Question

(y+zx)p-(x+yz)q=x^2-y^2 general solution​

Answer 1

$(y + zx)p - (x + yz)q = x^{2} + y ^{2}$

$py + pzx - qx - qyz = x {}^{2} + y {}^{2}$

$x {}^{2} + y {}^{2} + qx - py + qyz - pzx = 0$

I think you ask about the general equation, so here it is.

Making sure this will help you...

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