Question

To find: The solution of the given initial value problem.

Transformation properties
ANSWERED
asked 2020-11-23
To find: The solution of the given initial value problem.

Answers (1)

2020-11-24

Given:
The system of equation is,
\({\left({D}-{4}\right)}{x}+{6}{y}={9}{e}^{{-{3}{t}}}{x}-{\left({D}-{1}\right)}{y}={5}{e}^{{-{3}{t}}}\)
The initial conditions are given as, \({x}{\left({0}\right)}=-{9}{y}{\left({0}\right)}={4}\)
Calculation: \({L}{e}{t}{X}{\left({s}\right)}={\mathcal{\text{L}}}{\left\lbrace{x}\right\rbrace}{\left({s}\right)}{\quad\text{and}\quad}{Y}{\left({s}\right)}={\mathcal{\text{L}}}{\left\lbrace{y}\right\rbrace}{\left({s}\right)}\)
The given differential equation is written as,
\({x}'-{4}{x}+{6}{y}={9}{e}^{{-{3}{t}}}\ldots{\left({1}\right)}\)
\({x}-{y}'+{y}={5}{e}^{{-{3}{t}}}\ldots{\left({2}\right)}\)
Take Laplace transform on both sides of equation (1) and (2) and apply linear property as,
\({\mathcal{\text{L}}}{\left\lbrace{x}'-{4}{x}+{6}{y}\right\rbrace}{\left({s}\right)}={\mathcal{\text{L}}}{\left\lbrace{9}{e}^{{-{3}{t}}}\right\rbrace}{\left({s}\right)}\)
\({\mathcal{\text{L}}}{\left\lbrace{x}'\right\rbrace}{\left({s}\right)}-{4}{\mathcal{\text{L}}}{\left\lbrace{x}\right\rbrace}{\left({s}\right)}+{6}{\mathcal{\text{L}}}{\left\lbrace{y}\right\rbrace}{\left({s}\right)}={9}{\mathcal{\text{L}}}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}{\left({s}\right)}\ldots{\left({3}\right)}\)
\(\displaystyle{\mathcal{\text{L}}}{\left\lbrace{x}-{y}'+{y}\right\rbrace}{\left({s}\right)}={\mathcal{\text{L}}}{\left\lbrace{5}{e}^{{-{3}{t}}}\right\rbrace}{\left({s}\right)}\)
\({\mathcal{\text{L}}}{\left\lbrace{x}\right\rbrace}{\left({s}\right)}-{\mathcal{\text{L}}}{\left\lbrace{y}'\right\rbrace}{\left({s}\right)}+{\mathcal{\text{L}}}{\left\lbrace{y}\right\rbrace}{\left({s}\right)}={5}{\mathcal{\text{L}}}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}{\left({s}\right)}\ldots{\left({4}\right)}\)
Use Laplace transforms formulas in equation (3) and (4) as,
\({s}{X}{\left({s}\right)}-{x}{\left({0}\right)}-{4}{X}{\left({s}\right)}+{6}{Y}{\left({s}\right)}=\frac{9}{{{s}+{3}}}\)
\({X}{\left({s}\right)}-{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}+{Y}{\left({s}\right)}=\frac{5}{{{s}+{3}}}\)
Substitute the initial conditions as,
\({s}{X}{\left({s}\right)}+{9}-{4}{X}{\left({s}\right)}+{6}{Y}{\left({s}\right)}=\frac{9}{{{s}+{3}}}\)
\({X}{\left({s}\right)}-{s}{Y}{\left({s}\right)}-{4}+{Y}{\left({s}\right)}=\frac{5}{{{s}+{3}}}\ldots{\left({5}\right)}\)
Simplify the system of equations (5) as,
\({\left({s}-{4}\right)}{X}{\left({s}\right)}+{6}{Y}{\left({s}\right)}=-{9}+\frac{9}{{{s}+{3}}}\ldots{\left({6}\right)}\)
