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I was wondering if it was possible to use convolution to solve $$y'' +3ty'-6y=\sin(t)$$ How would you use convolution on the $3y't$ part, or is that even allowed?

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  • $\begingroup$ I don't see the Convolution used. $\endgroup$
    – Andrew
    Nov 13 '17 at 5:39
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Well, we know that:

$$\mathscr{L}_t\left[\text{n}\cdot t\cdot\text{y}\space'\left(t\right)\right]_{\left(\text{s}\right)}=\text{n}\cdot\mathscr{L}_t\left[t\cdot\text{y}\space'\left(t\right)\right]_{\left(\text{s}\right)}\tag1$$

Now, using the 'frequency-domain derivative' property of the Laplace transform:

$$\text{n}\cdot\mathscr{L}_t\left[t\cdot\text{y}\space'\left(t\right)\right]_{\left(\text{s}\right)}=-\text{n}\cdot\frac{\text{d}}{\text{d}\text{s}}\left\{\mathscr{L}_t\left[\text{y}\space'\left(t\right)\right]_{\left(\text{s}\right)}\right\}\tag2$$

Now, using the 'derivative' property of the Laplace transform:

$$-\text{n}\cdot\frac{\text{d}}{\text{d}\text{s}}\left\{\mathscr{L}_t\left[\text{y}\space'\left(t\right)\right]_{\left(\text{s}\right)}\right\}=-\text{n}\cdot\frac{\text{d}}{\text{d}\text{s}}\left(\text{s}\cdot\text{Y}\left(\text{s}\right)-\text{y}\left(0\right)\right)=-\text{n}\cdot\left(\text{s}\cdot\text{Y}\space '\left(\text{s}\right)+\text{Y}\left(\text{s}\right)\right)\tag3$$

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