Question

the Laplace transform of F(t)= e^5t​

Answer 1

Step-by-step explanation:

There's a property -

If [math]L[f(t)] = F(s)[/math], then [math] L[t^{n} f(t)] = (-1)^{n} \frac{d^{n}}{ds^{n}} F(s) [/math].

So,

[math] L[t e^{-5t}] = (-1) \dfrac{d}{ds} L[e^{-5t}] [/math]

[math] = (-1) \frac{d}{ds} \frac{1}{s+5} [/math]

[math] = (-1). \frac{-1}{(s+5)^{2}} [/math]

[math] = \dfrac{1}{(s + 5)^2} [/math]

Hope you find this helpful :-)