I'll try to explain the idea of the "varioidaan vakiota" that is stated in your .pdf. The idea traces back to Legendre, which used it to solve linear ODEs.
We have the equation
y'-\frac{2xy}{1+x^2}=1+x^2
We first solve the homogeneous equation
y'-\frac{2xy}{1+x^2}=0
y'=\frac{2xy}{1+x^2}
\frac{y'}{y}=\frac{2x}{1+x^2}
$\log y = \log(1+x^2)+C_1$
$ y = C(1+x^2)$
What Legendre thinks now, which Spiegel says "in first sight, seems ridiculous is":
Let's assume $C$ is not constant, but rather variable. What will this new function $C(x)$ be so that
$y = C(x)(1+x^2)$
is a solution to our original equation?
So in your case we have that, differentiating produces
y' = C'(x)(1+x^2)+C(x) 2x
But from our equation we have that y'=\dfrac{2xy}{1+x^2}+(1+x^2) and that $ \dfrac{y}{1+x^2} = C(x)$
So plugging this in we have
y' = C'(x)(1+x^2)+\frac{2xy}{1+x^2}
So that, comparing to our original equation we have C'(x) = 1 or $C(x) = x+C_1$, so that our solution is
$y = (x+C_1)(1+x^2)$
Hope this helped clear out the "varioidaan vakiota" issue. For more infor refer to Spiegel's book on Differential Equations, page 202.
You have
(1+x^2)y' -2xy = (1+x^2)^2
or
$(1+x^2)\dfrac{dy} {dx} -2xy = (1+x^2)^2$
$\dfrac{dy} {dx} -\dfrac{2x}{1+x^2}y = 1+x^2$
You can solve this by the integrating factor $\exp\left(-\log\left(1+x^2\right)\right)=\dfrac{1}{1+x^2}$ which will give
$\dfrac{1}{1+x^2}\dfrac{dy} {dx} -\dfrac{1}{1+x^2}\dfrac{2x}{1+x^2}y = 1$
\left(\dfrac{y}{1+x^2}\right)' = 1
$\dfrac{y}{1+x^2} = x+C$ $y = (x+C)(1+x^2)$