Simple second order equation to be solved by reduction of order

Question

This is the question, given that; $y_1 = x$

$$ (x^2+1)y''-2xy' +2y = 0 $$

other form

$$ y''-\dfrac {2x}{(x^2+1)}y' +\dfrac {2}{(x^2+1)}y = 0 $$

using the formula

$y_2 = y_1 \int e^{-\int p(x)dx} / y_1^2 dx $

for $y_2$, I ended up with $y_2 = cx^2+c_1 $

But the solution according to wolframalpha says that it is $y_2 = (x-i)^2$

May anyone tell me where is the deal?

Answer 1

Although WA's solution does solve the DEQ, it looks like a bug, so let's see if we can prove it. Also, see the nice comment by @LutzLehmann above.

We want to use Reduction of Order, given that a solution is $y_1=x$, to solve

$$\tag 1 (x^2+1)y''-2xy' +2y = 0$$

We will use these examples as a guide, assume $y_2 = v x$, and calculating derivatives

$$y_2 = v x, ~~~~y_2' = v + v' x, ~~~~y_2'' = v'' x + 2 v'$$

Substituting into $(1)$

$$v''x(x^2+1) + 2 v' = 0$$

Let $w = v' \implies w' = v''$, so

$$w'x(x^2+1) + 2 w = 0$$

Solving for $w$

$$w = \left(\dfrac{1}{x^2} + 1 \right)$$

From $v' = w = \left(\dfrac{1}{x^2} + 1 \right)$, solve for $v$

$$v = x - \dfrac{1}{x}$$

We have $y_2 = v x$, hence

$$y_2 = x^2 - 1$$

We can now write

$$y(x) = c_1 y_1(x) + c_2 y_2(x) = c_1 x + c_2(x^2-1)$$

Checking

$$(x^2+1)y''-2xy' +2y = (x^2+1) (2c_2) -2x(c_1+2 c_2 x) + 2(c_1 x + c_2(x^2-1)) = 0 ~~~~ \Large\color{\green}{\checkmark}$$