Can you solve $\frac{dy}{dx}=(1-x)y^2+(2x-1)y-x$? $$\frac{dy}{dx}=(1-x)y^2+(2x-1)y-x$$
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...
