question 2\ncalculate the laplace transform of the following functions\n(2.1) $e^{-2t}cos4t$.\n(2.2) $(2t +…

Explanation:

Step1: Recall Laplace - transform formula for $e^{at}\cos(bt)$

The Laplace - transform formula for $e^{at}\cos(bt)$ is $L{e^{at}\cos(bt)}=\frac{s - a}{(s - a)^2+b^2}$. For $e^{-2t}\cos(4t)$, where $a=-2$ and $b = 4$, we have: $L{e^{-2t}\cos(4t)}=\frac{s+2}{(s + 2)^2+16}=\frac{s + 2}{s^{2}+4s+4 + 16}=\frac{s + 2}{s^{2}+4s + 20}$

Step2: Rewrite $(2t + 1)u(t - 1)$

First, rewrite $2t+1=2(t - 1)+3$. Then $(2t + 1)u(t - 1)=(2(t - 1)+3)u(t - 1)=2(t - 1)u(t - 1)+3u(t - 1)$. Using the formula $L{t^n u(t - a)}=e^{-as}\frac{n!}{s^{n + 1}}$ and $L{u(t - a)}=\frac{e^{-as}}{s}$, we get $L{(2t + 1)u(t - 1)}=2e^{-s}\frac{1!}{s^{2}}+3\frac{e^{-s}}{s}=e^{-s}(\frac{2}{s^{2}}+\frac{3}{s})=\frac{e^{-s}(3s + 2)}{s^{2}}$

Step3: Use the convolution theorem for $\int_{0}^{t}e^{-\tau}\cos\tau d\tau$

The Laplace - transform of $e^{-t}\cos t$ is $L{e^{-t}\cos t}=\frac{s + 1}{(s + 1)^2+1}=\frac{s + 1}{s^{2}+2s+2}$. By the convolution theorem $L{\int_{0}^{t}f(\tau)g(t-\tau)d\tau}=L{f(t)}L{g(t)}$, and since $\int_{0}^{t}e^{-\tau}\cos\tau d\tau$ is the convolution of $e^{-t}\cos t$ and $1$, and $L{1}=\frac{1}{s}$, we have $L{\int_{0}^{t}e^{-\tau}\cos\tau d\tau}=\frac{s + 1}{s(s^{2}+2s + 2)}$

Step4: Use the multiplication - by - $t$ property for $te^{2t}\sin(6t)$

The Laplace - transform of $\sin(6t)$ is $L{\sin(6t)}=\frac{6}{s^{2}+36}$, and by the property $L{t^n f(t)}=(-1)^n\frac{d^n}{ds^n}L{f(t)}$. For $n = 1$ and $f(t)=e^{2t}\sin(6t)$ (where $L{e^{2t}\sin(6t)}=\frac{6}{(s - 2)^{2}+36}=\frac{6}{s^{2}-4s + 40}$), we have: $\frac{d}{ds}\left(\frac{6}{s^{2}-4s + 40}\right)=6\frac{-(2s - 4)}{(s^{2}-4s + 40)^{2}}$, so $L{te^{2t}\sin(6t)}=\frac{12 - 12s}{(s^{2}-4s + 40)^{2}}$

Step5: Rewrite $t^{2}u(t - 2)$

First, rewrite $t^{2}=(t - 2+2)^{2}=(t - 2)^{2}+4(t - 2)+4$. Then $t^{2}u(t - 2)=((t - 2)^{2}+4(t - 2)+4)u(t - 2)$. Using the formula $L{t^n u(t - a)}=e^{-as}\frac{n!}{s^{n+1}}$, we get $L{t^{2}u(t - 2)}=e^{-2s}(\frac{2}{s^{3}}+\frac{4}{s^{2}}+\frac{4}{s})=\frac{e^{-2s}(4s^{2}+4s + 2)}{s^{3}}$

Step6: Use the convolution theorem for $t\int_{0}^{t}\tau e^{-\tau}d\tau$

The Laplace - transform of $\tau e^{-\tau}$ is $L{\tau e^{-\tau}}=\frac{1}{(s + 1)^{2}}$, and $L{\int_{0}^{t}\tau e^{-\tau}d\tau}=\frac{1}{s(s + 1)^{2}}$. By the property $L{t f(t)}=(-1)\frac{d}{ds}L{f(t)}$, we have $\frac{d}{ds}\left(\frac{1}{s(s + 1)^{2}}\right)=\frac{d}{ds}\left(\frac{1}{s(s^{2}+2s + 1)}\right)=\frac{d}{ds}\left(\frac{1}{s^{3}+2s^{2}+s}\right)$. Using the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = 1$ and $v=s^{3}+2s^{2}+s$, $u'=0$ and $v'=3s^{2}+4s + 1$. So $\frac{d}{ds}\left(\frac{1}{s^{3}+2s^{2}+s}\right)=-\frac{3s^{2}+4s + 1}{(s^{3}+2s^{2}+s)^{2}}$, and $L{t\int_{0}^{t}\tau e^{-\tau}d\tau}=\frac{3s^{2}+4s + 1}{(s^{3}+2s^{2}+s)^{2}}$

Answer:

(2.1) $\frac{s + 2}{s^{2}+4s + 20}$ (2.2) $\frac{e^{-s}(3s + 2)}{s^{2}}$ (2.3) $\frac{s + 1}{s(s^{2}+2s + 2)}$ (2.4) $\frac{12 - 12s}{(s^{2}-4s + 40)^{2}}$ (2.5) $\frac{e^{-2s}(4s^{2}+4s + 2)}{s^{3}}$ (2.6) $\frac{3s^{2}+4s + 1}{(s^{3}+2s^{2}+s)^{2}}$