How Do You Solve the Differential Equation y' = 2xy^2 at Point (1,3)?

[nhrock3]

y(x) has a point (1,3) so the tangent of y(x) in (x,y) point passes y axes in a point
2xy^2
find y?

how i tried:
y'=f(x,y)
y is the solution of the differential equation.
without knowing y we can find its slope by putting the values in f(x,y)
if we find a function which is tangent in every point the field then its a solution.

the line which has a slope of y'(x)=f(x,y) passes threw (0,2xy^2)
that is theory i know i don't know how to make it into a practice solution

 

[LCKurtz]

Try solving y' = 2xy² by separation of variables using x = 1, y = 3 to evaluate the constant.