# I got other method of solving

1. May 30, 2009

### transgalactic

U(t)=1
$$Vs(t)=V_0 U(t)$$
$$(Vc)' + \frac{1}{RC}Vc=\frac{1}{RC}Vc=\frac{1}{RC}V_0$$
$$(Vc)' + \frac{1}{RC}Vc(t)=\frac{1}{RC}V_0$$
given:
$$Vc(0^+)=0$$
from her i cant under anything regarding the term of use:
"
homogeneous solution is:
$$(V_ch)'+\frac{1}{RC}Vch=0$$
we guess a solution from the form of
$$V_ch=Ae^{st}$$
and substitute into the homogeneous equation:
$$\int Ae^{st} +\frac{1}{rc}Ae^{st}=0$$
"

these are only the first two steps but i cant understand why are they doing that

??

2. May 30, 2009

### HallsofIvy

I don't understand this. Did you mean to have those two "=" or is this a typo?

Okay, is this what you meant above?

Well, that's the "associated homogeneous equation", not yet a "solution".
You can rewrite it as $dV_{ch}/V_{ch}= (-1/RC)dt$ and integrate that.
$ln(V_{ch})= -t/RC+ K$ and taking exponentials of both sides, $V_{ch}= K_1e^{t/RC}$ where $K_1= e^K$.

Why an integral? $V_{ch}'$ is the derivative
$$(Ae^{st})'+ \frac{1}{rc}Ae^{st}= sAe^{st}+ \frac{1}{rc}Ae^{st}= 0$$
so the exponentials cancel leaving s+ 1/rc= 0. s= -1/rc and the solution becomes $Ae^{t/rc} just as before. That method is more often used with higher order differential equations where you cannot integrate as I did above. You can justify that method by arguing that for y(x) something like Ay"+ By'+ Cy= 0, a "linear equation with constant coefficients", in order that y and its derivatives cancel, to give 0, y' and y" must be the same "kind" of function as y. Exponentials do that nicely: the derivative of [itex]e^{ax}$ is $ae^{ax}$, the same exponential multiplied by a.

But you should be aware this is not based on any idea that a solution MUST be an exponential! For example, The equation y"= 0 has general solution y= Ax+ B and y"+ y= 0 has general solution y= A cos x+ B sin x. Since those solutions are indirectly related to exponentials, "trying" $e^{sx}$ can still lead you to them.

Last edited by a moderator: May 30, 2009
3. May 30, 2009

### transgalactic

U(t) is called the shock function
it equals 1 in this case
Vs=vo*U(t) (Vs the voltage of the source)

does this help ??