Solve the system of differential equations with by using Laplace transforms.

Monday, February 17, 2014

Solve the system of differential equations with by using Laplace transforms.

Hello!

For this problem, we need some properties of Laplace transform. They are:

$\displaystyle{f{'}}{\left({t}\right)}-\gt{s}{F}{\left({s}\right)}-{f{{\left({0}\right)}}}$

(here $\displaystyle{F}{\left({s}\right)}$ is the Laplace transform of $\displaystyle{f{{\left({t}\right)}}}$ ).

$\displaystyle{\sin{{\left({t}\right)}}}-\gt\frac{{1}}{{{{s}}^{{2}}+{1}}},$ $\displaystyle{\cos{{\left({t}\right)}}}-\gt\frac{{s}}{{{{s}}^{{2}}+{1}}},$

$\displaystyle{\sinh{{\left({t}\right)}}}-\gt\frac{{1}}{{{{s}}^{{2}}-{1}}},$ $\displaystyle{\cosh{{\left({t}\right)}}}-\gt\frac{{s}}{{{{s}}^{{2}}-{1}}}.$

From these properties we obtain

$\displaystyle{s}{X}{\left({s}\right)}={Y}{\left({s}\right)}+\frac{{1}}{{{{s}}^{{2}}+{1}}},$ $\displaystyle{s}{Y}{\left({s}\right)}={X}{\left({s}\right)}+\frac{{{2}{s}}}{{{{s}}^{{2}}+{1}}}.$

It is a linear system for $\displaystyle{X}$ and $\displaystyle{Y}$ (the Laplace transforms of $\displaystyle{x}$ and $\displaystyle{y}$ ).

Its solution is

$\displaystyle{X}{\left({s}\right)}=\frac{{{3}{s}}}{{{\left({{s}}^{{2}}+{1}\right)}{\left({{s}}^{{2}}-{1}\right)}}}=\frac{{3}}{{2}}{\left(\frac{{s}}{{{{s}}^{{2}}-{1}}}-\frac{{s}}{{{{s}}^{{2}}+{1}}}\right)},$

$\displaystyle{Y}{\left({s}\right)}=\frac{{{2}{{s}}^{{2}}+{1}}}{{{\left({{s}}^{{2}}+{1}\right)}{\left({{s}}^{{2}}-{1}\right)}}}=\frac{{1}}{{2}}{\left(\frac{{3}}{{{{s}}^{{2}}-{1}}}+\frac{{s}}{{{{s}}^{{2}}+{1}}}\right)}.$

Inverting Laplace transform we see that

$\displaystyle{x}{\left({t}\right)}=\frac{{3}}{{2}}{\left({\cosh{{\left({t}\right)}}}-{\cos{{\left({t}\right)}}}\right)},$ $\displaystyle{y}{\left({t}\right)}=\frac{{1}}{{2}}{\left({3}{\sinh{{\left({t}\right)}}}+{\sin{{\left({t}\right)}}}\right)}.$

This is the answer.

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Monday, February 17, 2014