# Transform the given initial value problem into an initial value problem for two first-order quations u"+0.25u'+4u=2cos(3t), u(0)= 1, u'(0)= -2

Question
Transform the given initial value problem into an initial value problem for two first-order quations $$u"+0.25u'+4u=2cos(3t)$$, u(0)= 1, u'(0)= -2

2021-03-03
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
$$x_1'= 0*x_1+1*x_2$$
$$x_2'= -4x_1-0.25x_2+2cos(3t)$$
$$x_1(0)= 1$$, $$x_2(0)= -2$$

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