1. Consider the differential equation $$\frac{dy}{dx} = \frac{1}{2} x (y - 4)^3.$$ We are asked to find the particular solution passing through given points and analyze the slope field. 2. For part (a), the slope field shows the slope at each point $(x,y)$ given by $$\frac{dy}{dx} = \frac{1}{2} x (y - 4)^3.$$ The solution curve passing through $(0,2)$ must satisfy this differential equation and the initial condition $y(0) = 2$. At $x=0$, the slope is zero because of the factor $x$ in the equation, so the curve is flat at $(0,2)$. 3. For part (b), find the particular solution with initial condition $f(1) = 0$. 4. The differential equation is separable: $$\frac{dy}{dx} = \frac{1}{2} x (y - 4)^3$$ Rewrite as: $$\frac{dy}{(y - 4)^3} = \frac{1}{2} x \, dx$$ 5. Integrate both sides: $$\int (y - 4)^{-3} dy = \int \frac{1}{2} x \, dx$$ 6. The left integral: $$\int (y - 4)^{-3} dy = \int u^{-3} du = \frac{u^{-2}}{-2} + C = -\frac{1}{2 (y - 4)^2} + C$$ 7. The right integral: $$\int \frac{1}{2} x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{x^2}{4} + C$$ 8. Equate the integrals (combine constants into one $C$): $$-\frac{1}{2 (y - 4)^2} = \frac{x^2}{4} + C$$ 9. Multiply both sides by $-2$: $$\cancel{-2} \cdot \left(-\frac{1}{2 (y - 4)^2}\right) = \cancel{-2} \cdot \left(\frac{x^2}{4} + C\right)$$ $$\frac{1}{(y - 4)^2} = -\frac{x^2}{2} - 2C$$ 10. Let $K = -2C$ (a constant), then: $$\frac{1}{(y - 4)^2} = K - \frac{x^2}{2}$$ 11. Use initial condition $f(1) = 0$: $$\frac{1}{(0 - 4)^2} = K - \frac{1^2}{2}$$ $$\frac{1}{16} = K - \frac{1}{2}$$ $$K = \frac{1}{16} + \frac{1}{2} = \frac{1}{16} + \frac{8}{16} = \frac{9}{16}$$ 12. Substitute $K$ back: $$\frac{1}{(y - 4)^2} = \frac{9}{16} - \frac{x^2}{2}$$ 13. Solve for $y$: $$ (y - 4)^2 = \frac{1}{\frac{9}{16} - \frac{x^2}{2}} = \frac{1}{\frac{9}{16} - \frac{8x^2}{16}} = \frac{1}{\frac{9 - 8x^2}{16}} = \frac{16}{9 - 8x^2}$$ 14. Taking square root: $$ y - 4 = \pm \frac{4}{\sqrt{9 - 8x^2}}$$ 15. Since $f(1) = 0$, check sign: $$0 - 4 = -4 = \pm \frac{4}{\sqrt{9 - 8}} = \pm \frac{4}{\sqrt{1}} = \pm 4$$ Negative sign matches, so: $$ y = 4 - \frac{4}{\sqrt{9 - 8x^2}}$$ 16. For problem 15, given: $$\frac{dy}{dx} = x + 2xy$$ Rewrite as: $$\frac{dy}{dx} = x(1 + 2y)$$ 17. Rearrange: $$\frac{dy}{dx} - 2xy = x$$ 18. This is a linear differential equation with integrating factor: $$\mu(x) = e^{-2 \int x dx} = e^{-x^2}$$ 19. Multiply both sides by $\mu(x)$: $$e^{-x^2} \frac{dy}{dx} - 2x e^{-x^2} y = x e^{-x^2}$$ 20. Left side is derivative: $$\frac{d}{dx} (y e^{-x^2}) = x e^{-x^2}$$ 21. Integrate both sides: $$y e^{-x^2} = \int x e^{-x^2} dx + C$$ 22. Use substitution $u = -x^2$, $du = -2x dx$, so: $$\int x e^{-x^2} dx = -\frac{1}{2} \int e^{u} du = -\frac{1}{2} e^{u} + C = -\frac{1}{2} e^{-x^2} + C$$ 23. Substitute back: $$y e^{-x^2} = -\frac{1}{2} e^{-x^2} + C$$ 24. Multiply both sides by $e^{x^2}$: $$y = -\frac{1}{2} + C e^{x^2}$$ 25. Use initial condition $y(0) = 1$: $$1 = -\frac{1}{2} + C e^{0} = -\frac{1}{2} + C$$ $$C = \frac{3}{2}$$ 26. Final solution: $$y = -\frac{1}{2} + \frac{3}{2} e^{x^2}$$ 27. This matches options (c) and (e), which are identical. Final answers: - Part (b) solution: $$y = 4 - \frac{4}{\sqrt{9 - 8x^2}}$$ - Problem 15 solution: $$y = \frac{3}{2} e^{x^2} - \frac{1}{2}$$