\partial f}{\partial p_4}dp_4$$
And from Eq.$(1)$ : $\frac{\partial f}{\partial p_4} =\frac{\partial f}{\partial \phi_3}\frac{\phi_3}{p_4}$
$$df=\frac{\partial f}{\partial \phi_3}d\phi_3+ \frac{\partial f}{\partial \phi_3}\frac{\phi_3}{p_4} dp_4=0$$
$$df=\frac{\partial f}{\partial \phi_3}\left(d\phi_3+ \frac{\phi_3}{p_4} dp_4\right)=0$$
$$d\phi_3+ \frac{\phi_3}{p_4} dp_4=0$$
$$\frac{d\phi_3}{\phi_3}+ \frac{dp_4}{p_4} =0$$
$$\ln|\phi_3|+\ln|p_4|=\text{constant}$$
Also, there is a small mistake in the second part :

So, both methods leads to the same result, which can be written equivalently :
$$p_4\phi_3=C$$
The general solution of the PDE is :
$$f(p_4\:,\:\phi_3)=F(p_4\phi_3)$$
that is any function of the product $p_4\phi_3$.
The arbitrary function $F$ has to be determined according to some boundary condition (not specified in the above wording of the question).
