Answer
a) $I(t)$ = $\frac{1}{100}(e^{-t}-e^{-11t})$
b) $\approx$ $0.00715$ amperes
c) as below
Work Step by Step
a)
$\frac{dI}{dt}+11I$ = $\frac{1}{10}e^{-t}$
$A(t)$ = $11$
$B(t)$ = $\frac{1}{10}e^{-t}$
$α(x)$ = $e^{11t}$
$(e^{11t}I)'$ = $\frac{1}{10}e^{10t}$
$e^{11t}I$ = $\frac{1}{100}e^{10t}+C$
$I(t)$ = $\frac{1}{100}e^{-t}+Ce^{-11t}$
$I(0)$ = $0$
$0$ = $\frac{1}{100}+C$
$C$ = $-\frac{1}{100}$
$I(t)$ = $\frac{1}{100}(e^{-t}-e^{-11t})$
b)
$I(t)$ = $\frac{1}{100}(e^{-t}-e^{-11t})$ = $0$
$t$ = $t_{m}$ = $\frac{1}{10}\ln{11}$ seconds
$I(t_{m})$ = $\frac{1}{100}(e^{-\frac{1}{10}\ln{11}}-e^{-\frac{11}{10}\ln{11}})$ $\approx$ $0.00715$ amperes
