Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 20156 by tawa tawa last updated on 22/Aug/17

$$\mathrm{Solve}:\mathrm{inverse}\mathrm{laplace}.{\mathrm{L}}^{-1}\left(\frac{\mathrm{s}}{{\mathrm{s}}^2+6\mathrm{s}+25}\right)$$

Answered by ajfour last updated on 23/Aug/17

$${\mathcal{L}}^{-1}\left(\frac{s}{s^2+6s+25}\right)={\mathcal{L}}^{-1}\left\{\frac{s}{{(s+3)}^2+4^2}\right\}$$$$={\mathcal{L}}^{-1}\left\{\frac{s+3}{{(s+3)}^2+4^2}\right\}-\frac{3}{4}{\mathcal{L}}^{-1}\left\{\frac{4}{{(s+3)}^2+4^2}\right\}$$$$=e^{-3t}\cos 4t-\frac{3}{4}{e}^{-3t}\sin 4t$$$$=\frac{e^{-3t}}{4}(4\cos 4t-3\sin 4t).$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com