why there is a minus in the power

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transgalactic

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when the two roots s1 and s2 that are different of the homogenius equation
in normal diff theory i do

$v_c (t)=Ae^{s_1t}+Be^{s_2t}$

why why in basic circuit theory we do
$v_c (t)=Ae^{-|s_1|t}+Be^{-|s_2|t}$
(this is a standard latex code in this forum its not working
so you can copy it into here
and you will get my clear equations

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dknguyen

Well-Known Member
You can always have negative signs or positive signs if you wanted to. It works either way and doesn't change anything.

It's just that we know from observing circuits that the voltage is decaying (which is why the exponent is negative in the solution). So we put the negative sign in there beforehand so when we solve for s1 and s2 we get a nice positive number, instead of a negative number. If we didn't put the negative sign there we would just get a negative number for s instead.

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transgalactic

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thanks  Tesla23

Member
when the two roots s1 and s2 that are different of the homogenius equation
in normal diff theory i do

$v_c (t)=Ae^{s_1t}+Be^{s_2t}$

why why in basic circuit theory we do
$v_c (t)=Ae^{-|s_1|t}+Be^{-|s_2|t}$

This is a meaningless question without any context. In all the circuit theory I have done I have never seen this, nor have I needed to do it to 'get the right sign'. Indeed when you analyse your circuit, if your equations give you a negative sign for s then your circuit is stable in that whatever excitation you have hit it with will die out, whereas if the sign of s is positive then your circuit is unstable (as may be the case if you have an oscillator or an unstable control loop). The signs are important, it is silly to ignore them.

To make latex work, use
Code:
[ latex] ... [ /latex]

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