1. **State the problem:** Solve the first differential equation given: $$\frac{dy}{dx} + x^2 y^2 = x$$. 2. **Identify the type of equation:** This is a first-order nonlinear ordinary differential equation. 3. **Rewrite the equation:** $$\frac{dy}{dx} = x - x^2 y^2$$ 4. **Separate variables if possible:** Rewrite as $$\frac{dy}{dx} = x - x^2 y^2 = x(1 - x y^2)$$ 5. **Try substitution:** Let us try to separate variables by rewriting: $$\frac{dy}{dx} = x - x^2 y^2$$ Rearranged: $$\frac{dy}{dx} + x^2 y^2 = x$$ This is not separable in the usual way. Consider substitution or an integrating factor. 6. **Try substitution:** Let $v = y^2$, then $y = \sqrt{v}$ and $$\frac{dy}{dx} = \frac{1}{2 \sqrt{v}} \frac{dv}{dx}$$ Substitute into the original equation: $$\frac{1}{2 \sqrt{v}} \frac{dv}{dx} + x^2 v = x$$ Multiply both sides by $2 \sqrt{v}$: $$\frac{dv}{dx} + 2 x^2 v \sqrt{v} = 2 x \sqrt{v}$$ 7. **Rewrite:** $$\frac{dv}{dx} = 2 x \sqrt{v} - 2 x^2 v \sqrt{v} = 2 x \sqrt{v} (1 - x v)$$ This is still complicated. Instead, try to solve the original equation by treating it as a Bernoulli equation. 8. **Bernoulli equation form:** $$\frac{dy}{dx} + P(x) y = Q(x) y^n$$ Our equation is: $$\frac{dy}{dx} + 0 \cdot y = x - x^2 y^2$$ Rewrite as: $$\frac{dy}{dx} = x - x^2 y^2$$ Or: $$\frac{dy}{dx} + x^2 y^2 = x$$ This is not in Bernoulli form directly. But if we write: $$\frac{dy}{dx} = x - x^2 y^2$$ Try substitution $z = y^{-1}$, then $y = z^{-1}$ and $$\frac{dy}{dx} = - z^{-2} \frac{dz}{dx}$$ Substitute into the equation: $$- z^{-2} \frac{dz}{dx} = x - x^2 y^2 = x - x^2 z^{-2}$$ Multiply both sides by $- z^{2}$: $$\frac{dz}{dx} = - x z^{2} + x^{2}$$ 9. **Rewrite:** $$\frac{dz}{dx} = x^{2} - x z^{2}$$ This is a nonlinear first-order ODE. 10. **Separate variables:** $$\frac{dz}{dx} + x z^{2} = x^{2}$$ This is a Bernoulli equation with $n=2$, $P(x) = x$, $Q(x) = x^{2}$. 11. **Solve Bernoulli equation:** Substitute $w = z^{1 - 2} = z^{-1}$, then $$\frac{dw}{dx} = - z^{-2} \frac{dz}{dx}$$ From the equation: $$\frac{dz}{dx} + x z^{2} = x^{2}$$ Multiply both sides by $- z^{-2}$: $$- z^{-2} \frac{dz}{dx} - x = - x^{2} z^{-2}$$ But $- z^{-2} \frac{dz}{dx} = \frac{dw}{dx}$, so $$\frac{dw}{dx} - x = - x^{2} w^{2}$$ This is complicated; however, since $w = z^{-1}$, the Bernoulli substitution should yield a linear ODE for $w$. 12. **Correct Bernoulli substitution:** For Bernoulli equation $$\frac{dz}{dx} + P(x) z = Q(x) z^{n}$$ with $n=2$, substitute $w = z^{1 - n} = z^{-1}$, then $$\frac{dw}{dx} = (1 - n) z^{-n} \frac{dz}{dx} = - z^{-2} \frac{dz}{dx}$$ Rewrite original equation: $$\frac{dz}{dx} = x^{2} - x z^{2}$$ Multiply both sides by $- z^{-2}$: $$- z^{-2} \frac{dz}{dx} = - z^{-2} x^{2} + x$$ But left side is $\frac{dw}{dx}$, so $$\frac{dw}{dx} = - x^{2} w^{2} + x$$ This is nonlinear in $w$; however, the Bernoulli substitution should yield a linear ODE. The discrepancy suggests a misstep. 13. **Alternative approach:** Return to step 8 and try substitution $z = y^{-1}$: Original equation: $$\frac{dy}{dx} + x^{2} y^{2} = x$$ Substitute $y = \frac{1}{z}$, then $$\frac{dy}{dx} = - \frac{1}{z^{2}} \frac{dz}{dx}$$ Substitute into the equation: $$- \frac{1}{z^{2}} \frac{dz}{dx} + x^{2} \frac{1}{z^{2}} = x$$ Multiply both sides by $z^{2}$: $$- \frac{dz}{dx} + x^{2} = x z^{2}$$ Rearranged: $$\frac{dz}{dx} = x^{2} - x z^{2}$$ This matches step 9. 14. **Separate variables:** $$\frac{dz}{dx} = x^{2} - x z^{2}$$ Rewrite as $$\frac{dz}{dx} + x z^{2} = x^{2}$$ 15. **Try substitution:** Let $u = z^{3}$, then $$\frac{du}{dx} = 3 z^{2} \frac{dz}{dx}$$ From the equation: $$\frac{dz}{dx} = x^{2} - x z^{2}$$ Multiply both sides by $3 z^{2}$: $$3 z^{2} \frac{dz}{dx} = 3 z^{2} x^{2} - 3 x z^{4}$$ But $3 z^{2} \frac{dz}{dx} = \frac{du}{dx}$ and $z^{4} = z^{3} z = u z$, so $$\frac{du}{dx} = 3 x^{2} z^{2} - 3 x u z$$ This is complicated; thus, try another method. 16. **Try to solve the equation numerically or by integrating factor:** Rewrite: $$\frac{dz}{dx} = x^{2} - x z^{2}$$ This is a Riccati equation. 17. **Summary:** The original equation is a Riccati equation and can be transformed into a second-order linear ODE by substitution. 18. **Final answer:** The solution requires advanced methods beyond this scope. The key step is substitution $z = y^{-1}$ leading to $$\frac{dz}{dx} = x^{2} - x z^{2}$$ which is a Riccati equation. **Slug:** first_equation **Subject:** differential_equations