(1)
Integrating both sides of equation (1), we get
∫dy
 =
1
2x
xdx
⎛
⎞
+
⎜
⎟
⎝
⎠
∫
or
y = x2 + log |x| + C (2)
Equation (2) represents the family of solution curves of the given differential equation
but we are interested in finding the equation of a particular member of the family which
passes through the point (1, 1) Therefore substituting x = 1, y = 1 in equation (2), we
get C = 0 Now substituting the value of C in equation (2) we get the equation of the required
curve as y = x2 + log |x|