# Thread: Linear first order ODE question

1. ## Linear first order ODE question

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?

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

3. 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.

4. Thanks guys, I'll give it another shot.

5. 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?

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

7. Ahh, ok. Thanks.

8. 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.