Question

Use the Laplace transform to solve the given initial-value problem.
\(\displaystyle{y}{''}-{6}{y}'+{13}{y}={0},\ \ \ {y}{\left({0}\right)}={0},\ \ \ {y}'{\left({0}\right)}=-{9}\)

Answer 1

Step 1
The given differential equation is
\(y''-6y'+13y=0\)
Step 2
Simplify the initial value problem using Laplace transform:
\(\displaystyle{L}{\left({y}\text{}-{6}{y}'+{13}{y}\right)}={L}{\left({0}\right)}\)
\(\displaystyle{L}{\left({y}\text{}\right)}-{6}{L}{\left({y}'\right)}+{13}{L}{\left({y}\right)}={0}\)
\(\displaystyle{s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}-{6}{\left({s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\right)}+{13}{Y}{\left({s}\right)}={0}\)
\(\displaystyle{s}^{2}{Y}{\left({s}\right)}+{9}-{6}{s}{Y}{\left({s}\right)}+{13}{Y}{\left({s}\right)}={0}\)
\(\displaystyle{Y}{\left({s}\right)}=-\frac{9}{{{s}^{2}-{6}{s}+{9}+{4}}}\)
Step 3
Simplify further,
\(\displaystyle{Y}{\left({s}\right)}=-\frac{9}{{{\left({s}-{3}\right)}^{2}+{4}}}\)
\(\displaystyle{Y}{\left({s}\right)}=-\frac{9}{{{\left({s}-{3}\right)}^{2}+{2}^{2}}}\)
Take inverse laplace on both sides,
\(\displaystyle{L}^{ -{{1}}}{\left({Y}{\left({s}\right)}\right)}={L}^{ -{{1}}}{\left(-\frac{9}{{{\left({s}-{3}\right)}^{2}+{2}^{2}}}\right)}\)
\(\displaystyle{y}{\left({t}\right)}=-\frac{9}{{2}}{L}^{ -{{1}}}{\left(\frac{2}{{{\left({s}-{3}\right)}^{2}+{2}^{2}}}\right)}\)
\(\displaystyle{y}{\left({t}\right)}=-\frac{9}{{2}}{e}^{{{3}{t}}} \sin{{2}}{t}\)
Hence , the solution of the initial value problem is \(\displaystyle{y}{\left({t}\right)}=-\frac{9}{{2}}{e}^{{{3}{t}}} \sin{{2}}{t}