`y' + (-1+x^2)y' - 2xy = 0` Solve the differential equation

To solve this differential equation, we'll try to separate the variables: move `y` and `y'` to the left side and `x` without `y` to the right:

`y'(1 + x^2 - 1) = 2 x y,`  or  `(y')/y = (2x) / x^2 = 2/x.`

Now we can integrate both sides with respect to `x` and obtain

`ln|y| = 2ln|x| + C,`

which is the same as

`y = Ce^(2 ln|x|) = C |x|^2 = C x^2,`

...

To solve this differential equation, we'll try to separate the variables: move `y` and `y'` to the left side and `x` without `y` to the right:

`y'(1 + x^2 - 1) = 2 x y,`  or  `(y')/y = (2x) / x^2 = 2/x.`

Now we can integrate both sides with respect to `x` and obtain

`ln|y| = 2ln|x| + C,`

which is the same as

`y = Ce^(2 ln|x|) = C |x|^2 = C x^2,`

where `C` is an arbitrary constant. This is the general solution.