1. **Problem 5:** Find the particular solution to the differential equation $$y\sqrt{1 - x^2} y' - x\sqrt{1 - y^2} = 0$$ with initial condition $$y(0) = 1$$. 2. **Rewrite the equation:** $$y\sqrt{1 - x^2} \frac{dy}{dx} = x\sqrt{1 - y^2}$$ 3. **Separate variables:** $$\frac{y}{\sqrt{1 - y^2}} dy = \frac{x}{\sqrt{1 - x^2}} dx$$ 4. **Integrate both sides:** Use substitution for each integral: - Let $$u = 1 - y^2$$, then $$du = -2y dy$$, so $$y dy = -\frac{du}{2}$$. - Similarly for $$x$$ side. Rewrite integrals: $$\int \frac{y}{\sqrt{1 - y^2}} dy = \int \frac{y}{\sqrt{u}} dy = -\frac{1}{2} \int u^{-1/2} du = -\frac{1}{2} \cdot 2 u^{1/2} + C = -\sqrt{1 - y^2} + C$$ Similarly, $$\int \frac{x}{\sqrt{1 - x^2}} dx = -\sqrt{1 - x^2} + C$$ 5. **Equate integrals:** $$-\sqrt{1 - y^2} = -\sqrt{1 - x^2} + C$$ 6. **Simplify:** $$\sqrt{1 - y^2} = \sqrt{1 - x^2} + C$$ 7. **Apply initial condition $$y(0) = 1$$:** $$\sqrt{1 - 1^2} = \sqrt{1 - 0^2} + C \Rightarrow 0 = 1 + C \Rightarrow C = -1$$ 8. **Final implicit solution:** $$\sqrt{1 - y^2} = \sqrt{1 - x^2} - 1$$ --- 9. **Problem 6:** Find the particular solution to $$(1 + x^2)(1 + y^2) = x y y'$$ with initial condition $$y(1) = 0$$. 10. **Rewrite:** $$x y \frac{dy}{dx} = (1 + x^2)(1 + y^2)$$ 11. **Separate variables:** $$\frac{y}{1 + y^2} dy = \frac{1 + x^2}{x} dx$$ 12. **Integrate both sides:** Left side: $$\int \frac{y}{1 + y^2} dy$$ Use substitution $$v = 1 + y^2, dv = 2y dy$$, so $$\int \frac{y}{1 + y^2} dy = \frac{1}{2} \int \frac{dv}{v} = \frac{1}{2} \ln|1 + y^2| + C$$ Right side: $$\int \frac{1 + x^2}{x} dx = \int \left(\frac{1}{x} + x\right) dx = \ln|x| + \frac{x^2}{2} + C$$ 13. **Equate integrals:** $$\frac{1}{2} \ln|1 + y^2| = \ln|x| + \frac{x^2}{2} + C$$ 14. **Multiply both sides by 2:** $$\ln|1 + y^2| = 2 \ln|x| + x^2 + 2C$$ 15. **Exponentiate both sides:** $$1 + y^2 = e^{2C} x^2 e^{x^2}$$ Let $$K = e^{2C}$$. 16. **Apply initial condition $$y(1) = 0$$:** $$1 + 0 = K \cdot 1^2 \cdot e^{1} \Rightarrow 1 = K e$$ So, $$K = e^{-1}$$ 17. **Final solution:** $$1 + y^2 = x^2 e^{x^2 - 1}$$ --- 18. **Problem 7:** Find the general solution to $$e^{2y} y' = x^3$$ 19. **Rewrite:** $$e^{2y} \frac{dy}{dx} = x^3$$ 20. **Separate variables:** $$e^{2y} dy = x^3 dx$$ 21. **Integrate both sides:** Left side: $$\int e^{2y} dy = \frac{1}{2} e^{2y} + C$$ Right side: $$\int x^3 dx = \frac{x^4}{4} + C$$ 22. **Equate integrals:** $$\frac{1}{2} e^{2y} = \frac{x^4}{4} + C$$ 23. **Multiply both sides by 2:** $$e^{2y} = \frac{x^4}{2} + K$$ where $$K = 2C$$. 24. **Solve for y:** $$2y = \ln\left(\frac{x^4}{2} + K\right) \Rightarrow y = \frac{1}{2} \ln\left(\frac{x^4}{2} + K\right)$$ --- 25. **Problem 8:** Given $$P(t) = C e^{k t}$$, with $$P(0) = 8$$ and the consumption triples every 4 years. 26. **Use initial condition:** $$P(0) = C e^{k \cdot 0} = C = 8$$ 27. **Use tripling condition:** $$P(4) = 3 P(0) = 3 \times 8 = 24$$ So, $$24 = 8 e^{4k} \Rightarrow 3 = e^{4k}$$ 28. **Solve for k:** $$4k = \ln 3 \Rightarrow k = \frac{\ln 3}{4}$$ 29. **Final values:** $$C = 8, \quad k = \frac{\ln 3}{4}$$