\({X}{\left({s}\right)}-{\left({s}-{1}\right)}{Y}{\left({s}\right)}=-{4}+\frac{5}{{{s}+{3}}}\ldots{\left({7}\right)}\)
Multiply equation (7) with (s - 4) and substract from equation (6) as follows,
image
\({Y}{\left({s}\right)}=\frac{{{4}{s}-{25}}}{{{\left({s}^{2}-{5}{s}+{10}\right)}}}+\frac{{{\left(-{5}{s}+{29}\right)}}}{{{\left({s}+{3}\right)}{\left({s}^{2}-{5}{s}+{10}\right)}}}\ldots{\left({8}\right)}\)
Reduce the expression \(\frac{{{\left(-{5}{s}+{29}\right)}}}{{{\left({s}+{3}\right)}{\left({s}^{2}-{5}{s}+{10}\right)}}}\) with the help of partial fractions as,
\(\frac{{{\left(-{5}{s}+{29}\right)}}}{{{\left({s}+{3}\right)}{\left({s}^{2}-{5}{s}+{10}\right)}}}=\frac{22}{{17}}\frac{1}{{{s}+{3}}}+\frac{{-{22}{s}+{91}}}{{{17}{\left({s}^{2}-{5}{s}+{10}\right)}}}\)
Therefore, equation (8) is given as,
Take inverse Laplace transform of Y (s) as,
Therefore, \({y}{\left({t}\right)}=\frac{46}{{17}}{e}^{{\frac{{{5}{t}}}{{2}}}} \cos{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{{{146}\sqrt{{15}}}}{{85}}{e}^{{\frac{{{5}{t}}}{{2}}}} \sin{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{22}{{17}}{e}^{{-{3}{t}}}\)
Differentiate y (t) with respect to t as,
\({y}'{\left({t}\right)}={\left(\frac{46}{{17}}\frac{5}{{2}}{e}^{{\frac{{{5}{t}}}{{2}}}} \cos{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{46}{{17}}\frac{\sqrt{{15}}}{{2}}{e}^{{\frac{{{5}{t}}}{{2}}}} \sin{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{{{146}\sqrt{{15}}}}{{85}}\frac{5}{{2}}{e}^{{\frac{{{5}{t}}}{{2}}}} \sin{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{{{146}\sqrt{{15}}}}{{85}}\frac{\sqrt{{15}}}{{2}}{e}^{{\frac{{{5}{t}}}{{2}}}} \cos{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{66}{{17}}{e}^{{-{3}{t}}}\right)}\)
Solve equation (2) for x as, \({x}-{y}'+{y}={5}{e}^{{-{3}{t}}}{x}={5}{e}^{{-{3}{t}}}+{y}'-{y}\)
Substitute the value of y' (t) and y (t) as,
Therefore, \({x}{\left({t}\right)}=-\frac{150}{{17}}{e}^{{\frac{{{5}{t}}}{{2}}}} \cos{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{{{334}\sqrt{{15}}}}{{85}}{e}^{{\frac{{{5}{t}}}{{2}}}} \sin{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{3}{{17}}{e}^{{-{3}{t}}}\)
Conclusion:
Hence, the solution of the given invitial value problem is
\({x}{\left({t}\right)}=-\frac{150}{{17}}{e}^{{\frac{{{5}{t}}}{{2}}}} \cos{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{{{334} \sqrt {{15}}}}{{85}}e^{\frac{5t}{2}} sin{{\left (\frac{\sqrt15}{2}t)\right)}} - \frac{3}{17} e^{-3t}\)
\({y}{\left({t}\right)}=\frac{46}{{17}}{e}^{{\frac{{{5}{t}}}{{2}}}} \cos{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{{{146}\sqrt{{15}}}}{{85}}e^{\frac{5t}{2}}\)

\(\sin{{\left(\frac{\sqrt{{15}}}{{2}}{t}\right)}}-\frac{22}{{17}}{e}^{{-{3}{t}}}\)

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