What Occurs in the Differential Equation I = C*(dv/dt) as t Approaches Zero?

Andrew123
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What happens in the differential equation: I = C*(dv/dt) when we limit t ---> 0 and what is the working? This isn't really a homework Q its just me wanting to know the theory behind this. TY in advance for looking and helping.
 
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I(t) and V(t) are both functions of time if C is capacitance. This is just a relation between I and V. You can't 'solve' it unless you describe the physical situation.
 
Andrew123 said:
What happens in the differential equation: I = C*(dv/dt) when we limit t ---> 0 and what is the working? This isn't really a homework Q its just me wanting to know the theory behind this. TY in advance for looking and helping.
It's not clear what you are saying. Take the limit of what as t goes to 0? Both sides of the equation? If so, are I and C constants or functions of t? If constants then dv/dt is a constant and so taking the limit as t goes to 0 doesn't change anything- you would still have I= C(dv/dt). If they are continuous functions of t, then taking the limit as t goes to 0 gives I(0)= C(0)(dv/dt) where dv/dt is now evaluated at t= 0.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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