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Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y"+y=g(t) , y(0)=-4 , y'(0)=0 where g{{left({t}right)}}={leftlbracebegin{matrix}{t}&{t}<{4}{5}&{t}>{4}end{matrix}right.} Y(s)-?

Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y"+y=g(t) , y(0)=-4 , y'(0)=0 where g{{left({t}right)}}={leftlbracebegin{matrix}{t}&{t}<{4}{5}&{t}>{4}end{matrix}right.} Y(s)-?

Question
Laplace transform
asked 2020-11-16
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
\(y"+y=g(t) , y(0)=-4 , y'(0)=0\)
where \(g{{\left({t}\right)}}={\left\lbrace\begin{matrix}{t}&{t}<{4}\\{5}&{t}>{4}\end{matrix}\right.}\)
Y(s)-?

Answers (1)

2020-11-17
Step 1 Basics
Here we use Second Shifting Property and Laplace Transform of Derivative as follows
Step 1
Laplace transform of derivative
\({L}{\left\lbrace{f}\text{}{\left({t}\right)}\right\rbrace}={s}^{2}{L}{\left\lbrace{f}\right\rbrace}-{\mathsf{{{0}}}}-{f}'{\left({0}\right)}\)
second shifting property
\({L}{\left\lbrace f{{\left({t}-{t}_{{0}}\right)}}\right\rbrace}={e}^{{-{s}{t}_{{0}}}}{L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}\)
Step 2 Explanation
\(y"+y=g(t) , y(0)=-4 , y'(0)=0\)
where \(g{{\left({t}\right)}}={\left\lbrace\begin{matrix}{t}&{t}<{4}\\{5}&{t}>{4}\end{matrix}\right.}\)
Taking Laplace transform both side of above equation , we get
\({L}{\left\lbrace{y}\text{}\right\rbrace}+{L}{\left\lbrace{y}\right\rbrace}={L}{\left\lbrace g{{\left({t}\right)}}\right\rbrace}\)
\({\left[{s}^{2}{L}{\left\lbrace{y}{\left({t}\right)}\right\rbrace}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\right]}+{L}{\left\lbrace{y}\right\rbrace}={L}{\left\lbrace{t}{u}{\left({t}\right)}+{5}{u}{\left({t}-{u}\right)}\right]}\)
where u(t) is the step function
\({\left[\text{use }\ {L}{\left\lbrace{t}^{n}{u}{\left({n}\right)}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}}\right]}\)
shifting property as \({L}{\left\lbrace f{{\left({t}-{t}_{{0}}\right)}}\right\rbrace}={e}^{{-{s}{t}_{{0}}}}{L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}\)
\({s}^{2}{L}{\left\lbrace{y}\right\rbrace}+{4}{s}+{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{s}^{2}}+{e}^{{-{4}{s}}}{\left(\frac{5}{{s}}\right)}\)
\({L}{\left\lbrace{y}\right\rbrace}{\left[{s}^{2}+{1}\right]}=\frac{1}{{s}^{2}}+\frac{{{5}{e}^{{-{4}{s}}}}}{{s}}-{4}{s}\)
\({L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}^{2}{\left({s}^{2}+{1}\right)}}}+\frac{{{5}{e}^{{-{4}{s}}}}}{{{s}{\left({s}^{2}+{1}\right)}}}-{4}\frac{s}{{{s}^{2}+{1}}}\)
Answer
\({y}{\left({s}\right)}=\frac{1}{{{s}^{2}{\left({s}^{2}+{1}\right)}}}+\frac{{{5}{e}^{{-{4}{s}}}}}{{{s}{\left({s}^{2}+{1}\right)}}}-{4}\frac{s}{{{s}^{2}+{1}}}\)
0

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