Question

Use Theorem 7.4.1. THEOREM 7.4.1 Derivatives of Transforms If F(s) = ℒ{f(t)} and n = 1, 2, 3, . . . , then ℒ{tnf(t)} = (−1)n dn dsn F(s). Evaluate the given Laplace transform. (Write your answer as a function of s.) ℒ{te2t sin "3t"}'

Answer 1

Answer:

$L\left(te^{2t }sin3t\right)=\frac{6s-12}{(s^2-4s+13)^2}$.

Step-by-step explanation:

If F(s)= L{f(t)}

Then $L\left\{(t^nf(t)\right\}=(-1)^n\frac{\mathrm{d^n}F(s)}{\mathrm{d^n}s}$

$L\left\{te^{2t}sin3t\right\}$

f(t)=$e^{2t}sin3t$

$L\left\{e^{at}sinbt\right\}=\frac{b}{(s-a)^2+b^2}$

Therefore,$L\left\{e^{2t}sin3t\right\}=\frac{3}{(s-2)^2+(3)^2}$

$L\left\{e^{2t}sin3t\right\}=\frac{3}{s^2-4s+13}$

$L\left\{te^{2t}sin3t\right}=-\frac{\mathrm{d}F(s)}{\mathrm{d}s}$

=-$\frac{\mathrm{d}e^{2t}sin3t}{\mathrm{d}s}$

$L\left\{te^{2t}sin3t\right\}$

$=\frac{3(2s-4)}{(s^2-4s+13)^2}$

$L\left\{te^{2t}sin3t\right\}=\frac{6s-12}{(s^2-4s+13)^2}$.