Obtain the first two terms of the asymptotic (large x) solution for each of the 2 real solutions of $$u''+\left ( 1-\frac{\gamma}{x^{2}} \right )u=0$$ in $$x> 4$$
Now, For large x, the above equation reduces to $$u_{0}''+u_{0}=0$$ Solving gives $$u_{0}=C_{1}cos\left ( x \right )+C_{2}Sin\left ( x \right )$$ converting to exponential form yields $$\left\{\left\{u_{0}\to \frac{1}{2} c_1 e^{-i x}+\frac{1}{2} c_1 e^{i x}+\frac{1}{2} i c_2 e^{-i x}-\frac{1}{2} i c_2 e^{i x}\right\}\right\}$$
Dropping the imaginary and the non-decaying terms we get
$$u_{0}=\frac{1}{2} c_1 e^{-i x}+\ $$ Now, the 'real' solution to the original equation is of the form $$u=u_{0}+u_{1} where u_{0}\gg u_{1}$$
Any help is appreciated. What am I missing?