(2)
or
dy
dx  =
dv
v
x dx
+
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MATHEMATICS
418
or
dv
v
+x dx
 =
2
cos
sin
sin
cos
v
v
v
v
v
v
v
+
−
(using (1) and (2))
or
dv
x dx  =
sin2 cos
cos
v
v
v
v
v
−
or
sin
coscos
v
v
v
dv
v
−v
⎛
⎞
⎜
⎟
⎝
⎠
 = 2 dx
x
Therefore
sin
coscos
v
v
v dv
v
−v
⎛
⎞
⎜
⎟
⎝
⎠
∫
 =
1
2
xdx
∫
or
1
tanv dv
vdv
−
∫
∫
 =
1
2
xdx
∫
or
log sec
log | |
v
v
−
 =
1
2log |
|
log | C |
x +
or
logsec2
v xv
 = log |C1|
or
secv2
v x  = ± C1 (3)
Replacing v by y
x  in equation (3), we get
2
sec
(
)
y
x
y
x
x
⎛
⎞
⎜
⎟
⎝
⎠
⎛
⎞
⎜
⎟
⎝
⎠
 = C where, C = ± C1
or
sec y
x
⎛
⎞
⎜
⎟
⎝
⎠  = C xy
which is the general solution of the given differential equation Example 28 Solve the differential equation
(tan–1y  – x) dy = (1 + y2) dx Solution The given differential equation can be written as
2
1
dx
x
dy
y
+
+
 =
1
2
1tan
y
y
−
+