Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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

Specifically:

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

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

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

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

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

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

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

$$y = C x^b$$

share|cite|improve this answer
    
Hehe, yes, I was almost there after the second edit. Thanks :D – Valtteri Feb 6 '13 at 20:22
    
Sorry if I spoiled it for you, but I have to keep myself honest sometimes. – Ron Gordon Feb 6 '13 at 20:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.