a) laplace transform of 4t^2 sin 3t + e^-2t + tb)......................... te^-2t cos 3t
Accepted Solution
A:
Answer:a.L(s)=[tex](-24)\frac{81-s^4}{(s^2+9)^4}+\frac{1}{s+2}+\frac{1}{s+1}[/tex]b.[tex]L(s)=-\frac{-s^2-4s+5}{(s^2+4s+13)^2}[/tex]Step-by-step explanation:We have to find the laplace transform of each functiona.[tex]4t^2sin3t+e^{-2t}+t[/tex][tex]L(t^nf(t))=(-1)^n\frac{d^nF(s)}{ds^n}[/tex][tex]L(t^n)=\frac{n!}{s^n+1}[/tex][tex]L(e^{at})=\frac{1}{s-a}[/tex][tex]L(sinat)=\frac{a}{s^2+a^2}[/tex][tex]L(e^{at}cosbt)=\frac{s-a}{(s-a)^2+b^2}[/tex]Apply the formula Then we get [tex]12(-1)^2\frac{d^2(\frac{1}{s^2+9})}{ds^2}+\frac{1}{s+2}+\frac{1}{s+1}[/tex][tex](-24)\frac{s^4+81+18s^2-2s^4-18s^2}{(s^2+9)^4}+\frac{1}{s+2}+\frac{1}{s+1}[/tex]L(s)=[tex](-24)\frac{81-s^4}{(s^2+9)^4}+\frac{1}{s+2}+\frac{1}{s+1}[/tex]b.[tex]te^{-2t}cos3t[/tex]f(t)=[tex]e^{-2t}cos3t[/tex][tex]F(s)=\frac{s+2}{(s+2)^2+9}=\frac{s+2}{s^2+4s+13}[/tex][tex]L(s)=-\frac{dF(s)}{ds}=-\frac{s^2+4s+13-(2s+4)(s+2)}{(s^2+4s+13)^2}[/tex][tex]L(s)=-\frac{5-s^2-4s}{(s^2+4s+13)^2}[/tex]