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

& u(x,0)=x \\

& {{u}_{t}}(x,0)=1.

\end{aligned}

$

Okay first I should set $v(x,t)=u(x,t)-\dfrac16 t^3,$ then $u(x,t)=v(x,t)+\dfrac16 t^3$ so $u_{tt}=v_{tt}+t$ and $u_{xx}=v_{xx}$ so $v_{tt}+t=v_{xx}+t\implies v_{tt}=v_{xx},$ and $u(x,0)=v(x,0)$ and $u_t(x,0)=v_t(x,0),$ so I need to solve

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

& v(x,0)=x \\

& {{v}_{t}}(x,0)=1.

\end{aligned}$

which is a simple application of the formula and then once found $v$ the problem is solved! Is it correct?