Let \(u=x_1\) and \(u'=x_2\)

\(=> x_1'=u'\) and \(x_2'= u"\)

or \(x_1'=x_2\) and \(x_2'=u"\)

\(:' u"+0.25u'+4u=2cos(3t)\)

\(=> u"= 2cos(3t)-0.25u'-4u\) \((:' u=x_1, u'=x_2)\)

\(=> x_2'= 2cos(3t)-0.25x_2-4x_1\)

\(x_1'= x_2\)

\(x_2'= 2cos(3t)-4x_1-0.25x_2\)

or

\(x_1'= 0*x_1+x_2\)

\(x_2'= -4x_1-0.25x_2+2cos(3t)\)

\(:' u(0)=1 => x_1(0)=1\) (because \(u=x_1\))

\(u'(0)=-2 => x_2(0)= -2\) (because \(u'=x_2\))

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

\(x_1'= 0*x_1+1*x_2\)

\(x_2'= -4x_1-0.25x_2+2cos(3t)\)

\(x_1(0)= 1\), \(x_2(0)= -2\)

\(=> x_1'=u'\) and \(x_2'= u"\)

or \(x_1'=x_2\) and \(x_2'=u"\)

\(:' u"+0.25u'+4u=2cos(3t)\)

\(=> u"= 2cos(3t)-0.25u'-4u\) \((:' u=x_1, u'=x_2)\)

\(=> x_2'= 2cos(3t)-0.25x_2-4x_1\)

\(x_1'= x_2\)

\(x_2'= 2cos(3t)-4x_1-0.25x_2\)

or

\(x_1'= 0*x_1+x_2\)

\(x_2'= -4x_1-0.25x_2+2cos(3t)\)

\(:' u(0)=1 => x_1(0)=1\) (because \(u=x_1\))

\(u'(0)=-2 => x_2(0)= -2\) (because \(u'=x_2\))

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

\(x_1'= 0*x_1+1*x_2\)

\(x_2'= -4x_1-0.25x_2+2cos(3t)\)

\(x_1(0)= 1\), \(x_2(0)= -2\)