Since $(x^2-1)(-\delta_0)=\delta_0$, write $u'=-\delta_0+w$, where $(x^2-1)w=0$. The polynomial has two simple zeros. The <distribution annihilated by a function with simple zeros> lemma gives
$$
w=a\delta_{-1}+b\delta_1.
$$
Integrating with the <Heaviside step function> yields
$$
\boxed{u=-H(x)+aH(x+1)+bH(x-1)+c,\qquad a,b,c\in\mathbb C.}
$$
This includes every solution: its derivative has exactly the required delta terms, and any two primitives differ only by a constant distribution.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 327 / 2 / iii / Solution

If a smooth function $a$ on $\mathbb R$ has finitely many zeros $r_j$, all simple, then $aw=0$ exactly when $w=\sum_jc_j\delta_{r_j}$. Away from the zeros, divide a <test function> by the nonvanishing $a$ to show $w$ vanishes. Near $r_j$, write $a(x)=(x-r_j)b_j(x)$ with smooth nonzero $b_j$. The <kernel of multiplication by a coordinate>, after translation and multiplication by $b_j$, says that the local distribution is a multiple of $\delta_{r_j}$. A <partition of unity> gives the global sum. Simplicity of the zeros excludes derivatives of deltas.

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