1. **Problem statement:** Solve the ordinary differential equation (ODE) $$y' = y - y^3$$ with the initial condition $$y(0) = R > 0$$. 2. **Identify the type of ODE:** This is a separable differential equation because the right side can be factored and expressed as a function of $$y$$ alone. 3. **Rewrite the equation:** $$\frac{dy}{dx} = y - y^3 = y(1 - y^2)$$ 4. **Separate variables:** $$\frac{dy}{y(1 - y^2)} = dx$$ 5. **Partial fraction decomposition:** We want to express $$\frac{1}{y(1 - y^2)}$$ as partial fractions: $$\frac{1}{y(1 - y^2)} = \frac{A}{y} + \frac{B y + C}{1 - y^2}$$ Since $$1 - y^2 = (1 - y)(1 + y)$$, we can also write: $$\frac{1}{y(1 - y)(1 + y)} = \frac{A}{y} + \frac{B}{1 - y} + \frac{C}{1 + y}$$ 6. **Find coefficients:** Multiply both sides by $$y(1 - y)(1 + y)$$: $$1 = A(1 - y)(1 + y) + B y (1 + y) + C y (1 - y)$$ Simplify: $$1 = A(1 - y^2) + B y (1 + y) + C y (1 - y)$$ Expand terms: $$1 = A - A y^2 + B y + B y^2 + C y - C y^2$$ Group by powers of $$y$$: $$1 = A + y(B + C) + y^2(-A + B - C)$$ Equate coefficients: - Constant term: $$A = 1$$ - Coefficient of $$y$$: $$B + C = 0$$ - Coefficient of $$y^2$$: $$-A + B - C = 0$$ Substitute $$A=1$$: $$B + C = 0$$ $$-1 + B - C = 0$$ From the first: $$C = -B$$ Substitute into second: $$-1 + B - (-B) = -1 + 2B = 0 \implies 2B = 1 \implies B = \frac{1}{2}$$ Then $$C = -\frac{1}{2}$$ 7. **Rewrite integral:** $$\int \frac{dy}{y(1 - y^2)} = \int \left( \frac{1}{y} + \frac{1/2}{1 - y} - \frac{1/2}{1 + y} \right) dy$$ 8. **Integrate each term:** $$\int \frac{1}{y} dy = \ln|y|$$ $$\int \frac{1/2}{1 - y} dy = -\frac{1}{2} \ln|1 - y|$$ $$\int \frac{-1/2}{1 + y} dy = -\frac{1}{2} \ln|1 + y|$$ 9. **Combine integrals:** $$\int \frac{dy}{y(1 - y^2)} = \ln|y| - \frac{1}{2} \ln|1 - y| - \frac{1}{2} \ln|1 + y| + C$$ 10. **Set equal to $$x + C$$:** $$\ln|y| - \frac{1}{2} \ln|1 - y| - \frac{1}{2} \ln|1 + y| = x + C$$ 11. **Exponentiate both sides:** $$\frac{|y|}{\sqrt{|1 - y| \cdot |1 + y|}} = K e^{x}$$ where $$K = e^{C}$$. 12. **Simplify denominator:** $$\sqrt{|1 - y| \cdot |1 + y|} = \sqrt{|1 - y^2|}$$ 13. **Since $$y(0) = R > 0$$ and for $$|y| < 1$$, the expression is positive, so we write:** $$\frac{y}{\sqrt{1 - y^2}} = K e^{x}$$ 14. **Apply initial condition $$x=0, y=R$$:** $$\frac{R}{\sqrt{1 - R^2}} = K$$ 15. **Final implicit solution:** $$\frac{y}{\sqrt{1 - y^2}} = \frac{R}{\sqrt{1 - R^2}} e^{x}$$ 16. **Solve explicitly for $$y$$:** Let $$S = \frac{R}{\sqrt{1 - R^2}} e^{x}$$, then $$y = \frac{S}{\sqrt{1 + S^2}}$$ **Summary:** The explicit solution to the ODE $$y' = y - y^3$$ with $$y(0) = R > 0$$ is $$y(x) = \frac{\frac{R}{\sqrt{1 - R^2}} e^{x}}{\sqrt{1 + \left(\frac{R}{\sqrt{1 - R^2}} e^{x}\right)^2}} = \frac{R e^{x}}{\sqrt{1 - R^2 + R^2 e^{2x}}}$$