# Use the Laplace transform to solve the given initial-value problem {y}{''}+{2}{y}'+{y}={0},{y}{left({0}right)}={1},{y}'{left({0}right)}={1}

Question
Laplace transform
Use the Laplace transform to solve the given initial-value problem
$${y}{''}+{2}{y}'+{y}={0},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$

2020-11-01
Step 1
It is given that the initial value problem is $${y}{''}+{2}{y}'+{y}={0},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$
Step 2
Take Laplace transformation as,
$${L}{\left\lbrace{y}\text{}\right\rbrace}+{2}{L}{\left\lbrace{y}'\right\rbrace}+{L}{\left\lbrace{y}\right\rbrace}={L}{\left\lbrace{0}\right\rbrace}$$
$${s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{2}{\left[{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\right]}+{Y}{\left({s}\right)}={0}$$
$${s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{2}{s}{Y}{\left({s}\right)}-{2}{y}{\left({0}\right)}+{Y}{\left({s}\right)}={0}$$
$${s}^{2}{Y}{\left({s}\right)}-{s}-{1}+{2}{s}{Y}{\left({s}\right)}-{2}+{Y}{\left({s}\right)}={0}$$
$${Y}{\left({s}\right)}{\left({s}^{2}+{2}{s}+{1}\right)}-{s}-{3}={0}$$
$${Y}{\left({s}\right)}{\left({s}^{2}+{2}{s}+{1}\right)}={s}+{3}$$
Step 3
On further simplification,
$${Y}{\left({s}\right)}=\frac{{{s}+{3}}}{{\left({s}+{1}\right)}^{2}}$$
$${y}{\left({t}\right)}={L}^{ -{{1}}}{\left\lbrace\frac{{{s}+{3}}}{{\left({s}+{1}\right)}^{2}}\right\rbrace}$$
$${y}{\left({t}\right)}={L}^{ -{{1}}}{\left\lbrace\frac{2}{{\left({s}+{1}\right)}^{2}}+\frac{1}{{{s}+{1}}}\right\rbrace}\ \ \ \ \ \ \ {\left(\text{because }\ \frac{{{s}+{3}}}{{\left({s}+{1}\right)}^{2}}=\frac{2}{{\left({s}+{1}\right)}^{2}}+\frac{1}{{{s}+{1}}}\right)}$$
$${y}{\left({t}\right)}={L}^{ -{{1}}}{\left\lbrace\frac{2}{{\left({s}+{1}\right)}^{2}}\right\rbrace}+{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}+{1}}}\right\rbrace}$$
$${y}{\left({t}\right)}=\frac{{{2}{t}}}{{e}^{t}}+{e}^{{-{t}}}$$
$${y}{\left({t}\right)}={e}^{{-{t}}}{\left({2}{t}+{1}\right)}$$
Step 4
The solution of the initial value problem is $${y}{\left({t}\right)}={e}^{{-{t}}}{\left({2}{t}+{1}\right)}$$

### Relevant Questions

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