I need to find transformation $F(s)$ for: $f(t)=tsin(ωt)$
Could somebody please help me?
First of all, use the property:
$$\mathcal{L}_x\left[x\text{y}\left(x\right)\right]_{\left(\text{s}\right)}=-\frac{\text{d}}{\text{ds}}\left(\mathcal{L}_x\left[\text{y}\left(x\right)\right]_{\left(\text{s}\right)}\right)\tag1$$
Now, use:
$$\mathcal{L}_x\left[\sin\left(\omega x\right)\right]_{\left(\text{s}\right)}=\frac{\omega}{\text{s}^2+\omega^2}\tag2$$
So, you will get:
$$\mathcal{L}_x\left[x\sin\left(\omega x\right)\right]_{\left(\text{s}\right)}=-\frac{\partial}{\partial\text{s}}\left(\frac{\omega}{\text{s}^2+\omega^2}\right)\tag3$$
Copyright © 2021 JogjaFile Inc.
First of all, use the property:
$$\mathcal{L}_x\left[x\text{y}\left(x\right)\right]_{\left(\text{s}\right)}=-\frac{\text{d}}{\text{ds}}\left(\mathcal{L}_x\left[\text{y}\left(x\right)\right]_{\left(\text{s}\right)}\right)\tag1$$
Now, use:
$$\mathcal{L}_x\left[\sin\left(\omega x\right)\right]_{\left(\text{s}\right)}=\frac{\omega}{\text{s}^2+\omega^2}\tag2$$
So, you will get:
$$\mathcal{L}_x\left[x\sin\left(\omega x\right)\right]_{\left(\text{s}\right)}=-\frac{\partial}{\partial\text{s}}\left(\frac{\omega}{\text{s}^2+\omega^2}\right)\tag3$$