CoormaBak9

2020-12-17

use the Laplace transform to solve the given initial-value problem.$y"+{y}^{\prime }-2y=10{e}^{-t},y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$

Latisha Oneil

Step 1
Given $y"+{y}^{\prime }-2y=10{e}^{-t}-\left(1\right)$
and $y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$
Using the Laplace transform, (1)
${S}^{2}y\left(s\right)-Sy\left(0\right)-{y}^{\prime }\left(0\right)+sy\left(s\right)-y\left(0\right)-2y\left(s\right)=\frac{10}{S+1}$
Step 2
$⇒\left({S}^{2}+S-2\right)y\left(S\right).-S.0-1-0=\frac{10}{s+1}$
$⇒\left({S}^{2}+S-2\right)y\left(S\right)=1+\frac{10}{s+1}$
$\to Y\left(s\right)=\frac{1}{{S}^{2}+S-2}+\frac{10}{\left(S+1\right)\left({S}^{2}+S-2\right)}$
$=\frac{1}{\left(S+2\right)\left(S-1\right)}+\frac{10}{\left(S+1\right)\left(S-1\right)\left(S+2\right)}$
$=-\frac{1}{3\left(S+2\right)}+\frac{1}{3\left(S-1\right)}+\frac{5}{3\left(S-1\right)}-\frac{5}{S+1}+\frac{10}{3\left(S+2\right)}$
$Y\left(s\right)=\frac{3}{S+2}+\frac{2}{S-1}-\frac{5}{S+1}$
$\therefore y\left(t\right)={e}^{-2t}+{e}^{t}-{e}^{-t}$ for t>0

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