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Let's start with equation with two parameters $$y=a x^b.$$ Then we calculate $y'=a b x^{b-1}$ and solve $a=y/x^b$ from original. Substitute that to the derivative and $$y'=b \frac{y}{x}.$$ Then differentiate again and substitute $b=x y'/y$ and we get $$y''=\frac{y'^2}{y}-\frac{y'}{x}.$$ How to solve this "properly", without knowing where the final form came from?

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Hint: $(d/dx)(y'/y) = (y y'' - (y')^2)/y^2$

Specifically:

$$y y'' - (y')^2 = -\frac{y y'}{x}$$

$$\frac{y y'' - (y')^2}{y^2} = -\frac{y'}{x y}$$

$$\frac{d}{dx} \left ( \frac{y'}{y} \right ) = -\frac{1}{x} \frac{y'}{y}$$

$$\frac{d}{dx} \log{\left ( \frac{y'}{y} \right )} = -\frac{1}{x}$$

$$\log{\left ( \frac{y'}{y} \right )} = -\log{x} + K = \log{\left ( \frac{1}{x} \right )} + K$$

$$\frac{y'}{y} = \frac{d}{dx} \log{y} = \frac{b}{x}$$

$$\log{y} = b \log{x} + K' = \log{x^b} + K'$$

$$y = C x^b$$

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  • $\begingroup$ Hehe, yes, I was almost there after the second edit. Thanks :D $\endgroup$
    – Valtteri
    Commented Feb 6, 2013 at 20:22
  • $\begingroup$ Sorry if I spoiled it for you, but I have to keep myself honest sometimes. $\endgroup$
    – Ron Gordon
    Commented Feb 6, 2013 at 20:23

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