III.Find L-15²-15s-11)(s+1)(s-2)3.docs.google.com1-et-t²e²t+4te²t+e²tOption 12- et + t²e²t - 4te²te²tOption 213-e-3t-t²e-2t2te²t+e-²t10:01

We have to find the inverse Laplace of 5s2−15s−11s+1s−23 that is L−15s2−15s−11s+1s−23. First of…

Answered: Find 1-e¯¹ −²t²e²t + 4te²t + ½e²t…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

III.
Find L-15²-15s-11)
(s+1)(s-2)3.
docs.google.com
1-et-t²e²t+4te²t+e²t
Option 1
2- et + t²e²t - 4te²te²t
Option 2
1
3-e-3t-t²e-2t2te²t+e-²t
10:01

Expert Solution

Step 1

We have to find the inverse Laplace of $\frac{5 s^{2} - 15 s - 11}{(s + 1) {(s - 2)}^{3}}$ that is $L^{- 1} \{\frac{5 s^{2} - 15 s - 11}{(s + 1) {(s - 2)}^{3}}\}$ .

First of all, we will take a partial fraction $\frac{5 s^{2} - 15 s - 11}{(s + 1) {(s - 2)}^{3}}$ , which is given below

$\frac{5 s^{2} - 15 s - 11}{(s + 1) {(s - 2)}^{3}} = - \frac{1}{3 (s + 1)} + \frac{1}{3 (s - 2)} + \frac{4}{{(s - 2)}^{2}} - \frac{7}{{(s - 2)}^{3}}$

Hence,

$\begin{matrix} L^{- 1} \{\frac{5 s^{2} - 15 s - 11}{(s + 1) {(s - 2)}^{3}}\} = L^{- 1} \{- \frac{1}{3 (s + 1)} + \frac{1}{3 (s - 2)} + \frac{4}{{(s - 2)}^{2}} - \frac{7}{{(s - 2)}^{3}}\} \\ = L^{- 1} \{- \frac{1}{3 (s + 1)}\} + L^{- 1} \{\frac{1}{3 (s - 2)}\} + L^{- 1} \{\frac{4}{{(s - 2)}^{2}}\} + L^{- 1} \{- \frac{7}{{(s - 2)}^{3}}\} \\ = - \frac{1}{3} L^{- 1} \{\frac{1}{(s + 1)}\} + \frac{1}{3} L^{- 1} \{\frac{1}{(s - 2)}\} + 4 L^{- 1} \{\frac{1}{{(s - 2)}^{2}}\} - 7 L^{- 1} \{\frac{1}{{(s - 2)}^{3}}\} \end{matrix}$

Hence, $L^{- 1} \{\frac{5 s^{2} - 15 s - 11}{(s + 1) {(s - 2)}^{3}}\} = - \frac{1}{3} L^{- 1} \{\frac{1}{(s + 1)}\} + \frac{1}{3} L^{- 1} \{\frac{1}{(s - 2)}\} + 4 L^{- 1} \{\frac{1}{{(s - 2)}^{2}}\} - 7 L^{- 1} \{\frac{1}{{(s - 2)}^{3}}\} . . . . . . . (1)$ .