# Differential Equation Basic - please explain the detail of this step

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?

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

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 $g(x) = x^3$.

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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.

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+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
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 $$(f(x)\cdot g(x))'=f'(x)g(x) + f(x)g'(x).$$

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