1. **State the problem:** Solve the differential equation $$\frac{dy}{dp} = B(6400y - y^2)$$ where $B$ is a constant. 2. **Rewrite the equation:** Factor the right side: $$\frac{dy}{dp} = B y (6400 - y)$$ 3. **Separate variables:** $$\frac{dy}{y(6400 - y)} = B \, dp$$ 4. **Partial fraction decomposition:** Express $$\frac{1}{y(6400 - y)}$$ as $$\frac{A}{y} + \frac{C}{6400 - y}$$. Multiply both sides by $$y(6400 - y)$$: $$1 = A(6400 - y) + C y = 6400 A - A y + C y$$ Group terms: $$1 = 6400 A + y(C - A)$$ Equate coefficients: - Constant term: $$6400 A = 1 \implies A = \frac{1}{6400}$$ - Coefficient of $$y$$: $$C - A = 0 \implies C = A = \frac{1}{6400}$$ 5. **Rewrite integral:** $$\int \frac{dy}{y(6400 - y)} = \int \left( \frac{1}{6400 y} + \frac{1}{6400 (6400 - y)} \right) dy$$ 6. **Integrate both sides:** $$\int \left( \frac{1}{6400 y} + \frac{1}{6400 (6400 - y)} \right) dy = \int B \, dp$$ Calculate each integral: - $$\int \frac{1}{6400 y} dy = \frac{1}{6400} \ln|y|$$ - $$\int \frac{1}{6400 (6400 - y)} dy = -\frac{1}{6400} \ln|6400 - y|$$ (because derivative of denominator is $$-1$$) So, $$\frac{1}{6400} \ln|y| - \frac{1}{6400} \ln|6400 - y| = B p + C$$ 7. **Combine logarithms:** $$\frac{1}{6400} \ln \left| \frac{y}{6400 - y} \right| = B p + C$$ Multiply both sides by 6400: $$\ln \left| \frac{y}{6400 - y} \right| = 6400 B p + 6400 C$$ Let $$C_1 = 6400 C$$ (a constant), then $$\ln \left| \frac{y}{6400 - y} \right| = 6400 B p + C_1$$ 8. **Exponentiate both sides:** $$\left| \frac{y}{6400 - y} \right| = e^{6400 B p + C_1} = C_2 e^{6400 B p}$$ where $$C_2 = e^{C_1} > 0$$ 9. **Solve for $$y$$:** Assuming positive values, $$\frac{y}{6400 - y} = C_2 e^{6400 B p}$$ Multiply both sides by $$6400 - y$$: $$y = C_2 e^{6400 B p} (6400 - y) = 6400 C_2 e^{6400 B p} - C_2 e^{6400 B p} y$$ Bring terms with $$y$$ to one side: $$y + C_2 e^{6400 B p} y = 6400 C_2 e^{6400 B p}$$ Factor $$y$$: $$y (1 + C_2 e^{6400 B p}) = 6400 C_2 e^{6400 B p}$$ Divide both sides: $$y = \frac{6400 C_2 e^{6400 B p}}{1 + C_2 e^{6400 B p}}$$ 10. **Final solution:** $$\boxed{y(p) = \frac{6400 C e^{6400 B p}}{1 + C e^{6400 B p}}}$$ where $$C$$ is an arbitrary constant determined by initial conditions. This is the general solution to the given differential equation.