Linear first order ODE question

**Glitch** · December 29th 2010, 06:34 PM

The question:

$x^2 \frac{dy}{dx} -xy = y$

I'm not sure where to start. I tried getting it in a form where I can calculate the integrating factor, but nothing I do seems to work. Any suggestions?

**pickslides** · December 29th 2010, 06:39 PM

add $xy$ to both sides, then divide $x^2$ to both sides, does it separate?

**Prove It** · December 29th 2010, 06:41 PM

Originally Posted by Glitch

The question:

$x^2 \frac{dy}{dx} -xy = y$

I'm not sure where to start. I tried getting it in a form where I can calculate the integrating factor, but nothing I do seems to work. Any suggestions?

To make it linear

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

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

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

Go from here.

**Glitch** · December 29th 2010, 06:44 PM

Thanks guys, I'll give it another shot.

**Glitch** · December 29th 2010, 06:48 PM

Originally Posted by Prove It

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

If I divide both sides by x^2, wouldn't I lose part of the solution? Otherwise I could divide both sides by the entire LHS and get 1 = 0. Or am I missing something?

**Prove It** · December 29th 2010, 06:51 PM

You can if you assume that $\displaystyle x \neq 0$ . It's a standard technique.

**Glitch** · December 29th 2010, 06:53 PM

Ahh, ok. Thanks.

**Prove It** · December 29th 2010, 06:56 PM

I suppose that you could always write it like this...

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

$\displaystyle x^2\left(\frac{dy}{dx} - x^{-1}y - x^{-2}y\right) = 0$

$\displaystyle x^2\left[\frac{dy}{dx} - (x^{-1} + x^{-2})\,y\right] = 0$ .

You can still use the Integrating Factor Method and then solve the equation using NFL.

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