Given differential equation is ,

\(u''+0.5u'+2u=3\sin t\)

Let, \(u'=v\) then,

\((u')'+0.5u'+2u=3\sin t\)

\(\Rightarrow v'+0.5v+2u=3\sin t\)

\(\Rightarrow v'=3\sin t-2u-0.5v\)

Hence,

\begin{cases}u'=v\\v'=3\sin t-2u-0.5v\end{cases}\)

Given system of first order differential equations.

\(x_1"=x_2\)

\(x_2'=-kx_1-\gamma x_2+F(t)\)

\(\Rightarrow x_1''=-kx-\gamma x_2+F(t)\)

Hence, the required second order differential equation is

\(x_1''=-kx-\gamma x_2+F(t)\)

\(u''+0.5u'+2u=3\sin t\)

Let, \(u'=v\) then,

\((u')'+0.5u'+2u=3\sin t\)

\(\Rightarrow v'+0.5v+2u=3\sin t\)

\(\Rightarrow v'=3\sin t-2u-0.5v\)

Hence,

\begin{cases}u'=v\\v'=3\sin t-2u-0.5v\end{cases}\)

Given system of first order differential equations.

\(x_1"=x_2\)

\(x_2'=-kx_1-\gamma x_2+F(t)\)

\(\Rightarrow x_1''=-kx-\gamma x_2+F(t)\)

Hence, the required second order differential equation is

\(x_1''=-kx-\gamma x_2+F(t)\)