Solve the following initial value problems using Laplace Transforms: displaystylefrac{{{d}^{2}{y}}}{{{left.{d}{x}right.}^{2}}}+{25}{y}={t} y(0)=0 y'(0)=0.04

Solve the following initial value problems using Laplace Transforms: displaystylefrac{{{d}^{2}{y}}}{{{left.{d}{x}right.}^{2}}}+{25}{y}={t} y(0)=0 y'(0)=0.04

Question
Laplace transform
asked 2021-02-26
Solve the following initial value problems using Laplace Transforms:
\(\displaystyle\frac{{{d}^{2}{y}}}{{{\left.{d}{x}\right.}^{2}}}+{25}{y}={t}\)
\(y(0)=0\)
\(y'(0)=0.04\)

Answers (1)

2021-02-27
Step 1
The given IVP is given as follows.
\(\displaystyle\frac{{{d}^{2}{y}}}{{{\left.{d}{x}\right.}^{2}}}+{25}{y}={t},\ \ \ {y}{\left({0}\right)}={0},\ \ \ {y}'{\left({0}\right)}={0.04}\)
Apply the laplace transform on both sides of the differential equation as follows.
\(\displaystyle{L}{\left\lbrace{y}{''}\right\rbrace}+{25}{L}{\left\lbrace{y}\right\rbrace}={L}{\left\lbrace{t}\right\rbrace}\)
\(\displaystyle{s}^{2}{L}{\left\lbrace{y}\right\rbrace}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{25}{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{s}^{2}}\)
\(\displaystyle{s}^{2}{L}{\left\lbrace{y}\right\rbrace}+{25}{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{s}^{2}}+{0.04}\)
\(\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}^{2}{\left({s}^{2}+{25}\right)}}}+\frac{{{0.04}}}{{{s}^{2}+{25}}}\)
\(\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{25}{s}^{2}}}-\frac{1}{{{25}{\left({s}^{2}+{25}\right)}}}+\frac{{{0.04}}}{{{s}^{2}+{25}}}\)
Step 2
Apply inverse laplace transforms on both sides.
\(\displaystyle{y}{\left({t}\right)}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{25}{s}^{2}}}\right\rbrace}-{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{25}{\left({s}^{2}+{25}\right)}}}\right\rbrace}+{L}^{ -{{1}}}{\left\lbrace\frac{{{0.04}}}{{{s}^{2}+{25}}}\right\rbrace}\)
\(\displaystyle=\frac{1}{{25}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{s}^{2}}\right\rbrace}-\frac{1}{{125}}{L}^{ -{{1}}}{\left\lbrace\frac{5}{{{s}^{2}+{5}^{2}}}\right\rbrace}+{0.008}{L}^{ -{{1}}}{\left\lbrace\frac{5}{{{s}^{2}+{52}}}\right\rbrace}\)
\(\displaystyle=\frac{t}{{25}}-\frac{1}{{125}} \sin{{\left({5}{t}\right)}}+{0.008} \sin{{\left({5}{t}\right)}}\)
\(\displaystyle=\frac{t}{{25}}\)
0

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