use the Laplace transform to solve the given initial-value problem. y"+y=f(t) y(0)=0 , y'(0)=1 where displaystyle f{{left({t}right)}}={leftlbracebegin{matrix}{1}&{0}le{t}

use the Laplace transform to solve the given initial-value problem. y"+y=f(t) y(0)=0 , y'(0)=1 where displaystyle f{{left({t}right)}}={leftlbracebegin{matrix}{1}&{0}le{t}<frac{pi}{{2}}{0}&{t}gefrac{pi}{{2}}end{matrix}right.}

Question
Laplace transform
asked 2021-03-04
use the Laplace transform to solve the given initial-value problem.
\(y"+y=f(t)\)
\(y(0)=0 , y'(0)=1\) where
\(\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}\)</span>

Answers (1)

2021-03-05
Step 1
Given, the initial value problem
\(y"+y=f(t)\)
\(y(0)=0 , y'(0)=1\) where
\(\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}\)</span>
step 2
The Laplace transform of the two derivatives
\(\displaystyle{L}{\left\lbrace{y}'\right\rbrace}={s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\)
\(\displaystyle{L}{\left\lbrace{y}\text{}\right\rbrace}={s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\)
\(\displaystyle{L}{\left\lbrace{1}\right\rbrace}=\frac{1}{{s}}\)
\(\displaystyle{L}{\left\lbrace{e}^{{{a}{t}}}\right\rbrace}=\frac{1}{{{s}-{a}}}\)
Where y(0) and y’(0) are initial conditions.
Step 3
Now,
when \(\displaystyle{0}\le{t}\le\frac{\pi}{{2}}\)
\(y"+y'=1 , y(0)=0 . y'(0)=1\)
\(\displaystyle{s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}={1}\)
\(\displaystyle{\left\lbrace{s}^{2}+{s}\right\rbrace}{Y}{\left({s}\right)}-{s}\cdot{0}-{1}-{0}={1}\)
\(\displaystyle{\left\lbrace{s}^{2}+{s}\right\rbrace}{Y}{\left({s}\right)}={2}\)
\(\displaystyle{Y}{\left({s}\right)}=\frac{2}{{{s}^{2}+{s}}}=\frac{2}{{{s}{\left({s}+{1}\right)}}}={2}{\left\lbrace\frac{1}{{s}}-\frac{1}{{{s}+{1}}}\right\rbrace}\)
\(\displaystyle{Y}{\left({s}\right)}={2}{\left\lbrace{1}-{e}^{{-{t}}}\right\rbrace}\)
when \(\displaystyle{t}\ge\frac{\pi}{{2}}\)
\(y"+y'=0, y(0)=0 , y'(0)=1\)
\(\displaystyle{s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}={1}\)
\(\displaystyle{\left\lbrace{s}^{2}+{s}\right\rbrace}{Y}{\left({s}\right)}-{s}\cdot{0}-{1}-{0}={0}\)
\(\displaystyle{\left\lbrace{s}^{2}+{s}\right\rbrace}{Y}{\left({s}\right)}={1}\)
\(\displaystyle{Y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{s}}}=\frac{1}{{{s}{\left({s}+{1}\right)}}}={\left\lbrace\frac{1}{{s}}-\frac{1}{{{s}+{1}}}\right\rbrace}\)
\(\displaystyle{Y}{\left({s}\right)}={\left\lbrace{1}-{e}^{{-{t}}}\right\rbrace}\)
Step 4
Hence the solution is
\(\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{2}{\left\lbrace{1}-{e}^{{-{t}}}\right\rbrace}&{0}\le{t}<\frac{\pi}{{2}}\\{1}-{e}^{{-{t}}}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}\)</span>
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