= Solution
Multiply the <first-order linear differential equation> by the <integrating factor> $e^{t/\alpha}$. Then
$$
\frac d{dt}(e^{t/\alpha}y)=\frac1\alpha e^{t/\alpha}f(t).
$$
If $y\to0$ in the remote past, then $e^{t/\alpha}y\to0$ there. Integration from $-\infty$ gives the necessary formula
$$
\boxed{y(t)=\frac1\alpha\int_{-\infty}^t e^{-(t-s)/\alpha}f(s)\,ds.}
$$
When this integral exists and tends to zero as $t\to-\infty$, differentiating verifies the equation and the past condition. Two such solutions differ by $Ce^{-t/\alpha}$, which diverges at $-\infty$ unless $C=0$, proving uniqueness. This is the <causal response of a first-order relaxation equation>, a <convolution> with its decaying <impulse response>.
The phrase “general function” implicitly requires compatibility with the past condition: it does not guarantee existence for arbitrary $f$. A sufficient condition is bounded forcing with $f(t)\to0$ as $t\to-\infty$; after setting $r=t-s$, the integral is bounded by the supremum of $|f|$ before $t$. For example, $f\equiv1$ gives $y=1+Ce^{-t/\alpha}$, so no solution has the stated past limit. Both forcings below vanish in the past and meet the intended causal condition, the impulse being interpreted distributionally.
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