Answer:
A=-3/2
B=3
c=3/2
a=-2
Step-by-step explanation:
Knowing that for [tex]t>0[/tex] [tex]i(t)=2\left(1-e^{-2t}\right) A[/tex], [tex]i(t)=0[/tex] when t<0 and using the definition of charge
[tex]q(t)=\int_{-\infty}^0i(t')dt'+\int_0^t i(t')dt'=0+\int_0^t i(t')dt'[/tex]
The first term corresponds to q(0), the charge accumulated before t=0, in this case it luckily gives zero so we don't have to worry about it.
Let's proceed and integrate [tex]i(t)[/tex] then when t>0
[tex]i(t)=3\int_0^t \left[ \int_0^tdt'-\inte_0^t e^{-2t'}dt' \right]dt'=3\left[ t+\frac{1}{2}\left( e^{-2t}-1 \right) \right]=3t+\frac{3}{2}e^{-2t}-\frac{3}{2}\,\, C[/tex]
It is clear that:
A=-3/2
B=3
c=3/2
a=-2
