Maths C4 - Partial Fractions... HELP??

So I'm stuck on part (C).



I have no idea how to do the long algebraic division part with this one

How do I do long algebraic division from... $\frac{x^3 - x^2 - x - 3}{x^2 - x}$???

How do I do long algebraic division from... $\frac{x^3 - x^2 - x - 3}{x^2 - x}$???

So I'm stuck on part (C).



I have no idea how to do the long algebraic division part with this one

How do I do long algebraic division from... $\frac{x^3 - x^2 - x - 3}{x^2 - x}$???

OH YES HERE WE GO!

Well attempt it first so we can see where you're going wrong

It's pretty hard to explain, but this https://www.youtube.com/watch?v=JZda1TjCHlU is a good example of doing so, can you replicate the example for yours?

False alarm. After having a little break I realised I was being quite silly. I managed to do the long algebraic division part so now I can move on from there and finish the rest of the question

Yup! A nice way to do this without fiddly division is like so:

$\displaystyle \frac{x^3-x^2-x-3}{x(x-1)} = \frac{x^2(x-1) - x - 3}{x(x-1)} = \frac{x^2}{x(x-1)} - \frac{x+3}{x(x-1)} = x - \frac{x+3}{x(x-1)}$

But it's just a clever parlor trick really.

Ok so I'm doing a C4 past paper from January 2007 but am stuck on question 4...

I've managed to find that for part (a) the partial fraction is...

$\frac{2x-1}{(x-1)(2x-3)} = \frac{-1}{(x-1)} + \frac{4}{(2x-3)}$

but I dont understand what is meant by the "general solution" bit for part (b). Does it just mean find the original equation from differential equation given?? ...


Can someone help me please?

When you integrate something like $\displaystyle \int 2x .dx = x^2+c$ you get a whole family of particular solutions depending on what your constant is. By the "general" solution we are referring to $x^2+c$ explicitly where the constant is arbitrary (and not to be forgotten about). The next part asks you to find a particular solution, and these are found only if a certain condition is given; such as $y(2)=10$ from your Q which allows you solve for the particular constant