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I have the equation:

$\frac{d}{dx}(x^2\frac{dy}{dx})-6x \neq0$

How do you get R? Do you use chain rule or product rule? I cant see how it is done? Or implicit derivation? What is u and v in that case?

$R=x^2y''+2xy'-6x$

The same with:

$\frac{d}{dx}(k\frac{dy}{dx})+Q \neq 0$

I cant see the trick that is done.

R=???

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    $\begingroup$ What does R represent? $\endgroup$ – Kaynex Sep 22 '16 at 22:58
  • $\begingroup$ I know differential equations, but I never saw differential inequalities. $\endgroup$ – Peter Sep 22 '16 at 22:59
  • $\begingroup$ R represent a residual, used in galerkin method $\endgroup$ – Jensy1990 Sep 22 '16 at 23:03
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Product rule:

$$\frac{\text{d}}{\text{d}x}\left(x^2 \frac{\text{d}y}{\text{d}x}\right) = \frac{\text{d}x^2}{\text{d}x}\frac{\text{d}y}{\text{d}x} + x^2\frac{\text{d}}{\text{d}x}\frac{\text{d}y}{\text{d}x} = 2x\frac{\text{d}y}{\text{d}x} + \frac{\text{d}^2y}{\text{d}x^2} = 2xy' + x^2y''$$

And in general:

$$\frac{\text{d}}{\text{d}x}\left(K \frac{\text{d}y}{\text{d}x}\right) + Q = \frac{\text{d}K}{\text{d}x}\frac{\text{d}y}{\text{d}x} + K\frac{\text{d}}{\text{d}x}\frac{\text{d}y}{\text{d}x} + Q = K'y + Ky'' + Q$$

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