# Thread: second order x function 14

1. ## second order x function 14

x^2y''+xy'-0.25=0
how to solve it?

2. $\displaystyle x^2\,\frac{d^2y}{dx^2}+ x\,\frac{dy}{dx} - \frac{1}{4} = 0$

$\displaystyle \frac{d^2y}{dx^2} + \frac{1}{x}\,\frac{dy}{dx} - \frac{1}{4x^2} = 0$

$\displaystyle \frac{d^2y}{dx^2} + \frac{1}{x}\,\frac{dy}{dx} = \frac{1}{4x^2}$.

Now let $\displaystyle u = \frac{dy}{dx}$ so that $\displaystyle \frac{du}{dx} = \frac{d^2y}{dx^2}$ and the DE becomes

$\displaystyle \frac{du}{dx} + \frac{1}{x}\,u = \frac{1}{4x^2}$.

This is first order linear, so use the Integrating Factor method to find $\displaystyle u$, which you can then use to find $\displaystyle y$.