The Student Room Group
Reply 1
dyy\dfrac{dy}{y} means
Unparseable latex formula:

\dfrac{1}{y}\mbox{d}y

, meaning that
Unparseable latex formula:

\displaystyle \int \dfrac{1}{y}\ \mbox{d}y = \int \dfrac{2x - 1}{(x-1)(2x-3)}\ \mbox{d}x

.

You get it into that format by separating the variables; that is, dividing through by (2x - 3)(x - 1), dividing through by y, and multiplying through by dx. You integrate the RHS by using the result from above, so
Unparseable latex formula:

\displaystyle \int \dfrac{2x - 1}{(x-1)(2x-3)}\ \mbox{d}x = \int \left( \dfrac{-1}{x - 1} + \dfrac{4}{2x - 3} \right) \mbox{d}x

, which you can do using logs.
HitTheSwitch
question 4 on C4 jan07 edexcell

you know that
(2x - 1)/(x-1)(2x-3) = -1/(x-1) + 4/(2x-3)

solve
(2x - 3)(x-1)(dy/dx) = (2x-1)y


Re-arrange to get, 1/y dy = (2x-1)/(x-1)(2x-3) dx

Then substitute in: (2x - 1)/(x-1)(2x-3) = -1/(x-1) + 4/(2x-3)

Then integrate on either side.

1ydy=1(x1)+4(2x3)dx \int \frac{1}{y}\, dy = \int \frac{-1}{(x-1)} + \frac{4}{(2x-3)}\, dx
Reply 3
:O why can u multiply the "dx"? surely all that is is "with respect to x", is it something u can manipulate??
HitTheSwitch
:O why can u multiply the "dx"? surely all that is is "with respect to x", is it something u can manipulate??


It isn't really multiplying by dx, but what is really being done ends up with the same result.
Reply 5
Mathematician!
It isn't really multiplying by dx, but what is really being done ends up with the same result.


ok... so for the moment i can just act like i am really multiplying by dy and dx and it will come out with the same answer?
Reply 6
HitTheSwitch
:O why can u multiply the "dx"? surely all that is is "with respect to x", is it something u can manipulate??

Technically what you're actually doing is, without multiplying by dx, integrating everything with respect to x. That then gives you:

Unparseable latex formula:

\newline \displaystyle \int \dfrac{1}{y}\ \dfrac{\mbox{d}y}{\mbox{d}x}\ \mbox{d}x = \int \left( \dfrac{-1}{x-1} + \dfrac{4}{2x - 3}\ \right) \mbox{d}x



Which is equivalent to:

Unparseable latex formula:

\newline \displaystyle \int \dfrac{1}{y}\ \mbox{d}y = \int \left( \dfrac{-1}{x-1} + \dfrac{4}{2x - 3}\ \right) \mbox{d}x



However, pretending to multiply by dy or dx gives you the same answer, yes. It's technically bad practice, but it's fine for C4.
Reply 7
right I sort of get it now, cheers guys, rep given :smile:
HitTheSwitch
ok... so for the moment i can just act like i am really multiplying by dy and dx and it will come out with the same answer?


Correct. :smile: