<s>[INST] Solve the PDE $(3x+y-z)p+(x+y-z)q=2(z-y)$ Question: Solve the PDE $(3x+y-z)p+(x+y-z)q=2(z-y)$ by Lagrange's method of Solution.  
Progress: Lagrange's auxiliary equations are. 
$$\dfrac{dx}{3x+y-z}=\dfrac{dy}{x+y-z}=\dfrac{dz}{2(z-y)}$$ The three ratios are equal to $\dfrac{-dx+3dy+dz}{0}$. So we have $-dx+3dy+dz=0\implies x-3y-z=C_1$. How can I find the other one. It is given that other solution is $\dfrac{x-y+z}{\sqrt{x+y-z}}=C_2$.
How can I find this?
 [/INST] We can find that,
$$\text{Each ratio} = \frac{dx-dy+dz}{2x-2y+2z} = \frac{dx+dy-dz}{4x+4y-4z}$$
or
$$\frac{d(x-y+z)}{x-y+z} = \frac{d(x+y-z)}{2(x+y-z)}$$
On integrating,
$$\ln(x-y+z) = \frac{1}{2}\ln(x+y-z) +\ln C_2 \implies C_2 = \frac{x-y+z}{\sqrt{x+y-z}}$$
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