1. **Problem Statement:** Solve the differential equation solvable for $p=\frac{dy}{dx}$: $$x\left(\frac{dy}{dx}\right)^2 + (y - 1 - x^2) \frac{dy}{dx} - x(y - 1) = 0$$ 2. **Formula and Approach:** This is a quadratic equation in $p=\frac{dy}{dx}$: $$a p^2 + b p + c = 0$$ where $$a = x, \quad b = y - 1 - x^2, \quad c = -x(y - 1)$$ We solve for $p$ using the quadratic formula: $$p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (y - 1 - x^2)^2 - 4x(-x(y - 1)) = (y - 1 - x^2)^2 + 4x^2(y - 1)$$ 4. **Simplify the discriminant:** Expand $(y - 1 - x^2)^2$: $$= (y - 1)^2 - 2x^2(y - 1) + x^4$$ Add $4x^2(y - 1)$: $$\Delta = (y - 1)^2 - 2x^2(y - 1) + x^4 + 4x^2(y - 1) = (y - 1)^2 + 2x^2(y - 1) + x^4$$ Rewrite: $$\Delta = (y - 1)^2 + 2x^2(y - 1) + x^4 = \left(x^2 + y - 1\right)^2$$ 5. **Substitute back into formula for $p$:** $$p = \frac{-(y - 1 - x^2) \pm (x^2 + y - 1)}{2x}$$ 6. **Evaluate the two cases:** - For the plus sign: $$p = \frac{-(y - 1 - x^2) + (x^2 + y - 1)}{2x} = \frac{0}{2x} = 0$$ - For the minus sign: $$p = \frac{-(y - 1 - x^2) - (x^2 + y - 1)}{2x} = \frac{-y + 1 + x^2 - x^2 - y + 1}{2x} = \frac{2(1 - y)}{2x} = \frac{1 - y}{x}$$ 7. **Solve each differential equation:** - Case 1: $\frac{dy}{dx} = 0$ implies $y = C_1$ (constant). - Case 2: $\frac{dy}{dx} = \frac{1 - y}{x}$ Rewrite: $$\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x}$$ This is a linear first-order ODE. 8. **Integrating factor:** $$\mu(x) = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$$ Multiply both sides by $x$: $$x \frac{dy}{dx} + y = 1$$ Left side is derivative of $xy$: $$\frac{d}{dx}(xy) = 1$$ Integrate: $$xy = x + C_2$$ 9. **Solve for $y$:** $$y = 1 + \frac{C_2}{x}$$ 10. **Final family of solutions:** $$\boxed{y = C_1 \quad \text{or} \quad y = 1 + \frac{C_2}{x}}$$ --- **Slug:** solvable-for-p **Subject:** differential equations **Desmos:** {"latex":"y=1+\frac{C}{x}","features":{"intercepts":true,"extrema":true}} **q_count:** 6 (Note: Only the first problem is solved as per instructions.)