Let \(\displaystyle V = U'' \) so you have \(\displaystyle V' + \frac{1}{2} \xi V = 0\) (this separates)

\(\displaystyle

\frac{dV}{V} = - \frac{1}{2} \xi

\)

\(\displaystyle

\ln V = -\frac{1}{4} \xi^2 + ln c_1

\)

\(\displaystyle V = c_1 e^{-1/4 \xi^2}\)

so

\(\displaystyle

U''' = c_1 e^{-1/4 \xi^2}.

\)

Now integrate three times. Note. Once you have your final answer, substitute back into the original ODE to determine that it is fact a solution and determine whether there's any restriction on your contants of integration. Realize there should only be two arbitrary contants since the original ODE is second order.

Upon integrating you get

\(\displaystyle U=-8 c_1 e^{-1/4 \xi^2} +c_2 1/2 \xi^2 +c_3 \xi + c_4\)

\(\displaystyle U(0)=0\) gives \(\displaystyle -8 c_1 + c_4 =0\)

also \(\displaystyle U( \infty)=0\) gives \(\displaystyle c_4 + \infty =0\)

However I know that \(\displaystyle U= \xi^2 + 2\) is a soln

hence the \(\displaystyle c_1\) and \(\displaystyle c_3\) terms need to vanish

Im now stuck to achieve this.

Im not sure if as the modulus of xi tends to infinity then U = o(1/xi)

where o is ''little o''

and if it does whether this helps