We are given the first-order partial differential equation: [ 2xz - p x^2 - 2q xy + pq = 0, \quad p = \frac{\partial z}{\partial x}, ; q = \frac{\partial z}{\partial y}. ] Using Charpit's method, we start with the general system: [ \frac{dx}{F_p} = \frac{dy}{F_q} = \frac{dz}{pF_p + qF_q} = -\frac{dp}{F_x + p F_z} = -\frac{dq}{F_y + q F_z}. ] Here, (F(x, y, z, p, q) = 2xz - p x^2 - 2 q xy + pq). Compute partial derivatives: [ F_p = -x^2 + q, \quad F_q = -2xy + p, ] [ F_x = 2z - 2px - 2qy, \quad F_y = -2qx + 0 = -2qx, \quad F_z = 2x. ] The Charpit system is: [ \frac{dx}{-x^2 + q} = \frac{dy}{-2xy + p} = \frac{dz}{p(-x^2 + q) + q(-2xy + p)} = -\frac{dp}{(2z - 2px - 2qy) + 2x p} = -\frac{dq}{-2qx + 2x q}. ] Simplify some terms: * For (-\frac{dq}{F_y + q F_z}): [ F_y + q F_z = -2qx + 2x q = 0 \implies dq \text{ arbitrary along characteristics.} ] * For (-\frac{dp}{F_x + p F_z}): [ F_x + p F_z = (2z - 2px - 2qy) + 2x p = 2z - 2qy ] so (-\frac{dp}{2z - 2qy} = \frac{dx}{-x^2 + q}. ] We notice symmetry suggesting substitution (p = \alpha x + \beta y), (q = \gamma x + \delta y). Assume solution in form (z = xy + f(x) + g(y)). Then (p = \partial z/\partial x = y + f'(x)), (q = \partial z/\partial y = x + g'(y)). Substitute into PDE: [ 2x(xy + f + g) - (y + f') x^2 - 2(x + g') xy + (y + f')(x + g') = 0. ] Simplify term by term: