How Do Shock Functions Aid in Solving Differential Equations?

[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 can't 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 can't understand why are they doing that
the youtube solution differs a lot

??

 

[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 can't 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 [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.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> these are only the first two steps but i can't understand why are they doing that<br /> the youtube solution differs a lot<br /> <br /> ?? </div> </div> </blockquote> 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 $e^{ax}$ is $ae^{ax}$, the same exponential multiplied by a.<br /> <br /> 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.[/itex]

 

[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 ??