Question

Find the equation by applying the Laplace transform. displaystyle{y}^{{{left({4}right)}}}-{y}= sin{{h}}{t} y(0)=y'(0)=y"(0)=0 y'''(0)=1

Laplace transform
ANSWERED
asked 2021-02-10
Find the equation by applying the Laplace transform.
\(\displaystyle{y}^{{{\left({4}\right)}}}-{y}= \sin{{h}}{t}\)
\(y(0)=y'(0)=y"(0)=0\)
\(y'''(0)=1\)

Answers (1)

2021-02-11
Step 1
The given equation is,
\(\displaystyle{y}^{4}-{y}= \sinh{{t}}\ldots{\left({1}\right)}\)
The given initial condition is,
\(\displaystyle{y}{\left({0}\right)}={1}\ldots{\left({2}\right)}\)
\(\displaystyle{y}′{\left({0}\right)}={1}\ldots{\left({3}\right)}\)
\(\displaystyle{y}{''}{\left({0}\right)}={1}\ldots{\left({4}\right)}\)
\(\displaystyle{y}{'''}{\left({0}\right)}={1}\ldots{\left({5}\right)}\)
The Laplace transform is given as,
\(\displaystyle{L}{\left\lbrace{F}^{n}{\left({t}\right)}\right\rbrace}={s}^{n}{L}{\left\lbrace{F}{\left({t}\right)}\right\rbrace}-{s}^{{{n}-{1}}}{F}{\left({0}\right)}-{s}^{{{n}-{2}}}{F}′{\left({0}\right)}-\ldots-{F}^{{{n}-{1}}}{\left({0}\right)}\)
\(\displaystyle{L}{\left\lbrace \sin{{h}}{a}{t}\right\rbrace}=\frac{a}{{{s}^{2}-{a}^{2}}}\)
Step 2
The given equation (1) is,
\(\displaystyle{y}^{4}-{y}= \sinh{{t}}\)
Taking Laplace transform on both sides,
\(\displaystyle{L}{\left\lbrace{y}{''''}-{y}\right\rbrace}={L}{\left\lbrace \sin{{h}}{t}\right\rbrace}\)
\(\displaystyle{s}^{4}{L}{\left\lbrace{y}\right\rbrace}-{s}^{3}{y}{\left({0}\right)}-{s}^{2}{y}′{\left({0}\right)}-{s}{y}{''}{\left({0}\right)}-{y}{'''}{\left({0}\right)}-{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}^{2}-{1}}}\)
On putting the values of equation (2), (3), (4) & (5) in the above equation,
\(\displaystyle{s}^{4}{L}{\left\lbrace{y}\right\rbrace}-{s}^{3}{\left({1}\right)}-{s}^{2}{\left({1}\right)}-{s}{\left({1}\right)}-{\left({1}\right)}-{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}^{2}-{1}}}\)
\(\displaystyle{s}^{4}{L}{\left\lbrace{y}\right\rbrace}-{s}^{3}-{s}^{2}-{s}-{1}-{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}^{2}-{1}}}\)
\(\displaystyle{s}^{4}{L}{\left\lbrace{y}\right\rbrace}-{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}^{2}-{1}}}+{s}^{3}+{s}^{2}+{s}+{1}\)
\(\displaystyle{\left({s}^{4}-{1}\right)}{L}{\left\lbrace{y}\right\rbrace}=\frac{{{1}+{s}^{5}-{s}^{3}+{s}^{4}-{s}^{2}+{s}^{3}-{s}+{s}^{2}-{1}}}{{{s}^{2}-{1}}}\)
\(\displaystyle{\left({s}^{4}-{1}\right)}{L}{\left\lbrace{y}\right\rbrace}=\frac{{{s}^{5}+{s}^{4}-{s}}}{{{s}^{2}-{1}}}\)
\(\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{{{s}^{5}+{s}^{4}-{s}}}{{{\left({s}^{2}-{1}\right)}{\left({s}^{4}-{1}\right)}}}\ldots{\left({6}\right)}\)
Step 3
Let
\(\displaystyle\frac{{{s}^{5}+{s}^{4}-{s}}}{{{\left({s}^{2}-{1}\right)}{\left({s}^{4}-{1}\right)}}}=\frac{{{s}^{5}+{s}^{4}-{s}}}{{{\left({s}-{1}\right)}{\left({s}+{1}\right)}{\left({s}-{1}\right)}{\left({s}+{1}\right)}{\left({s}^{2}+{1}\right)}}}\)
\(\displaystyle=\frac{{{s}^{5}+{s}^{4}-{s}}}{{{\left({s}-{1}\right)}^{2}{\left({s}+{1}\right)}^{2}{\left({s}^{2}+{1}\right)}}}\)
