Let $u={x}_{1}$ and ${u}^{\prime}={x}_{2}$

$\Rightarrow {x}_{1}^{\prime}={u}^{\prime}$ and ${x}_{2}^{\prime}=u"$

or ${x}_{1}^{\prime}={x}_{2}$ and ${x}_{2}^{\prime}=u"$

${:}^{\prime}u"+0.25{u}^{\prime}+4u=2\mathrm{cos}(3t)$

$\Rightarrow u"=2\mathrm{cos}(3t)-0.25{u}^{\prime}-4u$ $({:}^{\prime}u={x}_{1},{u}^{\prime}={x}_{2})$

$\Rightarrow {x}_{2}^{\prime}=2\mathrm{cos}(3t)-0.25{x}_{2}-4{x}_{1}$

${x}_{1}^{\prime}={x}_{2}$

${x}_{2}^{\prime}=2\mathrm{cos}(3t)-4{x}_{1}-0.25{x}_{2}$

or

${x}_{1}^{\prime}=0\cdot {x}_{1}+{x}_{2}$

${x}_{2}^{\prime}=-4{x}_{1}-0.25{x}_{2}+2\mathrm{cos}(3t)$

${:}^{\prime}u(0)=1\Rightarrow {x}_{1}(0)=1$ (because $u={x}_{1}$)

${u}^{\prime}(0)=-2\Rightarrow {x}_{2}(0)=-2$ (because ${u}^{\prime}={x}_{2}$)

Hence the initial value problem for two first-order quations is

${x}_{1}^{\prime}=0\cdot {x}_{1}+1\cdot {x}_{2}$

${x}_{2}^{\prime}=-4{x}_{1}-0.25{x}_{2}+2\mathrm{cos}(3t)$

${x}_{1}(0)=1$, ${x}_{2}(0)=-2$