= Solution
Set $v=u''$. The resulting <second-order linear differential equation> is $v''+2v=4t^2$. A <polynomial> <particular solution> $v_p=at^2+b$ requires $2a=4$ and $2a+2b=0$, hence $v_p=2t^2-2$. The homogeneous <characteristic roots of a constant-coefficient differential equation> are $\pm i\sqrt2$, so
$$
v=A_1\cos(\sqrt2t)+B_1\sin(\sqrt2t)+2t^2-2.
$$
Integrating twice and renaming constants gives \b[the general real solution]
$$
\boxed{u(t)=A\cos(\sqrt2t)+B\sin(\sqrt2t)+Ct+D+\frac{t^4}{6}-t^2.}
$$
The four independent homogeneous terms account for the four initial data of the fourth-order <linear ordinary differential equation>.
For the substitution $u(t)=y(t^2)$, repeated use of the <chain rule>, with $x=t^2$, gives
$$
u''=2y'+4t^2y'',\qquad
u''''=12y''+48t^2y'''+16t^4y''''.
$$
Therefore
$$
u''''+2u''=4\bigl(4x^2y''''+12xy'''+(3+2x)y''+y'\bigr).
$$
For $x>0$, the map $t=\sqrt x$ is invertible, and the transformed equation is exactly the equation just solved. Hence \b[the general real solution on $x>0$ is]
$$
\boxed{y(x)=A\cos(\sqrt{2x})+B\sin(\sqrt{2x})+C\sqrt x+D+\frac{x^2}{6}-x.}
$$
There is no requirement that $u$ extend as an even function through $t=0$: only the branch $t>0$ is needed, so the sine and square-root terms must be retained.
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