(1)
Integrating both sides of equation (1), we get
y∫dy2
 =
4 x dx
− ∫
or
−y1
 = – 2x2 + C
or
y =
212
x −C (2)
Substituting y = 1 and x = 0 in equation (2), we get,  C = – 1 Now substituting the value of C in equation (2), we get the particular solution of the
given differential equation as 
212
1
y
x
=
+ Example 12 Find the equation of the curve passing through the point (1, 1) whose
differential equation is x dy = (2x2 + 1) dx (x ≠ 0)