Transform the given initial value problem into an initial value problem for two first-order quations...

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 $x_2^{\prime }=u"$
${:}^{\prime }u"+0.25u^{\prime }+4u=2\cos (3t)$
$\Rightarrow u"=2\cos (3t)-0.25u^{\prime }-4u$ $({:}^{\prime }u=x_1,u^{\prime }=x_2)$
$\Rightarrow x_2^{\prime }=2\cos (3t)-0.25x_2-4x_1$
$x_1^{\prime }=x_2$
$x_2^{\prime }=2\cos (3t)-4x_1-0.25x_2$
or
$x_1^{\prime }=0\cdot x_1+x_2$
$x_2^{\prime }=-4x_1-0.25x_2+2\cos (3t)$
${:}^{\prime }u(0)=1\Rightarrow x_1(0)=1$ (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
$x_1^{\prime }=0\cdot x_1+1\cdot x_2$
$x_2^{\prime }=-4x_1-0.25x_2+2\cos (3t)$
$x_1(0)=1$, $x_2(0)=-2$