Question

rdq/dt+q/c=v, q=1/r

Answer 1

Rearranging the equation\[r\frac{\mathrm dq}{\mathrm dt}+\frac qc=v\\
\frac{\mathrm dq}{\mathrm dt}+\frac q{cr}=\frac vr\]
an integrating factor can be found\[I(t)=\exp\int \frac{1}{rc}\mathrm dt=\exp (t/rc)=e^{t/rc}\]
Such that\[\frac{\mathrm d}{\mathrm dt}(qe^{t/rc})=\frac vre^{t/rc}\]
integrating
\[qe^{t/rc}=\int\frac vre^{t/rc}\mathrm dt\\
qe^{t/rc}=\frac{v}{r}\cdot rce^{t/rc}+C\\
q(t)=v+Ce^{-t/rc}\]

Answer 2

is q=1/r susposed to be an initial condition?