1. **State the problem:** We need to find the complete integral of the first-order PDE given by: $$b^2 + q^2 - 2px - 2qy + 2xy = 0$$ where $p = \frac{\partial z}{\partial x}$, $q = \frac{\partial z}{\partial y}$, and $b$ is a constant. 2. **Rewrite the PDE:** The PDE can be written as: $$b^2 + q^2 - 2px - 2qy + 2xy = 0$$ 3. **Charpit's method setup:** Charpit's equations are derived from the PDE $F(x,y,z,p,q) = 0$ where $$F = b^2 + q^2 - 2px - 2qy + 2xy$$ The Charpit system is: $$\frac{dx}{dt} = \frac{\partial F}{\partial p}, \quad \frac{dy}{dt} = \frac{\partial F}{\partial q}, \quad \frac{dz}{dt} = p \frac{\partial F}{\partial p} + q \frac{\partial F}{\partial q}, \quad \frac{dp}{dt} = -\frac{\partial F}{\partial x} - p \frac{\partial F}{\partial z}, \quad \frac{dq}{dt} = -\frac{\partial F}{\partial y} - q \frac{\partial F}{\partial z}$$ Since $F$ does not depend on $z$, $\frac{\partial F}{\partial z} = 0$. 4. **Calculate partial derivatives:** $$\frac{\partial F}{\partial p} = -2x$$ $$\frac{\partial F}{\partial q} = 2q - 2y$$ $$\frac{\partial F}{\partial x} = -2p + 2y$$ $$\frac{\partial F}{\partial y} = 2q - 2p$$ 5. **Write Charpit's system:** $$\frac{dx}{dt} = -2x$$ $$\frac{dy}{dt} = 2q - 2y$$ $$\frac{dz}{dt} = p(-2x) + q(2q - 2y) = -2px + 2q^2 - 2qy$$ $$\frac{dp}{dt} = -(-2p + 2y) = 2p - 2y$$ $$\frac{dq}{dt} = -(2q - 2p) = -2q + 2p$$ 6. **Solve the system:** From $\frac{dx}{dt} = -2x$, we get: $$x = C_1 e^{-2t}$$ From $\frac{dp}{dt} = 2p - 2y$ and $\frac{dy}{dt} = 2q - 2y$, and $\frac{dq}{dt} = -2q + 2p$, we solve the coupled system: Rewrite as: $$\frac{dp}{dt} = 2p - 2y$$ $$\frac{dy}{dt} = 2q - 2y$$ $$\frac{dq}{dt} = -2q + 2p$$ 7. **Find relations:** Add $\frac{dy}{dt} + \frac{dq}{dt}$: $$\frac{d}{dt}(y+q) = (2q - 2y) + (-2q + 2p) = -2y + 2p$$ But $\frac{dp}{dt} = 2p - 2y$ is the same as above, so: $$\frac{dp}{dt} = \frac{d}{dt}(y+q)$$ Integrate: $$p = y + q + C_2$$ 8. **Use PDE to express $b^2$:** Recall PDE: $$b^2 + q^2 - 2px - 2qy + 2xy = 0$$ Substitute $p = y + q + C_2$: $$b^2 + q^2 - 2x(y + q + C_2) - 2qy + 2xy = 0$$ Simplify: $$b^2 + q^2 - 2xy - 2xq - 2xC_2 - 2qy + 2xy = 0$$ $$b^2 + q^2 - 2xq - 2xC_2 - 2qy = 0$$ 9. **Group terms:** $$b^2 + q^2 - 2q(x + y) - 2xC_2 = 0$$ 10. **Solve quadratic in $q$:** $$q^2 - 2q(x + y) + (b^2 - 2xC_2) = 0$$ Use quadratic formula: $$q = \frac{2(x + y) \pm \sqrt{4(x + y)^2 - 4(b^2 - 2xC_2)}}{2} = (x + y) \pm \sqrt{(x + y)^2 - b^2 + 2xC_2}$$ 11. **Summary:** The complete integral depends on constants $C_1, C_2$ and is given implicitly by: $$x = C_1 e^{-2t}$$ $$p = y + q + C_2$$ $$q = (x + y) \pm \sqrt{(x + y)^2 - b^2 + 2xC_2}$$ 12. **Interpretation:** This system describes the complete integral of the PDE using Charpit's method with two arbitrary constants $C_1, C_2$. **Final answer:** The complete integral is given implicitly by the relations above involving $x,y,p,q$ and constants $C_1,C_2$.