Help solving an important ODE

Question

I could use some help solving the ODE $$(1+x^{2})y''+4xy'+2y=0$$

I have entered it into wolfram and it gave me the partial details of a solution. It read to rewrite $2=(4x)'-(x^{2}+1)''$ into the ODE.I gather we are to "reverse" the product rule. If would be helpful if someone can provide the remaining detail.Better yet if someone knows a better technique to approach to solve the ODE.

Answer 1

Hint $$(1+x^{2})y''+4xy'+2y=0$$ $$y''+(x^{2}y''+2xy')+2(xy'+y)=0$$ $$y''+(x^{2}y')'+2(xy)'=0$$ Integrate $$y'+(x^2y')+2xy=K$$ $$y'(1+x^2)+2xy=K$$ $$y'+(x^2y)'=K$$ integrate again $$y+(x^2y)=K_1x+K_2$$ $$y(1+x^2)=K_1x+K_2$$ $$\boxed{y(x)=\frac {K_1x+K_2}{(1+x^2)}}$$

Answer 2

If $v = 1+x^2$, the DE says $$ v y'' + 2 v' y' + v'' y = 0$$

Note that the left side is $(v y)''$.

The fact that this all works out so neatly should not lead to overconfidence: most second-order DE's with non-constant coefficients have non-elementary solution. For example, if you changed the last term from $2y$ to $y$ the solutions would not be elementary (they could be written in terms of hypergeometric functions).