Use Laplace transforms to solve the following initial value problem. x"+x=3 cos 3t , x(0)=1 , x'(0)=0

tabita57i 2021-09-26 Answered

Use Laplace transforms to solve the following initial value problem.
\(x''+x=3 \cos 3t , x(0)=1 , x'(0)=0\)
The solution is \(x(t)=?\)

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Expert Answer

cyhuddwyr9
Answered 2021-09-27 Author has 11209 answers

Step 1
The given differential equation is
\(\displaystyle{x}{''}+{x}={3}{\cos{{\left({3}{t}\right)}}}\)
Initial condition: \(\displaystyle{x}{\left({0}\right)}={1}{\quad\text{and}\quad}{x}'{\left({0}\right)}={0}\)
Solve the differential equation by using laplace transform.
step 2
Consider
\(\displaystyle{x}{''}+{x}={3}{\cos{{\left({3}{t}\right)}}}\)
Operate laplace operator on the both sides, then
\(L\left\{x"(t)\right\}+L\left\{x(t)\right\}=L\left\{3 \cos (3t)\right\}\)
\(\displaystyle{s}^{{2}}{x}{\left({s}\right)}-{s}{x}'{\left({0}\right)}-{x}'{\left({0}\right)}+{x}{\left({s}\right)}={\frac{{{3}{s}}}{{{s}^{{2}}+{9}}}}\)
\(\displaystyle{s}^{{2}}{x}{\left({s}\right)}+{x}{\left({s}\right)}-{1}={\frac{{{3}{s}}}{{{s}^{{2}}+{9}}}}\)
\(\displaystyle{\left({s}^{{2}}+{1}\right)}{x}{\left({s}\right)}={\frac{{{3}{s}}}{{{s}^{{2}}+{9}}}}+{1}\)
\(\displaystyle{x}{\left({s}\right)}={\frac{{{s}^{{2}}+{3}{s}+{9}}}{{{\left({s}^{{2}}+{9}\right)}{\left({s}^{{2}}+{1}\right)}}}}\)
Step 3
By using partial fraction method,
\(\displaystyle{x}{\left({s}\right)}={\frac{{{s}^{{2}}+{3}{s}+{9}}}{{{\left({s}^{{2}}+{9}\right)}{\left({s}^{{2}}+{1}\right)}}}}\)
\(\displaystyle={\frac{{{3}}}{{{8}}}}{\left({\frac{{{s}}}{{{\left({s}^{{2}}+{9}\right)}}}}\right)}+{\frac{{{3}}}{{{8}}}}{\left({\frac{{{s}}}{{{\left({s}^{{2}}+{1}\right)}}}}\right)}+{\frac{{{1}}}{{{\left({s}^{{2}}+{1}\right)}}}}\)
Operate laplace inverse on the both sides,then
\(L^{-1}\left\{x(s)\right\}=\frac{3}{8}L^{-1}\left(\frac{s}{(s^2+9)}\right)+\frac{3}{8}L^{-1}\left(\frac{s}{(s^2+1)}\right)+L^{-1}\left(\frac{1}{(s^2+1)}\right)\)
\(\displaystyle={\frac{{{3}}}{{{8}}}}{\cos{{\left({3}{t}\right)}}}+{\frac{{{3}}}{{{8}}}}{\cos{{\left({t}\right)}}}+{\sin{{\left({t}\right)}}}\)
\(\displaystyle\Rightarrow{x}{\left({t}\right)}={\frac{{{3}}}{{{8}}}}{\cos{{\left({3}{t}\right)}}}+{\frac{{{3}}}{{{8}}}}{\cos{{\left({t}\right)}}}+{\sin{{\left({t}\right)}}}\)
Hence,the required solution is
\(\displaystyle{x}{\left({t}\right)}={\frac{{{3}}}{{{8}}}}{\cos{{\left({3}{t}\right)}}}+{\frac{{{3}}}{{{8}}}}{\cos{{\left({t}\right)}}}+{\sin{{\left({t}\right)}}}\)

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