Use Laplace transforms to solve the following initial value problem y"-y'-6y=0 y(0)=1 y'(0)=-1

Question
Laplace transform
asked 2021-02-14
Use Laplace transforms to solve the following initial value problem
\(y"-y'-6y=0\)
\(y(0)=1\)
\(y'(0)=-1\)

Answers (1)

2021-02-15
Step 1
The given IVP is as follows.
\(y"-y'-6y=0\)
\(y(0)=1\)
\(y'(0)=-1\)
Apply Laplace transform on the IVP as follows.
\(y''-y'-6y=0\)
\(\displaystyle{L}{\left\lbrace{y}{''}\right\rbrace}-{L}{\left\lbrace{y}'\right\rbrace}-{L}{\left\lbrace{6}{y}\right\rbrace}={L}{\left\lbrace{0}\right\rbrace}\)
\(\displaystyle{s}^{2}{L}{\left\lbrace{y}\right\rbrace}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}-{s}{L}{\left\lbrace{y}\right\rbrace}+{y}{\left({0}\right)}-{6}{L}{\left\lbrace{y}\right\rbrace}={0}\)
\(\displaystyle{L}{\left\lbrace{y}\right\rbrace}{\left[{s}^{2}-{s}-{6}\right]}={s}{\left({1}\right)}+{\left(-{1}\right)}-{\left({1}\right)}\)
\(\displaystyle{L}{\left\lbrace{y}\right\rbrace}{\left[{s}^{2}-{s}-{6}\right]}={s}-{2}\)
\(\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{{{s}-{2}}}{{{s}^{2}-{s}-{6}}}\)
\(\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{{{s}-{2}}}{{{\left({s}-{3}\right)}{\left({s}+{2}\right)}}}\)
Step 2
Apply partial fraction on \(\displaystyle\frac{{{s}-{2}}}{{{\left({s}-{3}\right)}{\left({s}+{2}\right)}}}\) as follows.
\(\displaystyle\frac{{{s}-{2}}}{{{\left({s}-{3}\right)}{\left({s}+{2}\right)}}}=\frac{A}{{{s}-{3}}}+\frac{B}{{{s}+{2}}}\)
\(s-2=A(s+2)+B(s-3)\)
\(s-2=(A+B)s+(2A-3B)\)
\(A+B=1, 2A-3B=-2\)
\(A=1-B\)
\(2(1-B)-3B=-2\)
\(-5B=-4\)
\(\displaystyle{B}=\frac{4}{{5}}\)
\(\displaystyle{A}=\frac{1}{{5}}\)
\(\displaystyle\frac{{{s}-{2}}}{{{\left({s}-{3}\right)}{\left({s}+{2}\right)}}}=\frac{1}{{5}}{\left(\frac{1}{{{s}-{3}}}\right)}+\frac{4}{{5}}{\left(\frac{1}{{{s}+{2}}}\right)}\)
Obtain the solution of the IVP as follows.
\(\displaystyle{L}{\left\lbrace{y}\right\rbrace}={15}{\left({1}{s}-{3}\right)}+{45}{\left({1}{s}+{2}\right)}\)
Take inverse Laplace transforms as follows.
\(\displaystyle{L}^{ -{{1}}}{\left\lbrace{L}{\left\lbrace{y}\right\rbrace}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{5}}{\left(\frac{1}{{{s}-{3}}}\right)}+\frac{4}{{5}}{\left(\frac{1}{{{s}+{2}}}\right\rbrace}\right.}\)
\(\displaystyle{y}{\left({t}\right)}=\frac{1}{{5}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{3}}}\right\rbrace}+\frac{4}{{5}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}+{2}}}\right\rbrace}\)
\(\displaystyle{y}{\left({t}\right)}=\frac{1}{{5}}{e}^{{{3}{t}}}+\frac{4}{{5}}{e}^{{-{2}{t}}}\)
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