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I'm confused in solving linear constant coefficients differential equations (LCCDEs) by Laplace transform if initial conditions are given at time

  1. just before $t=0$
  2. just after $t=0$
  3. exactly at $t=0$

Is method of solving LCCDE by Laplace transform is same in all three cases or it's different?

I know it's more of mathematical question and I asked this question in mathematics stack exchange but I didn't get answers, so I ask it here as it is equally a problem of systems differential equation.enter image description here!

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Initial conditions are always given at $t=0^-$, because they define the state of the system before any input is applied, and - by definition - the input is applied at $t=0$. The state at $t=0^+$ is determined by the initial conditions as well as by the input signal.

The unilateral Laplace transform can be used to solve LCCDEs with initial conditions $y(0^-), y'(0^-),\ldots$ because of the definition

$$\mathcal{L}\{f(t)\}=F(s)=\int_{0^{\color{red} -}}^{\infty}f(t)e^{-st}dt\tag{1}$$

from which it follows that

$$\mathcal{L}\{f'(t)\}=sF(s)-y(0^-)\tag{2}$$

Note that it's common to write initial conditions as $y(0),y'(0),\ldots$, when actually $t=0^{-}$ is meant.

EDIT: Concerning the example in the book: if there is no Dirac delta impulse in the current $i(t)$ at $t=0$, the capacitor voltage $v_c(t)$ cannot jump at $t=0$. Consequently, $v_c(0^-)=v_c(0)=v_c(0^+)$ must hold.

In general, the initial conditions define values of the output signal and its derivatives right before the source signal is switched on. It can be the case that the limits of the output and its derivatives exist at $t=0$. If that is the case, it doesn't make a difference if we use $t=0^{-}$ or $t=0$ or $t=0^{+}$ because the corresponding function values are all the same. If the output signal or its derivatives are discontinuous at $t=0$, then the distinction becomes important, and the initial conditions define the values just before the discontinuity, i.e., at $t=0^{-}$.

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  • $\begingroup$ Thanks matt! I always solve any differential equation as you suggested for initial condition in Laplace domain but I got confused when I read network book by D Roy ,I add picture of that paragraph of book which puzzled me, $\endgroup$ – user215805 Aug 8 at 19:24
  • $\begingroup$ @user215805: I've added some information to my answer. $\endgroup$ – Matt L. Aug 9 at 8:40

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