1. **State the problem:** Solve the ordinary differential equation (ODE) given by $$y' = y - y^3$$. 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 like terms: $$1 = A + y(B + C) + y^2(-A + B - C)$$ Set coefficients equal: - 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 $$C = -B = -\frac{1}{2}$$ 7. **Rewrite integral:** $$\int \left( \frac{1}{y} + \frac{1/2}{1 - y} - \frac{1/2}{1 + y} \right) dy = \int dx$$ 8. **Integrate both sides:** $$\int \frac{1}{y} dy + \frac{1}{2} \int \frac{1}{1 - y} dy - \frac{1}{2} \int \frac{1}{1 + y} dy = x + C$$ Recall: $$\int \frac{1}{1 - y} dy = -\ln|1 - y|$$ $$\int \frac{1}{1 + y} dy = \ln|1 + y|$$ So, $$\ln|y| - \frac{1}{2} \ln|1 - y| - \frac{1}{2} \ln|1 + y| = x + C$$ 9. **Combine logarithms:** $$\ln|y| - \frac{1}{2} \ln|1 - y^2| = x + C$$ Rewrite as: $$\ln \left( \frac{|y|}{|1 - y^2|^{1/2}} \right) = x + C$$ 10. **Exponentiate both sides:** $$\frac{|y|}{\sqrt{|1 - y^2|}} = K e^{x}$$ where $$K = e^{C}$$ is an arbitrary positive constant. 11. **Final implicit solution:** $$\frac{y}{\sqrt{1 - y^2}} = \pm K e^{x}$$ This expresses the general solution implicitly. Depending on initial conditions, the sign and value of $$K$$ are determined. **Summary:** The solution to $$y' = y - y^3$$ is given implicitly by $$\frac{y}{\sqrt{1 - y^2}} = C e^{x}$$ where $$C$$ is an arbitrary constant determined by initial conditions.