To solve it, we make the substitution
x = vy (2)
Differentiating equation (2) with respect to y, we get
dx
dy  =
+
dv
v
y dy
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MATHEMATICS
404
Substituting the value of 
xand dx
dy  in equation (1), we get
dv
v
+y dy
 = 2
1
2
v
v ev
e
−
or
dv
y dy  = 2
1
2
v
v ev
v
e
− −
or
dv
y dy  =
2 ve1
−
or
2ev dv =
dy
y
−
or
2 ve
⋅dv
∫
 =
ydy
−∫
or
2 ev = – log |y| + C
and replacing v by 
x
y  , we get
2
yx
e  + log |y| = C (3)
Substituting x = 0 and y = 1 in equation (3), we get
2 e0 + log |1| = C ⇒ C = 2
Substituting the value of C in equation (3), we get
2
yx
e  + log |y | = 2
which is the particular solution of the given differential equation Example 18 Show that the family of curves for which the slope of the tangent at any
point (x, y) on it is  
2
2
2
x
y
+xy
, is given by x2 – y2 = cx