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Suppose we have that

$$\frac{dy}{dx} = -\frac{y}{x}.$$

Taking the derivative implicitly with respect to $x$, we can easily obtain

$$\frac{d^{2}y}{dx^{2}} = \frac{-\frac{dy}{dx}x + y}{x^{2}} = \frac{2y}{x^{2}}.$$

I figured that if I wanted to find $\frac{d^{2}y}{dx^{2}}$, I should be able to simply set $F = -y/x$ and get

$$\frac{d^{2}y}{dx^{2}} = - \frac{F_{x}}{F_{y}} = \frac{\frac{y}{x^{2}}}{\frac{1}{x}} = \frac{y}{x}$$

which clearly does not work. I tried taking the total derivative of

$$\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}\frac{dy}{dx} = 0$$

and it seems pretty clear that what I did was unlikely to work, but I don't understand why my intuition in this case would fail since $\frac{dy}{dx}$ is just a function that we can implicitly differentiate.

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if you wanted to go the long way..by setting $F = -y/x$ then you would have

$$ y'' = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}y' $$ or $$ y'' = \frac{y}{x^2} +\left(-\frac{1}{x}\right)\left(-\frac{y}{x}\right) = \frac{2y}{x^2} $$ now the problem is that you have the original result wrong it should not have a minus sign

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  • $\begingroup$ opps, too many negatives signs always confuses me $\endgroup$ – JessicaK Nov 26 '14 at 11:08
  • $\begingroup$ No worries! I do the exact same, but I always have to start from scratch even if I have already done 4 pages of calculations .. Damn OCD! $\endgroup$ – Chinny84 Nov 26 '14 at 11:10
  • $\begingroup$ After looking at these answers I completely understand what to do, but I can't help but feel disappointed more than anything that I cannot take $-F_{x}/F_{y}$ in this case. Thank you. $\endgroup$ – JessicaK Nov 26 '14 at 11:27
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If you take $$ \frac{dy}{dx} = -\frac{y}{x} = F(y,x)$$

This equation $$\frac{d^{2}y}{dx^{2}} = - \frac{F_{x}}{F_{y}}$$ is false. You should write

$$\frac{d^{2}y}{dx^{2}} = \frac {d}{dx}F(y(x),x) = \frac{\partial F}{\partial y} \frac{dy}{dx} +\frac{\partial F}{\partial x} = \left(-\frac 1x \right)\left(-\frac yx\right)+\frac{y}{x^2} = \frac{2y}{x^2}.$$

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