Answer:
[tex]L\left(te^{2t }sin3t\right)=\frac{6s-12}{(s^2-4s+13)^2}[/tex].
Step-by-step explanation:
If F(s)= L{f(t)}
Then [tex]L\left\{(t^nf(t)\right\}=(-1)^n\frac{\mathrm{d^n}F(s)}{\mathrm{d^n}s}[/tex]
[tex]L\left\{te^{2t}sin3t\right\}[/tex]
f(t)=[tex]e^{2t}sin3t[/tex]
[tex]L\left\{e^{at}sinbt\right\}=\frac{b}{(s-a)^2+b^2}[/tex]
Therefore,[tex] L\left\{e^{2t}sin3t\right\}=\frac{3}{(s-2)^2+(3)^2}[/tex]
[tex]L\left\{e^{2t}sin3t\right\}=\frac{3}{s^2-4s+13}[/tex]
[tex]L\left\{te^{2t}sin3t\right}=-\frac{\mathrm{d}F(s)}{\mathrm{d}s}[/tex]
=-[tex]\frac{\mathrm{d}e^{2t}sin3t}{\mathrm{d}s}[/tex]
[tex]L\left\{te^{2t}sin3t\right\}[/tex]
[tex]=\frac{3(2s-4)}{(s^2-4s+13)^2}[/tex]
[tex]L\left\{te^{2t}sin3t\right\}=\frac{6s-12}{(s^2-4s+13)^2}[/tex].
