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I am trying to understand the equations of the critically damped equations of a RLC circuit here

I see this part on page 16

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The part I do not understand is this one:

enter image description here

How can part of the equation be extracted and equaled to zero???

I mean, I understand what the author is trying to do but I cannot understand where this third line comes from.

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  • \$\begingroup\$ Look at the third line carefully. That's your answer.. \$\endgroup\$ – dirac16 Jul 27 '18 at 10:59
  • \$\begingroup\$ I am reading and it says nothing. \$\endgroup\$ – SpaceDog Jul 27 '18 at 11:01
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How can part of the equation be extracted and equaled to zero???

It isn't, y is equated to \$\dfrac{dv_{tr}}{dt}+\alpha v_{tr}\$ and that is the quantity within both square brackets above the equation you have a problem with.

enter image description here

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  • \$\begingroup\$ Sorry but you explanation is not hitting any neurons on my brain... I mean where this third line is coming from? \$\endgroup\$ – SpaceDog Jul 27 '18 at 11:03
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    \$\begingroup\$ Check out my addition @SpaceDog \$\endgroup\$ – Andy aka Jul 27 '18 at 11:08
  • \$\begingroup\$ now I see it. THANKSSSSSSSSSSSSSSSSSSSS!!!!!!!!!!!!!!!!!!!!!!!!!! \$\endgroup\$ – SpaceDog Jul 27 '18 at 11:20
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Andy’s answer is correct, but just to rephrase:

Suppose we take the stuff in the brackets, on the second line, and we call that y.

Then \$y=\dfrac{dv_{tr}}{dt}+\alpha v_{tr}\$

Now substitute both brackets of line two with y, and you have

\$\dfrac{dy}{dt}+\alpha y = 0\$

So, the third line is just the second line, re-written in terms of the y substitution.

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    \$\begingroup\$ HOLY COW! Now I see. I thought the guy was doing some black magic there, like the substitutions you do when you integrate complex stuff or something. DUH!!!! Sorry about that. My brain was melting and I was not able to see the obvious. THANKS!!!!!!!!!!! \$\endgroup\$ – SpaceDog Jul 27 '18 at 11:20

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