450 Higher Engineering Mathematics

Whenx=0,y=1 thus^12 ln1=ln1+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 circuit
containing resistanceRand inductanceLin series
with a constant voltage sourceEisgivenbythe
differential equationE−L
(
di
dt
)
=Ri.Solvethe
equation and findiin terms of timetgiven that
whent=0,i=0.
In theR−Lseries circuit shownin Fig.46.3, the supply
p.d.,E,isgivenby
E=VR+VL
VR=iRandVL=L
di
dt
Hence E=iR+L
di
dt
from which E−L
di
dt
=Ri
VR VL
R
E
i
L
Figure 46.3
Most electrical circuits can be reduced to a differential
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
−LRt
and E−Ri=Ee
−LRt
and
Ri=E−Ee
−LRt
.
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
i (1eRt/L )
(^0) Time t
E
R
E
R

Figure 46.4
Problem 13. For an adiabatic expansion of a gas
Cv
dp
p
+Cp
dV
V
= 0 ,