Question

Find Y(t) using Laplace transform.displaystyle{y}{''}+{2}{y}'+{5}{y}={e}^{{-{T}}} sin{{T}}y(0)=0 , y'(0)=1

Laplace transform
ANSWERED
asked 2021-01-13

Find Y(t) using Laplace transform.
\(\displaystyle{y}{''}+{2}{y}'+{5}{y}={e}^{{-{T}}} \sin{{T}}\)
\(y(0)=0 , y'(0)=1\)

Expert Answers (1)

2021-01-14
Step 1
Given differential equation is \(\displaystyle{y}{''}+{2}{y}'+{5}{y}={e}^{{-{T}}} \sin{{T}}\) \(y(0)=0 , y'(0)=1\)
\(\displaystyle{y}{\left({T}\right)}\leftrightarrow{Y}{\left({s}\right)}\)
\(\displaystyle{y}'{\left({T}\right)}\leftrightarrow{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}={s}{Y}{\left({s}\right)}\)
Step 2
\(\displaystyle{y}{''}{\left({T}\right)}\leftrightarrow{s}^{2}{Y}{\left({s}\right)}-{s}{y}'{\left({0}\right)}-{y}{\left({0}\right)}={s}^{2}{Y}{\left({s}\right)}^{{-{s}}}\)
since, \(y'(0)=1\)
\(\displaystyle \sin{{\left({T}\right)}}\leftrightarrow\frac{1}{{{s}^{2}+{1}}}\)
\(\displaystyle{e}^{{-{T}}} \sin{{T}}\leftrightarrow\frac{1}{{{\left({s}+{1}\right)}^{2}+{1}}}\)
Now, after applying Laplace transform , the given equation reduces to
\(\displaystyle{s}^{2}{Y}{\left({s}\right)}-{s}+{2}{\left({s}{Y}{\left({s}\right)}\right)}+{5}{Y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{1}}}\)
Hence, our result.
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