Answer
See below
Work Step by Step
Given: $R=3\Omega\\C=\frac{3}{10}H\\E(t)=10V$
When switch is closed in circuit, we have:
$E(t)-iR-\frac{q}{c}=0$
Let $i=\frac{dq}{dt}$, we get: $E(t)-\frac{dq}{dt}R-\frac{q}{c}=0$
Substitute :$10-3i-\frac{3}{i}\frac{di}{dt}=0\\
\rightarrow \frac{dq}{dt}+10i=\frac{100}{3}$
Integrating factor is: $I(t)=e^{\int 10dt}=e^{10t}$
The equation becomes: $\frac{d}{dt}(ie^{10t})=\frac{100}{3}e^{10t}$
Integrating both sides: $ie^{10t})=\frac{10}{3}e^{10t}+c$
Then, $i(t)=\frac{10}{3}+ce^{-10t}$
Since $i(0)=3A$, we have: $c+\frac{10}{3}=3\\
\rightarrow c=-\frac{1}{3}$
The current in the circuit is:
$i(t)=\frac{10}{3}-\frac{1}{3}e^{-10t}$
