Question

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

Answer 1

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 \({y}{\left({t}\right)}={e}^{{-{t}}}{\left({2}{t}+{1}\right)}\)