Question

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

First order differential equations
ANSWERED
asked 2021-03-02

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

Answers (1)

2021-03-03

Let \(u=x_1\) and \(u'=x_2\)
\(\Rightarrow x_1'=u'\) and \(x_2'= u"\)
or \(x_1'=x_2\) and \(x_2'=u"\)
\(:' u"+0.25u'+4u=2\cos(3t)\)
\(\Rightarrow u"= 2\cos(3t)-0.25u'-4u\) \((:' u=x_1, u'=x_2)\)
\(\Rightarrow x_2'= 2\cos(3t)-0.25x_2-4x_1\)
\(x_1'= x_2\)
\(x_2'= 2\cos(3t)-4x_1-0.25x_2\)
or
\(x_1'= 0 \cdot x_1+x_2\)
\(x_2'= -4x_1-0.25x_2+2\cos(3t)\)
\(:' u(0)=1 \Rightarrow x_1(0)=1\) (because \(u=x_1\))
\(u'(0)=-2 \Rightarrow x_2(0)= -2\) (because \(u'=x_2\))
Hence the initial value problem for two first-order quations is
\(x_1'= 0 \cdot x_1+1 \cdot x_2\)
\(x_2'= -4x_1-0.25x_2+2\cos(3t)\)
\(x_1(0)= 1\), \(x_2(0)= -2\)

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