An equation for derivation of gravitational waves polarization forms 
When I was reading Spacetime and Geometry An Introduction to General Relativity by Sean Carroll(Page 297,Equation 7.110), I couldn't solve this problem in the proper way to get an approximation.
Equation is $$\frac{\partial^{2}}{\partial t^{2}} S^{1}=\frac{1}{2} S^{1} \frac{\partial^{2}}{\partial t^{2}}\left(h_{+} e^{i k_{\sigma} x^{\sigma}}\right)$$
and the book said, "These can be immediately solved to yield, to lowest order:"
$$S^{1}=\left(1+\frac{1}{2} h_{+} e^{i k_{\sigma} x^{\sigma}}\right) S^{1}(0).$$
My ideal is rewrite this equation as
$$\frac{\partial^{2}}{\partial t^{2}} S^{1}+\frac{1}{2} S^{1} \left(h_{+} k_0^2e^{i k_{\sigma} x^{\sigma}}\right)=0$$
and use series to expand the term:
$$h_{+} k_0^2e^{i k_{\sigma} x^{\sigma}}$$
But I still have some questions on how to expand this term by series due to solution above.
 A: I find a way to solve this problem by expanding the solution by series, suppose the solution can be written down as:
$$S^1(t) = \sum^{+\infty}_{n=0}a_n \exp(i nk_\sigma x^\sigma)$$
$a_n$are numbers, the equation can be rewrite as:
$$k_0^2\sum^{+\infty}_{n=1}n^2a_n\exp(ink_\sigma x^\sigma)=\frac{1}{2} h_{+}k_0^2\sum^{+\infty}_{n=0}a_n\exp(i(n+1)k_\sigma x^\sigma)$$
$$\sum^{+\infty}_{n=0}(n+1)^2a_{n+1}\exp(i(n+1)k_\sigma x^\sigma)=\frac{1}{2} h_{+}\sum^{+\infty}_{n=0}a_n\exp(i(n+1)k_\sigma x^\sigma) $$
Since the bases $\{\exp(int)\}$ are linear independent, we get:
$$a_{n+1}=\frac{1}{2}\frac{a_n h_+}{(n+1)^2}$$
And in weak field, $|h_+|<<1$, we dropped higher order terms:
$$S^1(t) = a_0 + \frac{1}{2}a_0 h_+\exp(i k\sigma x^\sigma)$$
Use initial condition to get the $a_0 = S^1(0)$, thus we get the answer.
A: The comment "These can be immediately solved to yield, to lowest order" should hint a much simpler solution is possible. Write $S^1=\color{red}{S^1(0)}(1+\delta)$ with $\delta\ll1$ so$$\frac{\partial^2}{\partial t^2}\color{red}{S^1(0)}(1+\delta)=\frac12\color{red}{S^1(0)}\color{orange}{(1+\delta)}\frac{\partial^2}{\partial t^2}(h_+\exp ik_\sigma x^\sigma).$$Cancel the red factors and neglect the orange one, viz.$$\frac{\partial^2\delta}{\partial t^2}=\frac{\partial^2}{\partial t^2}\left(\frac12h_+\exp ik_\sigma x^\sigma\right)\implies\delta=A+Bt+\frac12h_+\exp ik_\sigma x^\sigma.$$Since $\delta\ll1$ for all $t$, $A=B=0$.
