Answer to Question #232313 in Math for abi

Question

L{e^2t(t-1)²}

Answer 1

$$F(s)=L\{f(t)\}=\displaystyle\int_{0}^{\infin}f(t)e^{-st}dt$$

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

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

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

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

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

$$=-\dfrac{e^{(2-s)t}}{s-2}(t-1)-\dfrac{e^{(2-s)t}}{(s-2)^2}+C_1$$

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

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

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

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

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

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

$$L\{e^{2t}(t-1)^2\}=\dfrac{1}{s-2}-\dfrac{2}{(s-2)^2}+\dfrac{2}{(s-2)^3}$$

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