= Solution
Odd elements of a <Grassmann algebra> anticommute, and in particular $\psi_i^2=0$. For variations with fixed endpoints, move each odd variation to the left before applying <integration by parts>. For example,
$$
\delta(\psi\dot\psi)=\delta\psi\dot\psi+\psi\delta\dot\psi
\quad\Longrightarrow\quad
\delta\int\psi\dot\psi\,dt=2\int\delta\psi\dot\psi\,dt.
$$
The interaction varies as $\delta_{\psi_1}(\psi_1\psi_2)=\delta\psi_1\psi_2$ and $\delta_{\psi_2}(\psi_1\psi_2)=-\delta\psi_2\psi_1$. Thus variation of the <action> gives the <Euler-Lagrange equations>
$$
\boxed{\ddot x=UU'+U''\psi_1\psi_2,\qquad \dot\psi_1=U'\psi_2,\qquad \dot\psi_2=U'\psi_1.}
$$
All products of $U$ and its derivatives commute with the odd variables. These signs are essential for this variant of <Grassmann-valued classical supersymmetric mechanics>.
The fermion bilinear is constant along a solution:
$$
\frac{d}{dt}(\psi_1\psi_2)=\dot\psi_1\psi_2+\psi_1\dot\psi_2=U'(\psi_2^2+\psi_1^2)=0.
$$
For the <energy>, direct differentiation now gives
$$
\dot E=\dot x(\ddot x-UU'-U''\psi_1\psi_2)-U'\frac{d}{dt}(\psi_1\psi_2)=0.
$$
Similarly the given <supercharge>, denoted $Q_1$, satisfies
$$
\begin{aligned}
\dot Q_1&=\ddot x\psi_1+\dot x\dot\psi_1-U'\dot x\psi_2-U\dot\psi_2\\
&=(\ddot x-UU')\psi_1=U''\psi_1\psi_2\psi_1=0.
\end{aligned}
$$
A second conserved <supercharge> is
$$
\boxed{Q_2=\dot x\psi_2-U\psi_1,}
$$
since $\dot Q_2=(\ddot x-UU')\psi_2=U''\psi_1\psi_2\psi_2=0$. The two charges are independent functions on the unrestricted <phase space>: their coefficient matrix on $(\psi_1,\psi_2)$ has <determinant> $\dot x^2-U^2$, and is invertible wherever its even body is nonzero. They can become dependent on a restricted family of trajectories.
Under $\dot x=U$ and $\psi_2=-\psi_1$, the bilinear $\psi_1\psi_2$ vanishes. Differentiating the first-order equation gives $\ddot x=U'\dot x=UU'$, which is the bosonic <Euler-Lagrange equation>. The first fermionic equation becomes $\dot\psi_1=-U'\psi_1$; differentiating $\psi_2=-\psi_1$ gives $\dot\psi_2=U'\psi_1$, so the second fermionic equation holds too. This establishes the <first-order zero-energy solution in Grassmann supersymmetric mechanics> without imposing an additional equation.
Let $x_0=x(0)$ and $\psi_{10}=\psi_1(0)$. For $U=1+cx$ and $c\ne0$, solve $d(x+1/c)/dt=c(x+1/c)$ and $\dot\psi_1=-c\psi_1$ to obtain
$$
\boxed{x(t)=e^{ct}x_0+\frac{e^{ct}-1}{c},\qquad \psi_1(t)=e^{-ct}\psi_{10},\qquad \psi_2(t)=-e^{-ct}\psi_{10}.}
$$
For $c=0$, the nonsingular limiting answer is $\boxed{x(t)=x_0+t,\ \psi_1(t)=\psi_{10},\ \psi_2(t)=-\psi_{10}}$. These formulas also hold for even initial positions in the <Grassmann algebra> by ordinary polynomial evaluation. Finally, $U(x(t))=(1+cx_0)e^{ct}$, so the conserved quantities are
$$
\boxed{E=0,\qquad Q_1=2(1+cx_0)\psi_{10},\qquad Q_2=-2(1+cx_0)\psi_{10}.}
$$
The individual odd <supercharges> generally do not vanish, although their sum vanishes on this restricted solution.
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