Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm looking through a solution of some problem. It has this step I don't quite understand. Please help me clarify.

Relevant equations:

  • $ u(x) = y^3 $
  • $\frac{dy}{dx} = \frac{1}{3y^2} \frac{du}{dx} $

Problem:

$$(1)\quad x^3 \frac{du}{dx} + 3x^2 u(x) = 6x $$

$$(2)\quad\quad \frac{d}{dx} (x^3 u(x)) = 6x $$ How do you get from [1] to [2]. Is think its not the first time I've seen something like this, is it a special property or pattern?

share|improve this question
    
It's the familiar Product Rule for differentiation. –  André Nicolas Jan 25 '13 at 4:40

3 Answers 3

That is the product rule at work; if you have two functions $f$ and $g$ then

$$(f g)' = f' g + f g'$$

In your case, $f(x) = u(x)$ and $g(x) = x^3$.

share|improve this answer

You can do it via this way similarly: $$x^3u'(x)+3x^2u(x)=6x$$ If you know the Differential of a function , you will see that the LHS of above OE has a form $$d\left(x^3u(x)\right)$$ Now you need just to solve $$d\left(x^3u(x)\right)=6x$$ by a simple integration of both sides.

share|improve this answer
    
+1 I like "process" oriented answers! –  amWhy Jan 27 '13 at 15:25
    
@amWhy: I saw how you edited. I'll try to do it for my other answers here. Thank you very much. –  Babak S. Jan 27 '13 at 15:28
1  
there's so much to learn about formatting; I'm always learning new tricks. I just copied your link, highlighted the terms "Differential of a function", clicked on the "chain-link" at the top left menu, and pasted the link into the pop-up asking for the URL. You can also do this [Differential of a function](http://en.wikipedia.org/wiki/Differential_of_a_function) and only the expression "Differential of a function" will show up - (blue, as a link) –  amWhy Jan 27 '13 at 15:32

This is known as the integrating factor trick. In your case, the integrating factor is simply one, so $$x^3(u(x))'+u(x)(x^3)'=(x^3\cdot u(x))'.$$ Remember that $$(f(x)\cdot g(x))'=f'(x)g(x) + f(x)g'(x).$$

share|improve this answer

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.