Circuit equation proof?

In a certain question managed to get to:





R*(dQ/dt) + (1/C)*Q = 0





In the answers it then says:





Q(t) = Q*e^-(t/RC)





I know this is true, obviously, but don't understand the jump between the two steps, does he integrate it, how?





Cheers!

Circuit equation proof?
The equation: -





R*(dQ/dt) + (1/C)*Q=0





May be rearranged to give:-





(1/Q)*(dQ/dt) = -1/(R*C)





taking the differential dQ/dt operator to either side of the equation gives two integrals: -





∫(1/Q).dQ = ∫-1/(R*C).dt





Solving, this gives: -





Loge(Q(t))+const' = -t/(R*C)+const





since


const''=Loge(const)





Loge(Q(t)*const'') = -t/(R*C)+const





Taking the exponential of each side of the equation gives: -





Q(t) = (1/const'')*e^-(t/(R*C))+const





at t=0 Q=Q therefore (1/const'') =Q and const=0





hence





Q(t) = Qe^-(t/(R*C))





Sorry about the first mistake!
Reply:i forget this was for a discharging capacitor instead of intergrating from 0 to Q you intergrate from Q0 to Q





subtract (1/c)Q on the other side


R(dQ/dt)=-Q/C


its a separable differential equation so you need the Q on the other side.


R(dQ/Q)(1/dt)=-1/C





(dQ/Q)(1/dt)=-1/RC


dQ/Q=-dt/RC


intergrate dQ from Q0 to Q and


intergrate dt from 0 to t so


lnQ/Q0=-t/RC


take the e on both sides


Q=Q0e^(-t/RC)


Q is a function of time so


Q(t)=Q0e^(-t/RC)





if you need any more help just ask