(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| Example 13 Find the equation of a curve passing through the point (–2, 3), given that
the slope of the tangent to the curve at any point (x, y) is 
2x2
y