Solving a Differential Equation with an Initial Condition of y(1) = 1

[prace]

I am asked to solve this DE with the initial condition of y(1) = 1.

$$(x+y)^2dx + (2xy + x^2-1)dy = 0$$

So, after working the problem out, I came to this as an answer:
$$F(x,y)=\frac{1}{3}x^3 + x^2y + xy^2-y$$

My question is what do I do with the initial condition. I assume that I am just suppossed to plug 1 in somewhere, but the syntax of the initial condition does not seem very intuitive to me. What does y(1) = 1 mean?

Thank you

 

[dextercioby]

They hint towards a x=x(y) solution.

Daniel.