(2)
Differentiating equation (2) with respect to x, we get
dy
dx  =
dv
v
+x dx (3)
Substituting the value of y and dy
dx  in equation (1), we get
dv
v
+x dx
 =
cos
1
vcos
vv
+
or
dv
x dx  =
cos
1
vcos
v
v
v
+ −
or
dv
x dx  =
1
cosv
or
cosv dv = dx
x
Therefore
∫cosv dv
 =
1 dx
x∫
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DIFFERENTIAL EQUATIONS
403
or
sin v = log |x| + log |C|
or
sin v = log |Cx|
Replacing v by y
x , we get
sin
y
x
⎛
⎞
⎜
⎟
⎝
⎠  = log |Cx|
which is the general solution of the differential equation (1) Example 17 Show that the differential equation 2
2
0
x
x
y
y
y e dx
y
x e
dy
⎛
⎞
⎜
⎟
+
−
=
⎝
⎠
is
homogeneous and find its particular solution, given that, x = 0 when y = 1 Solution The given differential equation can be written as
dx
dy  = 2
2
x
y
x
y
x e
y
y e
−