second order x function 14

**transgalactic** · November 20th 2010, 12:23 PM

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

**Prove It** · November 20th 2010, 02:49 PM

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

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

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

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 $\displaystyle y$ .

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