(a) We simply need to subtract 7x dx from both sides, then insert integral signs and integrate:

`dy=-7x\ dx`

`intdy=-int7x\ dx`

`y=-7/2x^2+K`

NOTE 1: We are now writing our (simple) example as a differential equation. Earlier, we would have written this example as a basic integral, like this:

`(dy)/(dx)+7x=0`

Then `(dy)/(dx)=-7x` and so `y=-int7x\ dx=-7/2x^2+K`

The answer is the same - the way of writing it, and thinking about it, is subtly different.


NOTE 2: `int dy` means `int1\ dy`, which gives us the answer `y`.

We could also have:

`intdt=t`

`intd theta=theta`

` int da=a`

and so on. We'll come across such integrals a lot in this section.


(b) We now use the information y(0) = 3 to find K.

The information means that at x = 0, y = 3. We substitute these values into the equation that we found in part (a), to find the particular solution.

`3=7/2(0)^2+K` gives K = 3.

So the particular solution is: `y=-7/2x^2+3`, an "n"-shaped parabola.