9709 P31 - Jun 2021 - Q10
2328
The variables x and t satisfy the differential equation \(\frac{dx}{dt} = x^2(1 + 2x)\), and \(x = 1\) when \(t = 0\).
Using partial fractions, solve the differential equation, obtaining an expression for t in terms of x.
Solution
1. Start with the differential equation \(\frac{dx}{dt} = x^2(1 + 2x)\).
2. Separate variables: \(\frac{1}{x^2(1 + 2x)} dx = dt\).
3. Use partial fractions to express \(\frac{1}{x^2(1 + 2x)}\) as \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{1+2x}\).
4. Solve for constants: \(A = -2\), \(B = 1\), \(C = 4\).
5. Integrate both sides: \(\int \left( \frac{-2}{x} + \frac{1}{x^2} + \frac{4}{1+2x} \right) dx = \int dt\).
6. Integrate: \(-2 \ln x - \frac{1}{x} + 2 \ln(1+2x) = t + C\).
7. Use initial condition \(x = 1\) when \(t = 0\) to find \(C\): \(C = 1\).
8. Solve for \(t\): \(t = -\frac{1}{x} + 2 \ln \left( \frac{1+2x}{3x} \right) + 1\).
