Question

A) laplace transform of 4t^2 sin 3t + e^-2t + t b)......................... te^-2t cos 3t

Answer 1

Answer:

a.L(s)=$(-24)\frac{81-s^4}{(s^2+9)^4}+\frac{1}{s+2}+\frac{1}{s+1}$

b.$L(s)=-\frac{-s^2-4s+5}{(s^2+4s+13)^2}$

Step-by-step explanation:

We have to find the laplace transform of each function

a.$4t^2sin3t+e^{-2t}+t$

$L(t^nf(t))=(-1)^n\frac{d^nF(s)}{ds^n}$

$L(t^n)=\frac{n!}{s^n+1}$

$L(e^{at})=\frac{1}{s-a}$

$L(sinat)=\frac{a}{s^2+a^2}$

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

Apply the formula

Then we get

$12(-1)^2\frac{d^2(\frac{1}{s^2+9})}{ds^2}+\frac{1}{s+2}+\frac{1}{s+1}$

$(-24)\frac{s^4+81+18s^2-2s^4-18s^2}{(s^2+9)^4}+\frac{1}{s+2}+\frac{1}{s+1}$

L(s)=$(-24)\frac{81-s^4}{(s^2+9)^4}+\frac{1}{s+2}+\frac{1}{s+1}$

b.$te^{-2t}cos3t$

f(t)=$e^{-2t}cos3t$

$F(s)=\frac{s+2}{(s+2)^2+9}=\frac{s+2}{s^2+4s+13}$

$L(s)=-\frac{dF(s)}{ds}=-\frac{s^2+4s+13-(2s+4)(s+2)}{(s^2+4s+13)^2}$

$L(s)=-\frac{5-s^2-4s}{(s^2+4s+13)^2}$