On taking partial fractions
\(\displaystyle\frac{{{s}^{5}+{s}^{4}-{s}}}{{{\left({s}-{1}\right)}^{2}{\left({s}+{1}\right)}^{2}{\left({s}^{2}+{1}\right)}}}=\frac{A}{{{s}-{1}}}+\frac{B}{{\left({s}-{1}\right)}^{2}}+\frac{C}{{{s}+{1}}}+\frac{D}{{\left({s}+{1}\right)}^{2}}+\frac{{{E}{s}+{F}}}{{{s}^{2}+{1}}}\ldots{\left({7}\right)}\)
Equation (7) gives,
\(\displaystyle{s}^{5}+{s}^{4}-{s}={A}{\left({s}-{1}\right)}{\left({s}+{1}\right)}^{2}{\left({s}^{2}+{1}\right)}+{B}{\left({s}+{1}\right)}^{2}{\left({s}^{2}+{1}\right)}+{C}{\left({s}-{1}\right)}^{2}{\left({s}+{1}\right)}{\left({s}^{2}+{1}\right)}+{D}{\left({s}-{1}\right)}^{2}{\left({s}^{2}+{1}\right)}+{\left({E}{s}+{F}\right)}{\left({s}-{1}\right)}^{2}{\left({s}+{1}\right)}^{2}\ldots{\left({8}\right)}\)
Step 4
Put s=1 in equation (8)
\(1+1-1=A(0)+B(1+1)2(1+1)+0+0+0\)
\(1=B(2^2)(2)\)
\(1=8B\)
\(\displaystyle{B}=\frac{1}{{8}}\)
Similarly on solving equation (8), by putting different values of s, the values of the constants will be obtained.
The values will be,
\(\displaystyle{A}=\frac{3}{{4}},{B}=\frac{1}{{8}},{C}=\frac{1}{{4}},{D}=\frac{1}{{8}},{E}={0},{F}=\frac{1}{{4}}\)
Now substitute these values in equation (7)
\(\displaystyle\frac{{{s}^{5}+{s}^{4}-{s}}}{{{\left({s}-{1}\right)}^{2}{\left({s}+{1}\right)}^{2}{\left({s}^{2}+{1}\right)}}}=\frac{3}{{4}}\cdot\frac{1}{{{s}-{1}}}+\frac{1}{{8}}\cdot\frac{1}{{\left({s}-{1}\right)}^{2}}+\frac{1}{{4}}\cdot\frac{1}{{{s}+{1}}}+\frac{1}{{8}}\cdot\frac{1}{{\left({s}+{1}\right)}^{2}}+\frac{{{\left({0}\right)}{s}+{\left(\frac{1}{{4}}\right)}}}{{{s}^{2}+{1}}}\)
\(\displaystyle=\frac{3}{{{4}{\left({s}-{1}\right)}}}+\frac{1}{{{8}{\left({s}-{1}\right)}^{2}}}+\frac{1}{{{4}{\left({s}+{1}\right)}}}+\frac{1}{{{8}{\left({s}+{1}\right)}^{2}}}+\frac{1}{{{4}{\left({s}^{2}+{1}\right)}}}\)
Step 5
Now substitute this value in equation (6)
\(\displaystyle{L}{\left\lbrace{y}\right\rbrace}=\frac{3}{{{4}{\left({s}-{1}\right)}}}+\frac{1}{{{8}{\left({s}-{1}\right)}^{2}}}+\frac{1}{{{4}{\left({s}+{1}\right)}}}+\frac{1}{{{8}{\left({s}+{1}\right)}^{2}}}+\frac{1}{{{4}{\left({s}^{2}+{1}\right)}}}\)
Now taking Laplace inverse,
\(\displaystyle{y}=\frac{3}{{4}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{1}}}\right\rbrace}+\frac{1}{{8}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{\left({s}-{1}\right)}^{2}}\right\rbrace}+\frac{1}{{4}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}+{1}}}\right\rbrace}+\frac{1}{{8}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{\left({s}+{1}\right)}^{2}}\right\rbrace}+\frac{1}{{4}}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}^{2}+{1}}}\right\rbrace}\)
\(\displaystyle{y}=\frac{{{3}{e}^{t}}}{{4}}+\frac{{{t}{e}^{t}}}{{8}}+\frac{{{e}^{{-{t}}}}}{{4}}+\frac{{{t}{e}^{{-{t}}}}}{{8}}+\frac{{ \sin{{t}}}}{{4}}\)
Answer \(\displaystyle{y}=\frac{{{3}{e}^{t}}}{{4}}+\frac{{{t}{e}^{t}}}{{8}}+\frac{{{e}^{{-{t}}}}}{{4}}+\frac{{{t}{e}^{{-{t}}}}}{{8}}+\frac{{ \sin{{t}}}}{{4}}\)
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