Solving PDE using Lagrange method of characteristics

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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.

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$$ -\frac{\partial f}{\partial \phi_3}\frac{\phi_3}{p_4} + \frac{\partial f}{\partial p_4} = 0 \tag 1$$

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 :

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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).