1. **Problem statement:** Solve the ordinary differential equation (ODE) $$y' = y - y^3$$ with the initial condition $$y(0) = R$$ where $$R > 0$$. 2. **Rewrite the ODE:** The equation can be written as $$\frac{dy}{dt} = y - y^3 = y(1 - y^2)$$. 3. **Separate variables:** We separate variables to integrate: $$\frac{dy}{y(1 - y^2)} = dt$$. 4. **Partial fraction decomposition:** Decompose the left side: $$\frac{1}{y(1 - y^2)} = \frac{A}{y} + \frac{B y + C}{1 - y^2}$$. Since $$1 - y^2 = (1 - y)(1 + y)$$, we can also write: $$\frac{1}{y(1 - y)(1 + y)} = \frac{A}{y} + \frac{B}{1 - y} + \frac{C}{1 + y}$$. 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 $$A=1$$: $$-1 + B - C = 0 \implies B - C = 1$$. From $$B + C = 0$$ and $$B - C = 1$$, add equations: $$2B = 1 \implies B = \frac{1}{2}$$. Then $$C = -B = -\frac{1}{2}$$. 5. **Rewrite integral:** $$\int \frac{dy}{y(1 - y^2)} = \int \left( \frac{1}{y} + \frac{1/2}{1 - y} - \frac{1/2}{1 + y} \right) dy = \int dt$$. 6. **Integrate each term:** $$\int \frac{1}{y} dy = \ln|y|$$ $$\int \frac{1/2}{1 - y} dy = -\frac{1}{2} \ln|1 - y|$$ (because derivative of $$1 - y$$ is $$-1$$) $$\int -\frac{1/2}{1 + y} dy = -\frac{1}{2} \ln|1 + y|$$ So the integral becomes: $$\ln|y| - \frac{1}{2} \ln|1 - y| - \frac{1}{2} \ln|1 + y| = t + C$$. 7. **Combine logarithms:** $$\ln \left( \frac{|y|}{\sqrt{|1 - y| \cdot |1 + y|}} \right) = t + C$$. Since $$\sqrt{|1 - y| \cdot |1 + y|} = \sqrt{|1 - y^2|}$$, we have: $$\ln \left( \frac{|y|}{\sqrt{|1 - y^2|}} \right) = t + C$$. 8. **Exponentiate both sides:** $$\frac{|y|}{\sqrt{|1 - y^2|}} = K e^{t}$$ where $$K = e^{C} > 0$$. 9. **Solve for $$y$$:** Square both sides: $$\frac{y^2}{|1 - y^2|} = K^2 e^{2t}$$. Since $$y^2 < 1$$ or $$y^2 > 1$$ changes the sign inside absolute value, but initial condition $$y(0) = R > 0$$ and typically solution stays in $$|y| < 1$$ for stability, assume $$1 - y^2 > 0$$: $$\frac{y^2}{1 - y^2} = K^2 e^{2t}$$. Rearranged: $$y^2 = K^2 e^{2t} (1 - y^2) = K^2 e^{2t} - K^2 e^{2t} y^2$$. Bring terms with $$y^2$$ together: $$y^2 + K^2 e^{2t} y^2 = K^2 e^{2t}$$ $$y^2 (1 + K^2 e^{2t}) = K^2 e^{2t}$$ $$y^2 = \frac{K^2 e^{2t}}{1 + K^2 e^{2t}}$$ Take positive root (since $$R > 0$$): $$y = \frac{K e^{t}}{\sqrt{1 + K^2 e^{2t}}}$$. 10. **Apply initial condition:** At $$t=0$$, $$y(0) = R$$: $$R = \frac{K}{\sqrt{1 + K^2}}$$. Square both sides: $$R^2 = \frac{K^2}{1 + K^2}$$ Rearranged: $$R^2 (1 + K^2) = K^2$$ $$R^2 + R^2 K^2 = K^2$$ $$R^2 = K^2 - R^2 K^2 = K^2 (1 - R^2)$$ $$K^2 = \frac{R^2}{1 - R^2}$$. Since $$R > 0$$ and $$R < 1$$ for this to be valid, we have: $$K = \frac{R}{\sqrt{1 - R^2}}$$. 11. **Final explicit solution:** $$\boxed{y(t) = \frac{\frac{R}{\sqrt{1 - R^2}} e^{t}}{\sqrt{1 + \left( \frac{R}{\sqrt{1 - R^2}} \right)^2 e^{2t}}} = \frac{R e^{t}}{\sqrt{1 - R^2 + R^2 e^{2t}}}}$$. This is the explicit solution for $$y' = y - y^3$$ with $$y(0) = R$$ and $$0 < R < 1$$. If $$R \geq 1$$, the solution behavior changes and requires separate analysis.