SOLUTION OF FIRST ORDER DIFFERENTIAL EQUATIONS BY SEPARATION OF VARIABLES 449
I
Integrating both sides gives:
1
2 ln (1+x
(^2) )=lny+c
When x=0, y=1 thus^12 ln 1=ln 1+c, from
which,c=0.
Hence the particular solution is^12 ln (1+x^2 )=lny
i.e. ln (1+x^2 )
1
(^2) =lny, from which, (1+x^2 )
1
(^2) =y.
Hence the equation of the curve isy=
√
(1+x^2 ).
Problem 12. The currentiin an electric cir-
cuit containing resistanceRand inductanceLin
series with a constant voltage sourceEis given
by the differential equationE−L
(
di
dt
)
=Ri.
Solve the equation and findiin terms of time
tgiven that whent=0,i=0.
In theR−Lseries circuit shown in Fig. 46.3, the
supply p.d.,E,isgivenby
E =VR+VL
VR=iRand VL=L
di
dt
Hence E =iR+L
di
dt
from which E −L
di
dt
=Ri
E
i
RL
VR VL
Figure 46.3
Most electrical circuits can be reduced to a differen-
tial equation.
RearrangingE−L
di
dt
=Rigives
di
dt
E−Ri
L
and separating the variables gives:
di
E−Ri
dt
L
Integrating both sides gives:
∫
di
E−Ri
∫
dt
L
Hence the general solution is:
−
1
R
ln (E−Ri)=
t
L
+c
(by making a substitution u=E−Ri, see
Chapter 39).
Whent=0,i=0, thus−
1
R
lnE=c
Thus the particular solution is:
−
1
R
ln (E−Ri)=
t
L
−
1
R
lnE
Transposing gives:
−
1
R
ln (E−Ri)+
1
R
lnE=
t
L
1
R
[lnE−ln (E−Ri)]=
t
L
ln
(
E
E−Ri
)
Rt
L
from which
E
E−Ri
=e
Rt
L
Hence
E−Ri
E
=e
−Rt
L and E−Ri=Ee
−Rt
L and
Ri=E−Ee
−Rt
L.
Hence current,
i=
E
R
(
1 −e
−Rt
L
)
,
which represents the law of growth of current in an
inductive circuit as shown in Fig. 46.4.
i
(^0) Time t
i = (^) RE (l−e-Rt/L)
E
R
Figure 46.4