I will say what Krizalid said, but in a more general way.
For a differential equation of the form: $\displaystyle y' + p(x)y = f(x)$
we define $\displaystyle \mu (x) = e^{\int p(x)~dx} = \mbox { exp} \left( \int p(x)~dx \right)$ as the integrating factor of the differential equation. We solve the differential equation by multiplying through by the integrating factor. so we get the new equation:
$\displaystyle \mu (x) y' + \mu (x) p(x) y = \mu (x) f(x)$
it will always be the case that the left hand side is the derivative we would get from applying the product rule. so the equation becomes:
$\displaystyle \left( \mu (x) y \right)' = \mu (x) f(x)$
$\displaystyle \Rightarrow \mu (x) y = \int \mu (x) f(x)~dx$
$\displaystyle \Rightarrow y = \frac { \int \mu (x) f(x) ~dx}{ \mu (x)}$ is our solution
remember that you will get an arbitrary constant from the integration, you have to divide this by $\displaystyle \mu (x)$ as well
after doing some reading on intergrating factors, from
ive used the intergrating factor, whereby f(x) = -4x and a(x) = 4
so g(x)= exp^(intergral of 4 dx)
so $\displaystyle g(x) = e^{4x}$
now using the formula y= 1/g(x).[intergral of f(x)g(x) dx]
so $\displaystyle y=1/e^{4x} [intergral -4xe^{4x}]
$
is this correct and how would i carry on from this?
yes, that is fine, and incidentally, that is exactly the process i described to you above. i tried to save you the trouble of trying to get it from you text book, sometimes those things are ... let's just say, I hate the text that i used for differential equations. the only thing i like about it is that it gives the answers for even as well as odd problems, which is rare for textbooks to do