The Laplace transform of $\mathcal{L}(te^t \cos t)$

Question

How do I find it?

I know that $\mathcal{L}(e^t \cos t) =\frac{s-1}{(s-1)^2+1^2}$ but what is it when multiplied by $t$, as written in the title?

Answer 1

You need the relation

$$\mathcal{L}\{tf(t)\}\Longleftrightarrow -F'(s)$$

i.e. multiplication in the time domain corresponds to differentiation in the $s$-domain (and a negative sign). Since you know $F(s)$, you can easily derive the result.

Answer 2

$$\text{Another approach: } \mathcal{L}(e^ttcost)=F(s-1)$$ $$\mathcal{L}(tcost)=-\frac{d}{ds}(\frac{s}{s^2+1})=\frac{s^2-1}{(s^2+1)^2}$$ $$\text{so the final answer is:}$$ $$\ F(s-1)=\frac{(s-1)^2-1}{[(s-1)^2+1]^2}$$