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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 $\cos$.

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  • $\begingroup$ $k^2+kx$ is not odd in $k$. I don't know if it's the only issue. $\endgroup$ – mathreadler Apr 23 at 6:18
  • $\begingroup$ @mathreadler $\sin(k^2+kx)$ is not odd, right? $\endgroup$ – Stuart-James Burney Apr 23 at 6:20
  • $\begingroup$ Since it seems your issue is resolved it would be appropriate to accept gcousin's answer. $\endgroup$ – Spencer Apr 23 at 6:40
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Your error is in the resolution of the differential equation for $Y(k)$.

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    $\begingroup$ Exactly, you should replace $k^2$ by $k^3/3$ in your solution. (check your current "solution") $\endgroup$ – gcousin Apr 23 at 5:17
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    $\begingroup$ k^2 is a function of k here, it should not be considered a constant (the equation is derivating with respect to k) , whence your error applying the method you mention. $\endgroup$ – gcousin Apr 23 at 5:36
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    $\begingroup$ @gcousin What does a "function of $k$" really mean ? If $k$ is a constant then $k^2$ is a constant. You can just let $c := k^2$. $\endgroup$ – Rebellos Apr 23 at 5:49
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    $\begingroup$ @Rebellos: please read my full comment, the variable is k in the differential equation. $\endgroup$ – gcousin Apr 23 at 6:06
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    $\begingroup$ @Stuart-JamesBurney Of course it does. Is this $$y'(k) - a(k)y(k) = 0$$ the same as this ? $$y'(x) - a(k)y(x) = 0 $$ Note that in the second case, $k$ is a constant for the differential operator. $\endgroup$ – Rebellos Apr 23 at 6:23

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