1. **Problem statement:** Find the differential equations of the space curves formed by the intersection of the two families of surfaces: $$u = x^2 + y^2 + z^2 = c_1$$ $$v = x + z = c_2$$ 2. **Formula and approach:** The space curves lie at the intersection of surfaces $u = c_1$ and $v = c_2$. The differential equations of these curves can be found by eliminating the parameters $c_1$ and $c_2$ from the total differentials: $$du = 2x dx + 2y dy + 2z dz = 0$$ $$dv = dx + dz = 0$$ 3. **Intermediate work:** From $dv=0$, we have: $$dx + dz = 0 \implies dz = -dx$$ Substitute $dz = -dx$ into $du=0$: $$2x dx + 2y dy + 2z (-dx) = 0$$ $$2x dx + 2y dy - 2z dx = 0$$ $$2(x - z) dx + 2y dy = 0$$ Divide both sides by 2: $$(x - z) dx + y dy = 0$$ 4. **Differential equation of the space curves:** Rearranged as: $$\frac{dy}{dx} = -\frac{x - z}{y}$$ 5. **Explanation:** This differential equation describes the slope of the curve in the $xy$-plane as a function of $x$, $y$, and $z$. Since $z$ is related to $x$ by $v = x + z = c_2$, we can express $z = c_2 - x$ if needed for further solving. **Final answer:** The differential equation of the space curves is $$\boxed{\frac{dy}{dx} = -\frac{x - z}{y}}$$