i got other method of solving..

[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
the youtube solution differs alot

??

[HallsofIvy]

U(t)=1

Vs(t)=V_0 U(t)


(Vc)' + \frac{1}{RC}Vc=\frac{1}{RC}Vc=\frac{1}{RC}V_0

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


(Vc)' + \frac{1}{RC}Vc(t)=\frac{1}{RC}V_0

Okay, is this what you meant above?

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

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.


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

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.

these are only the first two steps but i cant understand why are they doing that
the youtube solution differs alot

??
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.

[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 ??