Using Laplace Transform , solve the following differential equation {y}text{}-{4}{y}={e}^{{-{3}{t}}},{y}{left({0}right)}={0},{y}'{left({0}right)}={2} a) frac{14}{{20}}{e}^{{{2}{t}}}-frac{5}{{30}}{e}^{{-{2}{t}}}-frac{9}{{30}}{e}^{{-{6}{t}}} b) frac{11}{{20}}{e}^{{{2}{t}}}-frac{51}{{20}}{e}^{{-{2}{t}}}-frac{4}{{20}}{e}^{{-{3}{t}}} c) frac{14}{{15}}{e}^{{{2}{t}}}-frac{5}{{10}}{e}^{{-{2}{t}}}-frac{9}{{20}}{e}^{{-{3}{t}}} d) frac{14}{{20}}{e}^{{{2}{t}}}+frac{5}{{20}}{e}^{{-{2}{t}}}-frac{9}{{20}}{e}^{{-{3}{t}}}

Question
Laplace transform
asked 2020-12-01
Using Laplace Transform , solve the following differential equation
\({y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}\)
a) \(\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}\)
b) \(\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}\)
c) \(\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}\)
d) \(\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}\)

Answers (1)

2020-12-02
Step 1
As per guidelines, only typed answers are acceptable. so we will give only typed solution.
Given initial value problem is
\({y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}\)
Step 2
Applying Laplace transform on both sides
\({L}{\left\lbrace{y}\text{}-{4}{y}\right\rbrace}={L}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}\)
\(\Rightarrow{L}{\left\lbrace{y}\text{}\right\rbrace}-{4}{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}+{3}}}\)
\(\Rightarrow{s}^{2}{L}{\left\lbrace{y}\right\rbrace}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}-{4}{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{s}+{3}}}\)
\(\Rightarrow{\left({s}^{2}-{4}\right)}{L}{\left\lbrace{y}\right\rbrace}-{2}=\frac{1}{{{s}+{3}}}\)
\(\Rightarrow{L}{\left\lbrace{y}\right\rbrace}=\frac{1}{{{\left({s}+{3}\right)}{\left({s}^{2}-{4}\right)}}}+\frac{2}{{{s}^{2}-{4}}}\)
\(={\left(\frac{1}{{{5}{\left({s}+{3}\right)}}}+\frac{1}{{{20}{\left({s}-{2}\right)}}}-\frac{1}{{{4}{\left({s}+{2}\right)}}}\right)}+{\left(-\frac{1}{{{2}{\left({s}+{2}\right)}}}+\frac{1}{{{2}{\left({s}-{2}\right)}}}\right)}\)
\(=\frac{1}{{{5}{\left({s}+{3}\right)}}}+\frac{11}{{{20}{\left({s}-{2}\right)}}}-\frac{3}{{{4}{\left({s}+{2}\right)}}}\)
Taking inverse Laplace transform on both sides, we get
\({y}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{5}{\left({s}+{3}\right)}}}+\frac{11}{{{20}{\left({s}-{2}\right)}}}-\frac{3}{{{4}{\left({s}+{2}\right)}}}\right\rbrace}\)
\(={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{5}{\left({s}+{3}\right)}}}\right\rbrace}+{L}^{ -{{1}}}{\left\lbrace\frac{11}{{{20}{\left({s}-{2}\right)}}}\right\rbrace}-{L}^{ -{{1}}}{\left\lbrace\frac{3}{{{4}{\left({s}+{2}\right)}}}\right\rbrace}\)
\(=\frac{1}{{5}}{e}^{{-{3}{t}}}+\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{3}{{4}}{e}^{{-{2}{t}}}\)
Hence, solution is
\({y}{\left({t}\right)}=\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{3}{{4}}{e}^{{-{2}{t}}}+\frac{1}{{5}}{e}^{{-{3}{t}}}\ \text{ or }\ {y}{\left({t}\right)}=\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{15}{{20}}{e}^{{-{2}{t}}}+\frac{4}{{20}}{e}^{{-{3}{t}}}\)
Comparing the answer with options, none of the option is correct.
The option given are all wrong.
0

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