How Do Shock Functions Aid in Solving Differential Equations?

[transgalactic]

U(t)=1
<br /> Vs(t)=V_0 U(t)<br />
<br /> (Vc)&#039; + \frac{1}{RC}Vc=\frac{1}{RC}Vc=\frac{1}{RC}V_0<br />
<br /> (Vc)&#039; + \frac{1}{RC}Vc(t)=\frac{1}{RC}V_0<br />
given:
<br /> Vc(0^+)=0<br />
from her i can't under anything regarding the term of use:
"
homogeneous solution is:
<br /> (V_ch)&#039;+\frac{1}{RC}Vch=0<br />
we guess a solution from the form of
<br /> V_ch=Ae^{st}<br />
and substitute into the homogeneous equation:
<br /> \int Ae^{st} +\frac{1}{rc}Ae^{st}=0<br />
"

these are only the first two steps but i can't understand why are they doing that
the youtube solution differs alot

??

 

[HallsofIvy]

U(t)=1
<br /> Vs(t)=V_0 U(t)<br />
<br /> (Vc)&#039; + \frac{1}{RC}Vc=\frac{1}{RC}Vc=\frac{1}{RC}V_0<br />

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

<br /> (Vc)&#039; + \frac{1}{RC}Vc(t)=\frac{1}{RC}V_0<br />

Okay, is this what you meant above?

given:
<br /> Vc(0^+)=0<br />
from her i can't under anything regarding the term of use:
"
homogeneous solution is:
<br /> (V_ch)&#039;+\frac{1}{RC}Vch=0<br />

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
<br /> V_ch=Ae^{st}<br />
and substitute into the homogeneous equation:
<br /> \int Ae^{st} +\frac{1}{rc}Ae^{st}=0<br />

Why an integral? V_{ch}&#039; is the derivative
(Ae^{st})&#039;+ \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.<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&#039;t understand why are they doing that<br /> the youtube solution differs alot<br /> <br /> ?? </div> </div> </blockquote> You can justify that method by arguing that for y(x) something like Ay&quot;+ By&#039;+ Cy= 0, a &quot;linear equation with constant coefficients&quot;, in order that y and its derivatives cancel, to give 0, y&#039; and y&quot; must be the same &quot;kind&quot; 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&quot;= 0 has general solution y= Ax+ B and y&quot;+ y= 0 has general solution y= A cos x+ B sin x. Since those solutions are indirectly related to exponentials, &quot;trying&quot; 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 ??