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$\begin{aligned} & {{u}_{tt}}={{c}^{2}}{{u}_{xx}},\text{ }x\in \mathbb{R},\text{ }t>0, \\

& u(x,0)=f(x), \\

& {{u}_{t}}(x,0)=g(x).

\end{aligned}

$

So I have $\dfrac{{{\partial }^{2}}F(u)}{\partial {{t}^{2}}}=-{{c}^{2}}{{w}^{2}}F(u)$ which gives $F(u(x,w))(t)=c_1\cos(wct)+c_2\sin(wct)$ and $F(u(x,0))=F(f )$ (1) and $\dfrac{\partial F(u(x,0))}{\partial t}=g(x)$ (2). So by using (1) I get $F(u(x,0))=c_1=F( f)$ and $F_t(u(x,0))=c_2cw=g(x)$ so $F(u(x,w))=F( f)\cos(wct)+\dfrac{g(x)}{cw}\sin(wct).$

Well I want to know if I'm correct so far. After this, I'm having problems with the inverse!!