Answer to Question #343126 in Differential Equations for Tobias Felix

Question #343126

Find the Laplace transforms of the following function:

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


1
Expert's answer
2022-05-24T17:58:26-0400
F(s)=L(t2e2t+etcos(2t)+3)F(s)=L(t^2e^{-2t}+e^{-t}\cos(2t)+3)

=0est(t2e2t+etcos(2t)+3)dt=\displaystyle\int_{0}^{\infin}e^{-st}(t^2e^{-2t}+e^{-t}\cos(2t)+3)dt

t2e(s+2)tdt=t2e(s+2)ts+2+2s+2te(s+2)tdt\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

=t2e(s+2)ts+22te(s+2)t(s+2)2=-\dfrac{t^2e^{-(s+2)t}}{s+2}-\dfrac{2te^{-(s+2)t}}{(s+2)^2}


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

=t2e(s+2)ts+22te(s+2)t(s+2)22e(s+2)t(s+2)3+C=-\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(t2e2t)=limA0est(t2e2t)dtL(t^2e^{-2t})=\lim\limits_{A\to\infin}\displaystyle\int_{0}^{\infin}e^{-st}(t^2e^{-2t})dt

=[t2e(s+2)ts+22te(s+2)t(s+2)22e(s+2)t(s+2)3]A0=[-\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}

=000(002(s+2)3)=2(s+2)3=-0-0-0-(-0-0-\dfrac{2}{(s+2)^3})=\dfrac{2}{(s+2)^3}

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

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

+2e(s+1)tsin(2t)(s+1)24(s+1)2e(s+1)tcos(2t)dt+\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)e(s+1)tcos(2t)dt((s+1)^2+4)\int e^{-(s+1)t}\cos (2t)dt

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

e(s+1)tcos(2t)dt=e(s+1)tcos(2t)(s+1)s2+2s+5\int e^{-(s+1)t}\cos (2t)dt=-\dfrac{e^{-(s+1)t}\cos(2t)(s+1)}{s^2+2s+5}

+2e(s+1)tsin(2t)s2+s+5+C1+\dfrac{2e^{-(s+1)t}\sin(2t)}{s^2+s+5}+C_1


L(etcos(2t))=limA0est(etcos(2t))dtL(e^{-t}\cos(2t))=\lim\limits_{A\to\infin}\displaystyle\int_{0}^{\infin}e^{-st}(e^{-t}\cos(2t))dt

=[e(s+1)tcos(2t)(s+1)s2+2s+5+2e(s+1)tsin(2t)s2+s+5]A0=[-\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(s+1s2+2s+5+0)=s+1s2+2s+5=-0+0-(-\dfrac{s+1}{s^2+2s+5}+0)=\dfrac{s+1}{s^2+2s+5}



L(3)=limA0est(3)dtL(3)=\lim\limits_{A\to\infin}\displaystyle\int_{0}^{\infin}e^{-st}(3)dt

=[3ests+]A0=0+3s=3s=[-\dfrac{3e^{-st}}{s}+]\begin{matrix} A \\ 0 \end{matrix}=-0+\dfrac{3}{s}=\dfrac{3}{s}


L(t2e2t+etcos(2t)+3)L(t^2e^{-2t}+e^{-t}\cos(2t)+3)

=2(s+2)3+s+1s2+2s+5+3s=\dfrac{2}{(s+2)^3}+\dfrac{s+1}{s^2+2s+5}+\dfrac{3}{s}


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