Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 128093 by BHOOPENDRA last updated on 04/Jan/21

Answered by Dwaipayan Shikari last updated on 04/Jan/21

$$\mathcal{L}(e^{2t}+4t^3-2sin3t+3cos3t)$$$$={\int }_0^{\mathrm{\infty }}{e}^{2t-st}+{\int }_0^{\mathrm{\infty }}4t^3{e}^{-st}+i{\int }_0^{\mathrm{\infty }}{e}^{3it-st}-e^{-3it-st}+\frac{3}{2}{\int }_0^{\mathrm{\infty }}{e}^{3it-st}+e^{-3it+st}$$$$=\frac{1}{s-2}+\frac{4}{s^4}{\int }_0^{\mathrm{\infty }}{u}^3{e}^{-u}+i(\frac{1}{s-3i}-\frac{1}{s+3i})+\frac{3}{2}(\frac{1}{s-3i}+\frac{1}{s+3i})$$$$=\frac{1}{s-2}+\frac{4\mathrm{\Gamma }\left(4\right)}{s^4}-\frac{6}{s^2+9}+\frac{3s}{s^2+9}=\frac{1}{s-2}+\frac{24}{s^4}+\frac{3(s-2)}{s^2+9}$$

Commented by BHOOPENDRA last updated on 04/Jan/21

$$thankyou$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com