I have the following equation $ y & # 39; (x) = y (x) cdot y # (x) + (y # (x)) ^ 2 $ with the initial values $ y (1) = 0 $ and $ y (1) = -1 $.

I am looking for advice on the best way to tackle this particular problem.

This article suggests that a substitution v can reduce the problem in a certain way. So I leave $ v = y $ (x) as following:

begin {align *}

y & # 39; & # 39; (x) & = y (x) cdot y (x) + (y (x)) ^ 2 & (1) \

y & # 39; & # 39; (x) & = y (x) cdot v + v ^ 2 & (2) \

end {align *}

When I enter (2) in wolfram, the engine says that the equation is second-rate *linear* ordinary differential equation.

Can we use substitution from here to solve the problem? Or is there a better way to do that?