2.3.3.1 Existence and uniqueness analysis

\[\begin {aligned} x^{\prime \prime }+2 x^{\prime }+4 x&={\mathrm e}^{t} \cos \left (2 t \right )\\ x \left (0\right ) &= 0\\ x^{\prime }\left (0\right ) &= 1\\ \end {aligned}\]

This is a linear ODE. In canonical form it is written as

\begin{align*} x^{\prime \prime } + p(t)x^{\prime } + q(t) x &= F \end{align*}

Comparing the above to the given ODE shows that

\begin{align*} p(t) &=2\\ q(t) &=4\\ F &={\mathrm e}^{t} \cos \left (2 t \right ) \end{align*}

Hence the ode is

\begin{align*} x^{\prime \prime }+2 x^{\prime }+4 x = {\mathrm e}^{t} \cos \left (2 t \right ) \end{align*}

The domain of \(p(t)=2\) is

\begin{align*} \{-\infty <t <\infty \} \end{align*}

And the point \(t_0 = 0\) is inside this domain. The domain of \(q(t)=4\) is

\begin{align*} \{-\infty <t <\infty \} \end{align*}

And the point \(t_0 = 0\) is also inside this domain. The domain of \(F ={\mathrm e}^{t} \cos \left (2 t \right )\) is

\begin{align*} \{-\infty <t <\infty \} \end{align*}

And the point \(t_0 = 0\) is also inside this domain. Therefore solution exists and is unique.