 Clifland

2021-02-19

Use Laplace transform to solve the following initial-value problem
$y"+2{y}^{\prime }+y=0$
$y\left(0\right)=1,{y}^{\prime }\left(0\right)=1$
a) ${e}^{-t}+t{e}^{-t}$
b) ${e}^{t}+2t{e}^{t}$
c) ${e}^{-t}+2t{e}^{t}$
d) ${e}^{-t}+2t{e}^{-t}$
e) $2{e}^{-t}+2t{e}^{-t}$
f) Non of the above funblogC

Step 1
Given initial value problem,
${y}^{″}+2{y}^{\prime }+y=0$
$y\left(0\right)=1$
${y}^{\prime }\left(0\right)=1$
Step 2
Taking inverse Laplace transform,
$L\left[y{}^{″}+2{y}^{\prime }+y\right]=0$
$L\left[y{}^{″}\right]+2L\left[{y}^{\prime }\right]+L\left[y\right]=0$
Use the formula such that,
$L\left[y{}^{″}\right]={s}^{2}L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left[{y}^{\prime }\right]=sL\left[y\right]-y\left(0\right)$
Then,
${s}^{2}L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)+2\left[sL\left[y\right]-y\left(0\right)\right]+L\left[y\right]=0$
${s}^{2}L\left[y\right]-s-1+2\left[sL\left[y\right]-1\right]+L\left[y\right]=0$
${s}^{2}L\left[y\right]-s-1+2sL\left[y\right]-2+L\left[y\right]=0$
$\left({s}^{2}+2s+1\right)L\left[y\right]-s-3=0$
$L\left[y\right]=\frac{s+3}{{s}^{2}+2s+1}$
Step 3
Taking inverse Laplace transform of both sides,
$y={L}^{-1}\left[\frac{s+3}{{s}^{2}+2s+1}\right]$
$={L}^{-1}\left[\frac{s+1}{{\left(s+1\right)}^{2}}+\frac{2}{{\left(s+1\right)}^{2}}\right]$
$={L}^{-1}\left[\frac{1}{s+1}\right]+2{L}^{-1}\left[\frac{1}{{\left(s+1\right)}^{2}}\right]$
$={e}^{-t}+2t{e}^{-t}$
Step 4
Hence, the solution of given initial value problem is
$y\left(t\right)={e}^{-t}+2t{e}^{-t}$

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