1. **State the problem:** Find the particular solution of the differential equation $$x \frac{dy}{dx} = \frac{x^2}{y} + y^2$$ given the initial condition $x=1$, $y=4$. 2. **Rewrite the equation:** Divide both sides by $x$ (assuming $x \neq 0$) to isolate $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{1}{x} \left( \frac{x^2}{y} + y^2 \right) = \frac{x}{y} + \frac{y^2}{x}$$ 3. **Express the equation:** $$\frac{dy}{dx} = \frac{x}{y} + \frac{y^2}{x}$$ 4. **Rewrite the right side as a single fraction:** $$\frac{dy}{dx} = \frac{x^2 + y^3}{xy}$$ 5. **Separate variables if possible:** Multiply both sides by $xy$: $$xy \frac{dy}{dx} = x^2 + y^3$$ 6. **Rewrite as:** $$xy \frac{dy}{dx} - y^3 = x^2$$ 7. **Divide both sides by $y^3$ to simplify:** $$x y \frac{dy}{dx} \cdot \frac{1}{y^3} - y^3 \cdot \frac{1}{y^3} = x^2 \cdot \frac{1}{y^3}$$ $$x \frac{dy}{dx} \cdot \frac{1}{y^2} - 1 = \frac{x^2}{y^3}$$ 8. **Rewrite:** $$x \frac{1}{y^2} \frac{dy}{dx} = 1 + \frac{x^2}{y^3}$$ 9. **Substitute $v = y^{-2}$, so $y = v^{-1/2}$:** Calculate $\frac{dy}{dx}$ in terms of $v$: $$\frac{dy}{dx} = \frac{d}{dx} (v^{-1/2}) = -\frac{1}{2} v^{-3/2} \frac{dv}{dx}$$ 10. **Substitute into the equation:** $$x \cdot v \cdot \left(-\frac{1}{2} v^{-3/2} \frac{dv}{dx} \right) = 1 + \frac{x^2}{y^3}$$ Simplify $v \cdot v^{-3/2} = v^{-1/2}$: $$-\frac{1}{2} x v^{-1/2} \frac{dv}{dx} = 1 + \frac{x^2}{y^3}$$ 11. **Express $\frac{x^2}{y^3}$ in terms of $v$:** Since $y = v^{-1/2}$, then $$y^3 = (v^{-1/2})^3 = v^{-3/2}$$ So $$\frac{x^2}{y^3} = x^2 v^{3/2}$$ 12. **Rewrite the equation:** $$-\frac{1}{2} x v^{-1/2} \frac{dv}{dx} = 1 + x^2 v^{3/2}$$ 13. **Multiply both sides by $-2 v^{1/2} / x$ to isolate $\frac{dv}{dx}$:** $$\frac{dv}{dx} = -2 \frac{v^{1/2}}{x} - 2 x v^{2}$$ 14. **Rewrite:** $$\frac{dv}{dx} = -\frac{2 v^{1/2}}{x} - 2 x v^{2}$$ 15. **This is a nonlinear differential equation in $v$; solving analytically is complex. Instead, check initial condition:** Given $x=1$, $y=4$, then $$v = y^{-2} = 4^{-2} = \frac{1}{16}$$ 16. **Summary:** The substitution reduces the original equation to $$\frac{dv}{dx} = -\frac{2 v^{1/2}}{x} - 2 x v^{2}$$ with initial condition $v(1) = \frac{1}{16}$. 17. **The particular solution requires solving this nonlinear ODE with initial condition, which may need numerical methods.** **Final answer:** The particular solution satisfies $$\frac{dv}{dx} = -\frac{2 v^{1/2}}{x} - 2 x v^{2}, \quad v(1) = \frac{1}{16}$$ where $v = y^{-2}$. Slug: "particular solution" Subject: "differential equations" Desmos: {"latex":"x \frac{dy}{dx} = \frac{x^2}{y} + y^2","features":{"intercepts":true,"extrema":true}} q_count:1