Mathematical Tools for Physics

4—Differential Equations 100

Manipulate this into

L

dI

dt

=V 0 −IR, then L

dI

V 0 −IR

=dt

Now integrate this to get

L

∫

dI

V 0 −IR

=t+C, or −

L

R

ln(V 0 −IR) =t+C

Solve for the currentIto get
RI(t) =V 0 −e−(L/R)(t+C) (25)

Now does this make sense? Look at the dimensions and you see that itdoesn’t, at least not yet. The problem is
the logarithm on the preceding line where you see that its units don’t make sense either. How can this be? The
differential equation that you started with is correct, so how did the units get messed up? It goes back the the
standard equation for integration, ∫

dx/x= lnx+C

Ifxis a length for example, then the left side is dimensionless, but this right side is the logarithm of a length. It’s
a peculiarity of the logarithm that leads to this anomaly. You can write the constant of integration asC=−lnC′
whereC′is another arbitrary constant, then
∫
dx/x= lnx+C= lnx−lnC′= ln

x

C′

IfC′is a length this is perfectly sensible dimensionally. To see that the dimensions in Eq. ( 25 ) will work themselves
out (this time), put on some initial conditions. SetI(0) = 0so that the circuit starts with zero current.

R.0 =V 0 −e−(L/R)(0+C) implies e−(L/R)(C)=V 0

RI(t) =V 0 −V 0 e−Lt/R or I(t) = (1−e−Lt/R)V 0 /R

and somehow the units have worked themselves out. Logarithms do this, but you still better check. The current
in the circuit starts at zero and climbs gradually to its final valueI=V 0 /R.