<s>[INST] Solve the pde $(x^2-y^2-yz)p+ (x^2-y^2-zx)q=z(x-y)$ I'm solving this by Lagrange's method.
Lagrange's auxiliary equation is:
$\frac{dx}{x^2-y^2-yz}=\frac {dy}{x^2-y^2-zx}= \frac{dz}{z(x-y)}$
From the first two ratio and the last ratio:
$\frac{dx-dy}{z(x-y)}=\frac {dz}{z(x-y)}$
Solving which I get $z-x+y=c_1$
I'm unable to find another solution. Please help! 
 [/INST] since
$$\dfrac{dx}{x^2-y^2-yz}=\dfrac{dy}{x^2-y^2-xz}=\dfrac{dz}{z(x-y)}$$
$$\Longrightarrow \dfrac{xdx}{x^3-y^3-xyz}=\dfrac{ydy}{x^2y-y^3-xyz}=\dfrac{\dfrac{dz}{z}}{x-y}$$
$$\Longrightarrow\dfrac{xdx-ydy}{(x^3+y^3)-xy(x+y)}=\dfrac{\dfrac{dz}{z}}{x-y}$$
$$\Longrightarrow \dfrac{xdx-ydy}{(x+y)(x-y)^2}=\dfrac{\dfrac{dz}{z}}{x-y}$$
$$\Longrightarrow \dfrac{xdx-ydy}{x^2-y^2}=\dfrac{dz}{z}$$
$$\Longrightarrow\dfrac{d(x^2-y^2)}{2(x^2-y^2)}=\dfrac{dz}{z}$$
$$\Longrightarrow z^2=x^2-y^2+C_{2}$$
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