Set
$$
L=L_0f(u),
\qquad
u=\frac{t}{t_0},
\qquad
L_0=\frac{\eta^2}{\rho\sigma},
\qquad
t_0=\frac{\eta^3}{\rho\sigma^2}.
$$
Then $\dot L=(L_0/t_0)f'$ and $\ddot L=(L_0/t_0^2)f''$. Every term in the scaling equation has the common dimensions and magnitude $\rho^2\sigma^3/\eta^4$. Cancelling that factor yields the <dimensionless variable>[dimensionless] equation
$$
\alpha f''+\beta\frac{(f')^2}{f}
=\gamma\frac{f'}{f^2}+\frac{\delta}{f^2},
$$
or equivalently
$$
\alpha f^2f''+\beta f(f')^2
=\gamma f'+\delta.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-344/2/c/solution