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Find in the form of an integral, the solution of the equation $$\alpha\frac{dy}{dt}+y=f(t)$$ that satisfies $y\rightarrow 0$ as $t\rightarrow -\infty$. Here $f(t)$ is a general function and $\alpha$ is a positive constant.

I'm just a little confused as I get the solution $y=\frac1{\alpha}e^{-\frac1{\alpha}t}\int e^{\frac1{\alpha}t}f(t) \space dt$ and I don't know how to use the initial conditions- where can I put them in?

Thank you

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Note that we have

$$\frac{d}{dt}\left(e^{t/\alpha}y(t)\right)=\frac1\alpha e^{t/\alpha}f(t) \tag 1$$

with $\lim_{t\to -\infty}y(t)=0$. Integrating $(1)$ from $-\infty$ to $t$ yields

$$y(t)=\frac1\alpha e^{-t/\alpha}\int_{-\infty}^t e^{t'/\alpha}f(t')\,dt'$$

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    Ah got it. Thanks – 2017-01-12
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    You're welcome! My pleasure. And Happy Holidays! -Mark – 2017-01-12
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    You too. Just one thing though. I'm probably being incredibly thick here, but how do we know that this tends to $0$ as $t\rightarrow -\infty$? I can see that that the integral will tend to $0$ as the limits approach each other, but the $e^{-t/a}$ will tend to infinity, so how do we know that the integral tends to zero fast enough? – 2017-01-12
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    And that is a great point. Note that in order for the integral to converge, we must require $f(t)\to 0$ as $t\to -\infty$. Then, note that $$\lim_{t\to - \infty}\frac{\frac1\alpha \int_{-\infty}^t e^{-t'/\alpha}f(t')\,dt'}{e^{t/\alpha}}$$is of the indeterminate form $0/0$. Applying L'Hospital's Rule we have $$\lim_{t\to -\infty} \frac{\frac1\alpha e^{t/\alpha}f(t)}{\frac1\alpha e^{t/\alpha}}=\lim_{t\to -\infty}f(t)=0$$ – 2017-01-12
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    Excellent thank you very much – 2017-01-12