Answer to Question #343126 in Differential Equations for Tobias Felix

Question

Find the Laplace transforms of the following function:

8. L{t^2 e^-2t + e^-t cos2t + 3}

Answer 1

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

$$=\displaystyle\int_{0}^{\infin}e^{-st}(t^2e^{-2t}+e^{-t}\cos(2t)+3)dt$$

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

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

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

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

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

$$=[-\dfrac{t^2e^{-(s+2)t}}{s+2}-\dfrac{2te^{-(s+2)t}}{(s+2)^2}-\dfrac{2e^{-(s+2)t}}{(s+2)^3}]\begin{matrix} A \\ 0 \end{matrix}$$

$$=-0-0-0-(-0-0-\dfrac{2}{(s+2)^3})=\dfrac{2}{(s+2)^3}$$

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

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

$$+\dfrac{2e^{-(s+1)t}\sin(2t)}{(s+1)^2}-\dfrac{4}{(s+1)^2}\int e^{-(s+1)t}\cos (2t)dt$$

$$((s+1)^2+4)\int e^{-(s+1)t}\cos (2t)dt$$

$$=-e^{-(s+1)t}\cos(2t)(s+1)+2e^{-(s+1)t}\sin(2t)$$

$$\int e^{-(s+1)t}\cos (2t)dt=-\dfrac{e^{-(s+1)t}\cos(2t)(s+1)}{s^2+2s+5}$$

$$+\dfrac{2e^{-(s+1)t}\sin(2t)}{s^2+s+5}+C_1$$

$$L(e^{-t}\cos(2t))=\lim\limits_{A\to\infin}\displaystyle\int_{0}^{\infin}e^{-st}(e^{-t}\cos(2t))dt$$

$$=[-\dfrac{e^{-(s+1)t}\cos(2t)(s+1)}{s^2+2s+5}+\dfrac{2e^{-(s+1)t}\sin(2t)}{s^2+s+5}]\begin{matrix} A \\ 0 \end{matrix}$$

$$=-0+0-(-\dfrac{s+1}{s^2+2s+5}+0)=\dfrac{s+1}{s^2+2s+5}$$

$$L(3)=\lim\limits_{A\to\infin}\displaystyle\int_{0}^{\infin}e^{-st}(3)dt$$

$$=[-\dfrac{3e^{-st}}{s}+]\begin{matrix} A \\ 0 \end{matrix}=-0+\dfrac{3}{s}=\dfrac{3}{s}$$

$$L(t^2e^{-2t}+e^{-t}\cos(2t)+3)$$

$$=\dfrac{2}{(s+2)^3}+\dfrac{s+1}{s^2+2s+5}+\dfrac{3}{s}$$