This differential equation may be expressed as
`(y')/y = -x - 1`
and then integrated:
`ln|y| = -x^2/2 - x + C, or y = Ce^(-x^2/2 - x),`
where `C` is any constant.
To determine the specific constant, we use the given condition `y(-2) = 1,` which gives
`1 = C e^(-2 + 2) = C,`
so the final answer is `y(x) =e^(-x^2/2 - x).`
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