# Trying to understand the critically damped equations of a RLC circuit

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

I see this part on page 16 The part I do not understand is this one: 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.

• Look at the third line carefully. That's your answer.. – dirac16 Jul 27 '18 at 10:59
• I am reading and it says nothing. – Duck Jul 27 '18 at 11:01

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. • Sorry but you explanation is not hitting any neurons on my brain... I mean where this third line is coming from? – Duck Jul 27 '18 at 11:03
• Check out my addition @SpaceDog – Andy aka Jul 27 '18 at 11:08
• now I see it. THANKSSSSSSSSSSSSSSSSSSSS!!!!!!!!!!!!!!!!!!!!!!!!!! – Duck Jul 27 '18 at 11:20

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.

• 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!!!!!!!!!!! – Duck Jul 27 '18 at 11:20