I am trying to find $Y(k)$ of the equation $y''(x)-xy(x)=0$ and hence show that $$y(x)=\sqrt{\frac{2}{\pi}}\int_0^{\infty}\cos\left(\frac{k^3}{3}+kx\right) \ dk,$$ given $Y(0)=1$.
Here, we use the following definition of the Fourier transform: $$F(k)=\mathcal{F}(f(x))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ikx}f(x) \ dx.$$
It is easy to show that $$\mathcal{F}(xy(x))=i\frac{dY(k)}{dk},$$ where $Y(k)=\mathcal{F}(y(x))$. My working is as follows:
\begin{align} \mathcal{F}(y''(x))-\mathcal{F}(xy(x))&=0 \\ -k^2\mathcal{F}(y(x))-i\frac{dY(k)}{dk}&=0 \\ i\frac{dY(k)}{dk}+k^2Y(k)&=0 \\ \implies Y(k)&=Ae^{ik^2} \\ \implies Y(k)&=e^{ik^2} \\ y(x)&=\mathcal{F}^{-1}(e^{ik^2}) \\ y(x)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i(k^2+kx)} \ dk \\ y(x)&=\sqrt{\frac{2}{\pi}}\int_0^{\infty}\cos(k^2+kx) \ dk \ \ \text{(sine is odd)} \end{align} I don't know where/if I've made an error in the argument of $\cos$.
Your error is in the resolution of the differential equation for $Y(k)$.