(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I am working on a topic called differential equations, and Im stuck on some working out.

2. Relevant equations[/b]

Can someone please help me to factorise the following fraction (u-2)(-u^2+ 3u+ 2)

I have to integrate it, and I know I can without factorising, but it is so messy, my professor said to factorise the fraction so it looks a bit like a1/(bu+ c)+ a2/(du+ e)

Well thats what the professor said, I have no idea what it means... I can factorise as far as

= u/(-u^2 +3u +2) - 2/(-u^2 +3u +2)

And I know that -u^2 +3u +2 = (-u + 1)(u-2) +4

But not sure how to put it all together

Here is the working out for the whole question just in case you're interested

[tex]

\begin{align*}

y' &= \frac{y+2x}{y-2x} \\

Let y &= ux \\

\text{Then we have} y' &= \frac{ux + 2x}{ux - 2x} \\

&= \frac{u+2}{u-2} \\

\text{Now } y' &= \frac{dy}{dx} = \frac{d(ux)}{dx} = x \frac{dy}{dx} + u \\

\text{So we can say that} x \frac{du}{dx} + u& = \frac{u+2}{u-2} \\

x \frac{du}{dx} &= \frac{u+2}{u-2} - u \\

x \frac{du}{dx} &= \frac{u+2}{u-2} - \frac{u^2-2u}{u-2} \\

x \frac{du}{dx} &= \frac{u+2- u^2+2u}{u-2} \\

x \frac{du}{dx} &= \frac{-u^2+3u+2}{u-2} \\

\end{align*}

\bigskip

[/tex]

Then I have to integrate both sides of this equation...which is what I think I need to factorise in order to make it nice and neat

[tex]

\frac{u-2}{-u^2 + 3u + 2}du &= \frac{1}{x} dx

[/tex]

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# Homework Help: Can someone please help me factorise a fraction (u-2)(-u^2 +3u +2)

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