Circuit equation solve?

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 solve?
OK lets see





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


Dividing by R


(dQ/dt) + (1/RC)*Q = 0


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


let -1/RC = x dQ/dt = Q'


The equation can be written as


Q'/Q = x


Integratinf it we get


Q(t) = Qe^(tx)


Substitute x


Q(t) = Q e(-t/RC)
Reply:This is the which describes the charging and discharging of capacitor.





this is integration.


dQ/q=-dt/RC


log (Q)=-t/RC +log(Qo)


Q=Qoe^(t/RC)
Reply:http://web.njit.edu/phys_lab/Laboratory%...





http://www.wellesley.edu/Physics/phyllis...
Reply:R*(dQ/dt) + (1/C)*Q = 0


Dividing by R


(dQ/dt) + (1/RC)*Q = 0


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


let -1/RC = x dQ/dt = Q'


The equation can be written as


Q'/Q = x


Integratinf it we get


Q(t) = Qe^(tx)


Substitute x


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