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I got other method of solving

  1. May 30, 2009 #1
    U(t)=1
    [tex]
    Vs(t)=V_0 U(t)
    [/tex]
    [tex]
    (Vc)' + \frac{1}{RC}Vc=\frac{1}{RC}Vc=\frac{1}{RC}V_0
    [/tex]
    [tex]
    (Vc)' + \frac{1}{RC}Vc(t)=\frac{1}{RC}V_0
    [/tex]
    given:
    [tex]
    Vc(0^+)=0
    [/tex]
    from her i cant under anything regarding the term of use:
    "
    homogeneous solution is:
    [tex]
    (V_ch)'+\frac{1}{RC}Vch=0
    [/tex]
    we guess a solution from the form of
    [tex]
    V_ch=Ae^{st}
    [/tex]
    and substitute into the homogeneous equation:
    [tex]
    \int Ae^{st} +\frac{1}{rc}Ae^{st}=0
    [/tex]
    "

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

    ??
     
  2. jcsd
  3. May 30, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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 [itex]dV_{ch}/V_{ch}= (-1/RC)dt[/itex] and integrate that.
    [itex]ln(V_{ch})= -t/RC+ K[/itex] and taking exponentials of both sides, [itex]V_{ch}= K_1e^{t/RC}[/itex] where [itex]K_1= e^K[/itex].

    Why an integral? [itex]V_{ch}'[/itex] is the derivative
    [tex](Ae^{st})'+ \frac{1}{rc}Ae^{st}= sAe^{st}+ \frac{1}{rc}Ae^{st}= 0[/tex]
    so the exponentials cancel leaving s+ 1/rc= 0. s= -1/rc and the solution becomes [itex]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}[/itex] is [itex]ae^{ax}[/itex], 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" [itex]e^{sx}[/itex] can still lead you to them.
     
    Last edited: May 30, 2009
  4. May 30, 2009 #3
    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 ??
     
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