# Second Order ODE and integral of exponential divided by a polynomial

My original question was

Solve $$x^2y'' + 2y' - 2y = 0$$

First I noticed that $$x^2+2x+2$$ is a solution. Using order reduction, doing $$y = v(x)(x^2+2x+2)$$, I found that $$\int\frac{e^{2/x}}{(x^2+2x+2)^2}dx$$

Can someone help me in this part, please? The only method I found used complex numbers and the $$Ei$$ function, but I would like a solution in another way.

Thanks for attention.

## 1 Answer

$$x^2y'' + 2y' - 2y = 0$$ $$x^2y'' \color{red}{+ 2xy'-2xy'}+2y' - 2y = 0$$ $$(x^2y')' -2(xy)'+2y' = 0$$ $$x^2y' -2y(x-1) = C_1$$ $$v' +\dfrac 2{x^2}v = \dfrac {C_1}{x^4}$$ Where $$v=\dfrac y {x^2}$$: This is a first order DE that you can easily solve with integrating factor $$\mu=e^{-2/x}$$: $$(ve^{-2/x})' = \dfrac {C_1e^{-2/x}}{x^4}$$ $$ve^{-2/x} = C_1\int \dfrac {e^{-2/x}}{x^4}dx+C_2$$ For the integral substitute $$u=-\dfrac 1x$$: $$\int \dfrac {e^{-2/x}}{x^4}dx= \int {u^2e^{2u}}du$$ Integrate by part. You get the solution you already have: $$y_1=C_1(x^2+2x+2)$$ The second solution should be : $$y_2=C_2x^2e^{2/x}$$

• Nice solution! What a good trick! Awesome. Thanks Commented Mar 29, 2023 at 14:46
• You're welcome @TassandroCavalcanteLeitão Commented Mar 29, 2023 at 14:54