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Question 4 Your answer is INCORRECT. Give the Laplace transform of f(x) = -2x2 – 3 + 3e3* cos(2x) - 3xe2x » ©F6)=-* 1) OF(S) ==* * = 5-65+ 13 6- 00F(0) - 4,2-68–13 6-232 b) +- 3s 32-65 +13 ( e) - . 3(3-2) s2 - 65 + 13 - S (S-2)2 1) None of the above. Question 5 Your answer is CORRECT. Give the inverse Laplace transform of F(s) = S-4 s(8 + 3) + 4s - 2 S2 +9 9 a) f(x) = 2*** + 4 cos(33) – 2 sin(3x) b) f(x) = - + e* + 4 cos(38) – į sin(33) c) Of() = *e* +4 cos(3x) = -sin (3x) d) • f(x) =-**** +4cos(3x) – įsin(3x) e) Of(x) = *+ *e* + 4 cos(3x) – 2 sin(3x) f) None of the above.

Answer 1

$f(t)=-2t^2-3+3e^{3t}\cos \left(2t\right)-3te^{2t}$

take Laplace

$F(s)= L\left\{-2t^2-3+3e^{3t}\cos \left(2t\right)-3e^{2t}t\right\}$

$F(s)=-2L\left\{t^2\right\}-L\left\{3\right\}+3L\left\{e^{3t}\cos \left(2t\right)\right\}-3L\left\{te^{2t}\right\}$

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apply Laplace rule: $L\left\{t^n\right\}=\frac{n!}{s^{n+1}}$

$F(s)= -2\cdot \frac{2}{s^3}-\frac{3}{s} +3L\left\{e^{3t}\cos \left(2t\right)\right\}-3L\left\{te^{2t}\right\}$

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apply Laplace rule: $L\left\{e^{at}f\left(t\right)\right\}=F\left(s-a\right)$

$\mathrm{For\:}e^{3t}\cos \left(2t\right):\quad f\left(t\right)=\cos \left(2t\right),\:\quad \:a=3$

$L\left\{\cos \left(2t\right)\right\}=\frac{s}{s^2+2^2}$

$L\left\{\cos \left(2t\right)\right\}=\frac{s}{s^2+4}$

$L\left\{e^{3t}\cos \left(2t\right)\right\}= \frac{\left(s-3\right)}{\left(s-3\right)^2+4}$

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$F(s)=-2L\left\{t^2\right\}-L\left\{3\right\}+ 3\cdot \frac{s-3}{\left(s-3\right)^2+4}-3L\left\{te^{2t}\right\}$

apply Laplace rule: $\left(-1\right)^k\frac{d}{ds^k}\left(F\left(s\right)\right)$

$L\left\{e^{2t}\right\}=\frac{1}{s-2}$

as per rule take derivative

$L\left\{te^{2t}\right\}= \left(-1\right)^1\frac{d}{ds}\left(\frac{1}{s-2}\right)$

$L\left\{te^{2t}\right\}= \left(-1\right)\left(-\frac{1}{\left(s-2\right)^2}\right)$

$L\left\{te^{2t}\right\}=\frac{1}{\left(s-2\right)^2}$

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$F(s)=-2\cdot \frac{2}{s^3}-\frac{3}{s}+3\cdot \frac{s-3}{\left(s-3\right)^2+4}-3\cdot \frac{1}{\left(s-2\right)^2}$

$F(s)=-\frac{4}{s^3}-\frac{3}{s}+\frac{3\left(s-3\right)}{\left(s-3\right)^2+4}-\frac{3}{\left(s-2\right)^2}$

$F\left(s\right)=-\frac{4}{s^3}-\frac{3}{s}+\frac{3\left(s-3\right)}{s^2-6s+9+4}-\frac{3}{\left(s-2\right)^2}$

${\color{Red} F\left(s\right)=-\frac{4}{s^3}-\frac{3}{s}+\frac{3\left(s-3\right)}{s^2-6s+13}-\frac{3}{\left(s-2\right)^2}}$...............................option D

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$F(s)=\frac{s-4}{s\left(s+3\right)}+\frac{4s-2}{s^2+9}$

take inverse Laplace

$f(t)= L^{-1}\left\{\frac{s-4}{s\left(s+3\right)}+\frac{4s-2}{s^2+9}\right\}$

take partial fraction

$\frac{s-4}{s\left(s+3\right)}=\frac{a_0}{s}+\frac{a_1}{s+3}$

$s-4=a_0\left(s+3\right)+a_1s$

take s=-3

$\left(-3\right)-4=a_0\left(\left(-3\right)+3\right)+a_1\left(-3\right)$

$-7=-3a_1$

${\color{Blue} a_1=\frac{7}{3}}$

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taks s=0

$0-4=a_0\left(0+3\right)+a_1\cdot \:0$

$-4=3a_0$

${\color{Blue} a_0=-\frac{4}{3}}$

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$\frac{s-4}{s\left(s+3\right)}= \frac{\left(-\frac{4}{3}\right)}{s}+\frac{\frac{7}{3}}{s+3}$

$\frac{s-4}{s\left(s+3\right)}= \frac{7}{3\left(s+3\right)}-\frac{4}{3s}$

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$f(t)= L^{-1}\left\{\frac{7}{3\left(s+3\right)}-\frac{4}{3s}+\frac{4s-2}{s^2+9}\right\}$

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$f(t)=L^{-1}\left\{\frac{7}{3\left(s+3\right)}+\frac{4s}{s^2+9}-\frac{4}{3s}-\frac{2}{s^2+9}\right\}$

$f(t)=\frac{7}{3}L^{-1}\left\{\frac{1}{s+3}\right\}-\frac{4}{3}L^{-1}\left\{\frac{1}{s}\right\}+4L^{-1}\left\{\frac{s}{s^2+9}\right\}-2L^{-1}\left\{\frac{1}{s^2+9}\right\}$

$f(t)=\frac{7}{3}L^{-1}\left\{\frac{1}{s+3}\right\}-\frac{4}{3}L^{-1}\left\{\frac{1}{s}\right\}+4L^{-1}\left\{\frac{s}{s^2+3^2}\right\}-2L^{-1}\left\{\frac{1}{s^2+3^2}\right\}$

$f(t)=\frac{7}{3}L^{-1}\left\{\frac{1}{s+3}\right\}-\frac{4}{3}L^{-1}\left\{\frac{1}{s}\right\}+4L^{-1}\left\{\frac{s}{s^2+3^2}\right\}-2L^{-1}\left\{\frac{1}{3}\cdot \frac{3}{s^2+3^2}\right\}$

$f(t)=\frac{7}{3}L^{-1}\left\{\frac{1}{s+3}\right\}-\frac{4}{3}L^{-1}\left\{\frac{1}{s}\right\}+4L^{-1}\left\{\frac{s}{s^2+3^2}\right\}-2\cdot \frac{1}{3}\cdot L^{-1}\left\{\frac{3}{s^2+3^2}\right\}$

$f(t)= \frac{7}{3}e^{-3t}-\frac{4}{3}+4\cos \left(3t\right)-2\cdot \frac{1}{3}\sin \left(3t\right)$

$f(t)=\frac{7}{3}e^{-3t}-\frac{4}{3} +4\cos \left(3t\right)-\frac{2}{3}\sin \left(3t\right)$

${\color{Red} f(t)=-\frac{4}{3} +\frac{7}{3}e^{-3t}+4\cos \left(3t\right)-\frac{2}{3}\sin \left(3t\right)}$.....................option D