Question #232313

L{e2t(t-1)2}


1
Expert's answer
2021-09-02T07:25:35-0400
F(s)=L{f(t)}=0f(t)estdtF(s)=L\{f(t)\}=\displaystyle\int_{0}^{\infin}f(t)e^{-st}dt

=0e2t(t1)2estdt=\displaystyle\int_{0}^{\infin}e^{2t}(t-1)^2e^{-st}dt

e(2s)t(t1)2dt=e(2s)ts2(t1)2\int e^{(2-s)t}(t-1)^2dt=-\dfrac{e^{(2-s)t}}{s-2}(t-1)^2

+2s2e(2s)t(t1)dt+\dfrac{2}{s-2}\int e^{(2-s)t}(t-1)dt



e(2s)t(t1)dt=e(2s)ts2(t1)\int e^{(2-s)t}(t-1)dt=-\dfrac{e^{(2-s)t}}{s-2}(t-1)

+1s2e(2s)tdt+\dfrac{1}{s-2}\int e^{(2-s)t}dt

=e(2s)ts2(t1)e(2s)t(s2)2+C1=-\dfrac{e^{(2-s)t}}{s-2}(t-1)-\dfrac{e^{(2-s)t}}{(s-2)^2}+C_1


e(2s)t(t1)2dt=e(2s)ts2(t1)2\int e^{(2-s)t}(t-1)^2dt=-\dfrac{e^{(2-s)t}}{s-2}(t-1)^2

2e(2s)t(s2)2(t1)2e(2s)t(s2)3+C-\dfrac{2e^{(2-s)t}}{(s-2)^2}(t-1)-\dfrac{2e^{(2-s)t}}{(s-2)^3}+C


F(s)=L{f(t)}=limA0Ae(2s)t(1t)2dtF(s)=L\{f(t)\}=\lim\limits_{A\to \infin}\displaystyle\int_{0}^{A}e^{(2-s)t}(1-t)^2dt

=1s22(s2)2+2(s2)3=\dfrac{1}{s-2}-\dfrac{2}{(s-2)^2}+\dfrac{2}{(s-2)^3}

=s24s+42s+4+2(s2)3=\dfrac{s^2-4s+4-2s+4+2}{(s-2)^3}

=s26s+10(s2)3=\dfrac{s^2-6s+10}{(s-2)^3}

L{e2t(t1)2}=1s22(s2)2+2(s2)3L\{e^{2t}(t-1)^2\}=\dfrac{1}{s-2}-\dfrac{2}{(s-2)^2}+\dfrac{2}{(s-2)^3}

=s26s+10(s2)3=\dfrac{s^2-6s+10}{(s-2)^3}



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS