-7.1 transform u''+0.5u'+2u=0 into a system of first order eq

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SUMMARY

The discussion focuses on transforming the second-order differential equation \(u'' + 0.5u' + 2u = 0\) into a system of first-order equations. The transformation is achieved by defining \(u_1 = u\) and \(u_2 = u'\), leading to the equations \(u_1' = u_2\) and \(u_2' = -0.5u_2 - 2u_1\). A similar approach is applied to the non-homogeneous equation \(u'' + 0.5u' + 2u = 3\sin(t)\), resulting in the system \(u_1' = u_2\) and \(u_2' = 3\sin(t) - 0.5u_2 - 2u_1\).

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karush
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transform the given equation into a system of first order equation$$u''+0.5u'+2u=0$$ok from examples it looks all we do is get rid of some of the primes and this is done by substitutionso if $u_1=u$ and $u_2=u'_1$
then $u_2=u'$ and $u'_2=u''$
then we have $u'_2+0.5u_2 +2u_1 = 0$then isolate $u'_2$ thus $u'_2=-0.5u_2-2u_1$ok the next problem is$$u''+0.5u'+2u=3\sin t$$ so ?
 
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karush said:
transform the given equation into a system of first order equation$$u''+0.5u'+2u=0$$ok from examples it looks all we do is get rid of some of the primes and this is done by substitutionso if $u_1=u$ and $u_2=u'_1$
then $u_2=u'$ and $u'_2=u''$
then we have $u'_2+0.5u_2 +2u_1 = 0$then isolate $u'_2$ thus $u'_2=-0.5u_2-2u_1$
But you don't yet have a system of equations!
Your system of equations consist of the two equations
$u'_1= u_2$ and
$u'_2= -0.5u_2- 2u_1$.

ok the next problem is $$u''+0.5u'+2u=3\sin t$$ so ?
The same thing. Let $u_1= u$ and $u_2= u'$.
The given equation is $u_2'+ 0.5u_2+ 2u_1= 3\sin(t)$
which can be written $u_2'= 3\sin(t)- 0.5u_2- 2u_1$.

Your system of equations consist of the two equations
$u_1'= u_2$ and
$u_2'= 3\sin(t)- 0.5u_2- 2u_1$.
 
OK thank you much
I haven't been on the forum for a few days..

actually I plan on taking this class again in spring 2020 (UHM 307)
I just audited it last spring but really missed a lot of classes and homework
 

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