Question

Use the Laplace transform to solve the given initial-value problem. dy/dt-y=z,\ y(0)=0

Laplace transform
ANSWERED
asked 2021-08-22
Use the Laplace transform to solve the given initial-value problem.
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}-{y}={z},\ {y}{\left({0}\right)}={0}\)

Expert Answers (1)

2021-08-23

Step 1
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}-{y}={1},\ {y}{\left({0}\right)}={0}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}-{y}={1}\)
\(\mathcal{L}\left\{\frac{dy}{dt}-y\right\}=\mathcal{L}\{1\}\)
\(\mathcal{L}\left\{\frac{dy}{dt}\right\}-\mathcal{L}\{y\}=\frac{1}{s}\)
\(\displaystyle{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}-{Y}{\left({s}\right)}={\frac{{{1}}}{{{s}}}}\)
\(\displaystyle{s}{Y}{\left({s}\right)}-{0}-{Y}{\left({s}\right)}={\frac{{{1}}}{{{s}}}}\)
\(\displaystyle{s}{Y}{\left({s}\right)}-{Y}{\left({s}\right)}={\frac{{{1}}}{{{s}}}}\)
\(\displaystyle{\left({s}-{1}\right)}{Y}{\left({s}\right)}={\frac{{{1}}}{{{s}}}}\)
\(\displaystyle{Y}{\left({s}\right)}={\frac{{{1}}}{{{s}{\left({s}-{1}\right)}}}}\)
\(\displaystyle{y}{\left({t}\right)}={\mathcal{{{L}}}}^{{-{1}}}{\left\lbrace{Y}{\left({s}\right)}\right\rbrace}\)
\(=\mathcal{L}^{-1}\left\{\frac{1}{s(s-1)}\right\}\)
\(=\mathcal{L}^{-1}\left\{\frac{s-(s-1)}{s(s-1)}\right\}\)
\(=\mathcal{L}^{-1}\left\{\frac{s}{s(s-1)}-\frac{s-1}{s(s-1)}\right\}\)
\(=\mathcal{L}^{-1}\left\{\frac{1}{s-1}-\frac{1}{s}\right\}\)
\(=\mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\}-\mathcal{L}^{-1}\left\{\frac{1}{s}\right\}\)
\(\displaystyle={e}^{{{t}}}-{1}\)
That is
\(\displaystyle{y}{\left({t}\right)}={e}^{{{t}}}-{1}\)

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