1. **Problem 1: Solve the differential equation** Given: $$y' = y - \frac{2x}{y}, \quad y(0) = 1$$ 2. **Rewrite the equation:** $$\frac{dy}{dx} = y - \frac{2x}{y}$$ Multiply both sides by $y$ to clear the denominator: $$y \frac{dy}{dx} = y^2 - 2x$$ 3. **Rewrite as:** $$y \frac{dy}{dx} + 2x = y^2$$ 4. **Try to separate variables or find an integrating factor.** This is a nonlinear ODE. Let's try substitution. 5. **Multiply both sides by $dx$:** $$y \, dy = (y^2 - 2x) \, dx$$ 6. **Rewrite as:** $$y \, dy = y^2 \, dx - 2x \, dx$$ 7. **Try substitution $v = y^2$ so that $dv/dx = 2y dy/dx$:** From original equation: $$y' = y - \frac{2x}{y}$$ Multiply both sides by $2y$: $$2y y' = 2y^2 - 4x$$ But $2y y' = dv/dx$, so: $$\frac{dv}{dx} = 2v - 4x$$ 8. **This is a linear ODE in $v$: $$\frac{dv}{dx} - 2v = -4x$$** 9. **Find integrating factor:** $$\mu(x) = e^{-2x}$$ 10. **Multiply both sides by $\mu(x)$:** $$e^{-2x} \frac{dv}{dx} - 2 e^{-2x} v = -4x e^{-2x}$$ 11. **Left side is derivative:** $$\frac{d}{dx} (v e^{-2x}) = -4x e^{-2x}$$ 12. **Integrate both sides:** $$v e^{-2x} = \int -4x e^{-2x} dx + C$$ 13. **Integrate $$\int x e^{ax} dx$$ by parts:** Let $$I = \int x e^{-2x} dx$$ - Set $$u = x, dv = e^{-2x} dx$$ - Then $$du = dx, v = -\frac{1}{2} e^{-2x}$$ So, $$I = u v - \int v du = -\frac{x}{2} e^{-2x} + \frac{1}{2} \int e^{-2x} dx = -\frac{x}{2} e^{-2x} - \frac{1}{4} e^{-2x} + C$$ 14. **Therefore:** $$\int -4x e^{-2x} dx = -4 I = -4 \left(-\frac{x}{2} e^{-2x} - \frac{1}{4} e^{-2x}\right) + C = 2x e^{-2x} + e^{-2x} + C$$ 15. **Substitute back:** $$v e^{-2x} = 2x e^{-2x} + e^{-2x} + C$$ 16. **Multiply both sides by $e^{2x}$:** $$v = 2x + 1 + C e^{2x}$$ 17. **Recall $v = y^2$, so:** $$y^2 = 2x + 1 + C e^{2x}$$ 18. **Apply initial condition $y(0) = 1$:** $$1^2 = 2 \cdot 0 + 1 + C e^{0} \Rightarrow 1 = 1 + C \Rightarrow C = 0$$ 19. **Final solution:** $$y^2 = 2x + 1 \Rightarrow y = \pm \sqrt{2x + 1}$$ Since $y(0) = 1 > 0$, choose positive root: $$y = \sqrt{2x + 1}$$ --- "slug": "ode solution", "subject": "differential equations", "desmos": {"latex": "y=\sqrt{2x+1}","features": {"intercepts": true,"extrema": true}}, "q_count": 4 (Note: Only the first problem is solved as per instructions.)