$$\frac{dy}{dx}=(1-x)y^2+(2x-1)y-x$$ This is a form of riccati differential equation, which can be reduced to Bernoulli's equation if one particular solution is given. Here $y=1$ is a given solution. Hence the general solution must be in the form $y=1+z$ for some $z(x)$, but substituting this in main equation gives $$\frac{dz}{dx}+z^2x=z^2+z$$ Which is not in the form of Bernoulli equation. Please help !
Also $$2x^2\frac{dy}{dx}=(x-1)(y^2-x^2)+2xy$$ This is asked in the problem section of bernoulli's differential equation but i have no idea how to solve this.
Bernoulli's equation has form, $$ \frac{dy}{dx}+p(x)y=q(x)y^n $$
Now, consider this, $$ \frac{dz}{dx}+z^2x=z^2+z $$ This easily simplifies to, $$ \frac{dz}{dx}-z=(1-x^2)z^2 $$ where $p(x)=-1$ and $q(x)=1-x^2$ .
And similarly other one simplifies to, $$ 2x^2\frac{dy}{dx}=(x-1)(y^2-x^2)+2xy $$ $$ 2x^2\frac{dy}{dx}=xy^2-x^3-y^2+x^2+2xy $$ $$ \frac{dy}{dx}=\frac{xy^2}{2x^2}-\frac{x}{2}-\frac{y^2}{2x^2}+\frac{1}{2}+\frac{y}{x} $$ put $z=\frac{y}{x}$ and $\frac{dz}{dx}=\frac{-y}{x^2}+\frac{1}{x}\frac{dy}{dx}$ which is same as $\frac{dz}{dx}=\frac{-z}{x}+\frac{1}{x}\frac{dy}{dx}$ implies $\frac{dy}{dx}=x\frac{dz}{dx}+z$. Hence, the equation will be, $$ x\frac{dz}{dx}+z=\frac{xz^2}{2}-\frac{x}{2}-\frac{z^2}{2}+\frac{1}{2}+z $$ $$ 2x\frac{dz}{dx}=xz^2-x-z^2+1 $$ $$ 2x\frac{dz}{dx}=z^2(x-1)-(x-1) $$ $$ 2x\frac{dz}{dx}=(z^2-1)(x-1) $$ Finally, $$ \frac{2}{z^2-1}\frac{dz}{dx}=\frac{x-1}{x} $$ or $$ \frac{2}{z^2-1}dz=\frac{x-1}{x}dx $$ which can be solved by taking integration on both sides...