As the exam hint suggests, we make the substitution w = v − 1 / 2 {\displaystyle w=v^{-1/2}} so that
Under this substitution, t 2 w ′ + 2 t w − w 3 = 0 {\displaystyle t^{2}w'+2tw-w^{3}=0} becomes
t 2 ( − 1 2 v − 3 / 2 v ′ ) + 2 t v − 1 / 2 − v − 3 / 2 = 0 v ′ − 4 t v + 2 t 2 = 0 {\displaystyle {\begin{aligned}t^{2}(-{\frac {1}{2}}v^{-3/2}v')+2tv^{-1/2}-v^{-3/2}&=0\\v'-{\frac {4}{t}}v+{\frac {2}{t^{2}}}&=0\end{aligned}}}
where we multiplied the equation by − 2 v 3 / 2 t 2 . {\displaystyle -2{\frac {v^{3/2}}{t^{2}}}.}
This is now the same ODE as in part (a), so the solution is that v = 2 5 t + C t 4 {\displaystyle v={\frac {2}{5t}}+Ct^{4}} . Hence we have
so that
.
becomes
. Hence we have
.