Question

With the aid of Laplace Transform, solve the Initial Value Problem {y}text{}{left({t}right)}-{y}'{left({t}right)}={e}^{t} cos{{left({t}right)}}+ sin{{left({t}right)}},{y}{left({0}right)}={0},{y}'{left({0}right)}={0}

Laplace transform
ANSWERED
asked 2020-12-09
With the aid of Laplace Transform, solve the Initial Value Problem
\({y}\text{}{\left({t}\right)}-{y}'{\left({t}\right)}={e}^{t} \cos{{\left({t}\right)}}+ \sin{{\left({t}\right)}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={0}\)

Answers (1)

2020-12-10
Step 1
The Laplace Transform of a second order linear differential equation ,
\(y"+ay'+by=f(t) \text{ with } y(0)=A \text{ and } y'(0)=B\)
is given by,
\({Y}{\left({s}\right)}=\frac{{{\left({s}+{a}\right)}{A}+{B}}}{{{s}^{2}+{a}{s}+{b}}}+\frac{{{F}{\left({s}\right)}}}{{{s}^{2}+{a}{s}+{b}}}\)
Given,
\({y}\text{}{\left({t}\right)}-{y}'{\left({t}\right)}={e}^{t} \cos{{\left({t}\right)}}+ \sin{{\left({t}\right)}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={0}\)
Step 2
Applying Laplace Transform in the overall equation (1),
\({y}\text{}{\left({t}\right)}-{y}'{\left({t}\right)}={e}^{t} \cos{{\left({t}\right)}}+ \sin{{\left({t}\right)}}\)
\(\Rightarrow{L}{\left[{y}\text{}{\left({t}\right)},{s}\right]}-{L}{\left[{y}'{\left({t}\right)},{s}\right]}={L}{\left[{e}^{t} \cos{{\left({t}\right)}},{s}\right]}+{L}{\left[ \sin{{\left({t}\right)}},{s}\right]}\)
\(\Rightarrow{\left[{s}^{2}{Y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\right]}-{\left[{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}\right]}={L}{\left[ \cos{{\left({t}\right)}},{s}\right]}_{{s}}\to{s}-{1}+\frac{a}{{{s}^{2}+{1}^{2}}}\)
\(\Rightarrow{s}^{2}{Y}-{s}{Y}{\left({s}\right)}={\left[\frac{s}{{{s}^{2}+{1}}}\right]}_{{s}}\to{s}-{1}+\frac{1}{{{s}^{2}+{1}}}\)
\(\Rightarrow{\left({s}^{2}-{s}\right)}{Y}{\left({s}\right)}={\left[\frac{{{s}-{1}}}{{{\left({s}-{1}\right)}^{2}+{1}}}\right]}+\frac{1}{{{s}^{2}+{1}}}\)
\(\Rightarrow{s}{\left({s}-{1}\right)}{Y}{\left({s}\right)}={\left[\frac{{{s}-{1}}}{{{\left({s}-{1}\right)}^{2}+{1}}}\right]}+\frac{1}{{{s}^{2}+{1}}}\)
\(\Rightarrow{Y}{\left({s}\right)}=\frac{1}{{{s}{\left[{\left({s}-{1}\right)}^{2}+{1}\right]}}}+\frac{1}{{{s}{\left({s}-{1}\right)}{\left({s}^{2}+{1}\right)}}}\)
Step 3
The answer is :\({Y}{\left({s}\right)}=\frac{1}{{{s}{\left[{\left({s}-{1}\right)}^{2}+{1}\right]}}}+\frac{1}{{{s}{\left({s}-{1}\right)}{\left({s}^{2}+{1}\right)}}}\)
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