# How to solve this system of ODE's? [ <mtable rowspacing="4pt" columnspacing="1em">

Dale Tate 2022-06-24 Answered
How to solve this system of ODE's?
$\left[\begin{array}{c}{\stackrel{˙}{x}}_{1}\\ {\stackrel{˙}{x}}_{2}\end{array}\right]=\left[\begin{array}{cc}\mathrm{cos}t& -\mathrm{sin}t\\ \mathrm{sin}t& \mathrm{cos}t\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]$
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Angelo Murray
This is a linear system of the form
${x}^{\prime }=A\left(t\right)x,$
where $x\in {\mathbb{R}}^{2}$, $A\left(t\right)\in C\left(\mathbb{R};{\mathbb{R}}^{2×2}\right)$, and most important
$\begin{array}{}\text{(1)}& A\left(s\right)A\left(t\right)=A\left(t\right)A\left(s\right).\end{array}$
Satisfaction of $\left(1\right)$ implies that the solution of
${x}^{\prime }=A\left(t\right)x,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\left(0\right)={\xi }_{0},$
In our case
$A\left(t\right)=\left(\begin{array}{cc}\mathrm{cos}t& -\mathrm{sin}t\\ \mathrm{sin}t& \mathrm{cos}t\end{array}\right),$
then
${\int }_{0}^{t}A\left(s\right)\phantom{\rule{thinmathspace}{0ex}}ds=\left(\begin{array}{cc}\mathrm{sin}t& \mathrm{cos}t-1\\ 1-\mathrm{cos}t& \mathrm{sin}t\end{array}\right).$
Next we use use the fact that
$\mathrm{exp}\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)=\left(\begin{array}{cc}{\mathrm{e}}^{a}\mathrm{cos}b& -{\mathrm{e}}^{a}\mathrm{sin}b\\ {\mathrm{e}}^{a}\mathrm{sin}b& {\mathrm{e}}^{a}\mathrm{cos}b\end{array}\right),$
and finally we obtain that
$x\left(t\right)=\left(\begin{array}{cc}{\mathrm{e}}^{\mathrm{sin}t}\mathrm{cos}\left(1-\mathrm{cos}t\right)& -{\mathrm{e}}^{\mathrm{sin}t}\mathrm{sin}\left(1-\mathrm{cos}t\right)\\ {\mathrm{e}}^{\mathrm{sin}t}\mathrm{sin}\left(1-\mathrm{cos}t\right)& {\mathrm{e}}^{\mathrm{sin}t}\mathrm{cos}\left(1-\mathrm{cos}t\right)\end{array}\right)\phantom{\rule{thinmathspace}{0ex}}{\xi }_{0}.$