$ y''+\frac{2x}{1+x^2}y'+ \frac{y}{1+2x^2+x^4}=0$ I was told the following differential equation can be solved very easy.
$$ y''+\frac{2x}{1+x^2} y'+ \frac{y}{1+2x^2+x^4}=0$$
But I don't see the way. Do you see it?
 A: The equation can be rewritten as $$y'' + \dfrac{2x}{1+x^2}y' + \dfrac{y}{(1+x^2)^2} = 0$$
Multiplying throughout by $\sqrt{1+x^2}$, we get
$$\sqrt{1+x^2} y'' + \dfrac{2x}{\sqrt{1+x^2}}y' + \dfrac1{(1+x^2)^{3/2}}y = 0$$
Now let $h(x) = y(x) \sqrt{1+x^2}$. We then get
$$h'(x) = y'(x) \sqrt{1+x^2} + \dfrac{x}{\sqrt{1+x^2}}y(x)$$
$$h''(x) = y''(x) \sqrt{1+x^2} + \dfrac{2x}{\sqrt{1+x^2}} y'(x) + \dfrac1{(1+x^2)^{3/2}}y(x)$$
Hence, we get that
$$h''(x) = 0 \implies h(x) = c_1 + c_2x \implies y(x) = \dfrac{c_1 + c_2 x}{\sqrt{1+x^2}}$$

EDIT
The motivation for this method is the following. Essentially, we want to write the entire equation in the form $(g(x)y)''=0$. This gives us
$$gy'' + 2g'y' + g''y = 0$$
Hence, the goal is to find $g$ such that
$$\dfrac{g'}{g} = \dfrac{x}{1+x^2} \,\,\, \text{ and } \,\, \dfrac{g''}g = \dfrac1{(1+x^2)^2}$$
Solving $\dfrac{g'}{g} = \dfrac{x}{1+x^2}$, gives us $g = \sqrt{1+x^2}$. This also satisfies $\dfrac{g''}g = \dfrac1{(1+x^2)^2}$ and hence the entire equation can be rewritten as $(y \cdot \sqrt{1+x^2})''=0$.
A: This is a second order linear homogeneous differential equation. It's second order because the highest derivative in the dependent variable is degree 2. It's linear because $y$, $y'$, and $y''$ are combined linearly (with coefficients that are functions of $x$). It's homogeneous because it has no term that is only a function of $x$. Marvis's answer is great, with an observation about rewriting the equation with $\sqrt{1+x^2}$ that may have come about through experience or by a standard procedure. You might want to know that there are standard ways to solve any second order linear homogeneous differential equation.
Just type "how to solve a second order linear differential equation" into Google and see what comes up. 
