Can someone me factorise a fraction (u-2)(-u^2 +3u +2)

[laura_a]

Homework Statement

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

2. Homework 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 that's 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]

 

[HallsofIvy]

Homework Statement

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

2. Homework 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 that's 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]

That can't be factored using integer coefficients but you know that if an expression factors as (x-a)(x-b) then a and b must be roots of (x-a)(x-b)= 0. Use the quadratic formula to find the roots of x²- 3x- 2= 0.

 

[Defennder]

Try completing the square of [tex]-u^2+3u+2[/tex]. Then use a simple algebraic identity to decompose it into partial fractions.

 

[laura_a]

I think I've gotten a little closer, The question I'm trying to do is to solve a d.e. using change of variables, so once I get this part worked out I should be able to get it

So this is how far I've gone
Since [tex] u^2- 3 u -2= 0 [/tex]
has two roots
[tex] u_1 = (3+ 17^{0.5})/2, u_2= (3- 17^{0.5})/2 [/tex]

so

[tex] u^2- 3 u -2= (u - (3+ 17^{0.5})/2 ) (u- (3- 17^{0.5})/2 ) [/tex]

Originally I had
[tex] \frac{u-2}{-u+3u+2}du = \frac{1}{2}dx [/tex] - I'll call this equation (2)
So now I've got the LHS as

[tex] S = -u /((u - (3+ 17^{0.5})/2 ) +2 /(u- (3- 17^{0.5})/2 ) [/tex]
I changed the u-2 to -u+2 since I made a similar change on the denominator in order to solve it...

Anyhow, I integrated that and ended up with something really messy invloving logs

[tex] -2(u + 17^{0.5} ln(u-17^{0.5}-3)+3ln(u-17^{0.5}-3)-17^{0.5}-3+4ln(u+17^{0.5}-3) = ln(x) + C [/tex]

(I integrated both sides of equation (2) above)

Well the prob is now, I'm probably wrong anyway, but the next step is to solve for u, and then I have to sub back in the original change of varaible which was y=xu and end up with an expression for y in terms of x and C... before I go on, can anyone tell me if I should bother working through those logs, or is it wrong? Thanks :)

 

[arildno]

Let us take this as a GENERAL case, shal we?
We are to decompose:
[tex]\frac{u-u_{0}}{(u-u_{1})(u-u_{2})}=\frac{A}{u-u_{1}}+\frac{B}{u-u_{2}}[/tex]
where the unindexed u is our variable, the indexed u's known numbers, and A and B constants to be determined.
We therefore must have:
[tex]u-u_{0}=A(u-u_{2})+B(u-u_{1})[/tex], or by comparing coefficients, we get the system of equations:
[tex]A+B=1[/tex]
and
[tex]u_{2}A+u_{1}B=u_{0}[/tex]
whereby we arrive at the solutions:
[tex]A=\frac{u_{0}-u_{1}}{u_{2}-u_{1}},B=\frac{u_{2}-u_{0}}{u_{2}-u_{1}}[/tex]

This is much simpler than using specific numbers!