# Simple second order equation to be solved by reduction of order

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?

• Note that $(x-i)^2+2iy_1=x^2-1=y_2$. To find your specific error you would need to display the steps from the formula for $y_2$ to your proposed solution. May 8 at 13:30

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

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