Question

Solve the equation using Laplace transforms: y''+6y'+13y=0 y(0)=2, y'(0)=8

Answer 1

S

L(ysy(s)-y(0)

Ly y(s)sy(0)-y(0)

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6y13y = 0

take laplace

Ly6Ly+ 13L{y]= 0

(ssy(0y (0)6(sy(s)y(0)) +13y(s)0

here  y(0)=2, y'(0)=8

s2(s) s(2)8+6(sy(s) 2) 13y(s) 0

s2y(s) 2s 86sy(s) 12 13y(s) 0

s2(s)2s 20 6sy(s)13y(s) 0

ys6sy(s)+13y(s)2s20

y(s) (s6s 13) 2s 20

2s20 y(s)26s+ 13

2s 20 y(s)s+3)+ 4

y(s)2(s+3)14 2 (s3 4

$y(s)= 2\cdot \frac{s+3}{\left(s+3\right)^2+4}+14\cdot \frac{1}{\left(s+3\right)^2+4}$

take inverse laplace

$y(t)= L^{-1}\left\{2\cdot \frac{s+3}{\left(s+3\right)^2+4}+14\cdot \frac{1}{\left(s+3\right)^2+4}\right\}$

$y(t)= 2L^{-1}\left\{\frac{s+3}{\left(s+3\right)^2+4}\right\}+14L^{-1}\left\{\frac{1}{\left(s+3\right)^2+4}\right\}$

apply laplace rule  $\:L^{-1}\left\{F\left(s-a\right)\right\}=e^{at}f\left(t\right)$

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

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

$y(t)=2e^{-3t}L^{-1}\left\{\frac{s}{s^2+4}\right\}+14e^{-3t}L^{-1}\left\{\frac{1}{s^2+4}\right\}$

$y(t)=2e^{-3t}L^{-1}\left\{\frac{s}{s^2+2^2}\right\}+14e^{-3t}L^{-1}\left\{\frac{1}{s^2+2^2}\right\}$

$y(t)=2e^{-3t}L^{-1}\left\{\frac{s}{s^2+2^2}\right\}+14e^{-3t}L^{-1}\left\{\frac{1}{2}\cdot \frac{2}{s^2+2^2}\right\}$

1 3t 2 y(t) 2e3L1 s222 14e 1 2 2 22

$y(t)=2e^{-3t}\cos \left(2t\right)+14e^{-3t}\frac{1}{2}\sin \left(2t\right)$

${\color{Red} y(t)=2e^{-3t}\cos \left(2t\right)+7e^{-3t}\sin \left(2t\right)}$