It was the second part of the assignment where I have to get a differential equation describing all curves for which a tangent line in point $D=(x_o,y_0)$, together with a line through the origin $O=(0,0)$, makes a triangle with the $x$-axis (you can assume it's on the right hand side), and this triangle's surface is 1. So, I went by the triangle formulae $S(\triangle ABC)=\frac{(base)*height_{base}}{2}$, where the base is the length between the origin point and the cross section of the tangent line and $x$-axis, while the height of the triangle is actually $y_0$. From this I derived the differential equation which I don't know how to solve, it kind of smells like Bernoulli's equation but I can't quite get it to work:
$$xyy'-2y'=y^2$$
Any pointers would be much appreciated.
I tried solving this $$y^2-(xy-2)y'=0$$ by using Bernoulli's substitution $z=\frac{1}{y^{n-1}}$, and here $n=2$, but when I divide by $y^2$ after I introduce the substitution I get nowhere, or at least I can't see how. I think it does not work because Bernoulli's form is $y'=P(x)y+Q(x)y^n$, where $n\neq \{{0,1\}}$. And I don't see how to make it work.I don't understand how to implement the Abel equation method.