I'm fine for the first two parts of the question but for the last part, I'm just wondering how it was found in the solutions that y=2x was a suitable substitution to make so that the equation would work. Is there a method or should you be able to know by just looking at it or by trial and error? Thanks.

Hi all - there is now a STEP II matrices module available here: https://maths.org/step/step-ii-matrices-new

Hi,

Hello, I have a question concerning the following differential equation
$$x \frac{\mathrm dy}{\mathrm dx}=y+1$$

Separating variables and integrating:
$$\int \frac{1}{y+1} \mathrm dy=\int \frac 1x \mathrm dx$$
$$\ln|y+1|=\ln|x|+c$$

In the solutions to question 3a) in the Mixed Pure STEP I module, the following step is made:
$$y+1=Ax \,\,\,\,\, \text{(where} \ A=e^c).$$

Why is it possible to omit the modulus function?

Many thanks in advance.

Hello, I have a question on the coefficient of restitution.

A ball hits a wall with speed u and is reflected with speed v (one-dimensional motion). The coefficient of restitution of the collision is e. It (intuitively) seems clear that $$v=-eu.$$

If you are coming today (Thurs 5th July) or tomorrow (Fri 6th July), then come and have a a chat about STEP (or anything else) in the central core! You can also ask any questions on this forum...

Good luck to everyone sitting STEP III tomorrow!

Good luck to all the STEP II candidates for tomorrow!

Good luck to everyone sitting STEP I tomorrow!

Hope it goes well!

In the Hints for the question, the method for finding the upper bound on z^3 in part (ii) is given as using the fact that 1/3(4kz-z^4) is a square number. In the solutions it says that this is found by comparison with part (i), where the lower bound is found by 1/3(4z^3 -k^3).

I don't really understand how we go from the statement in part (i) to the statement in part (ii), it seems like they are, if equivalent in form, just arbitrarily chosen and I don't really get the methodology for deriving it as needed for the question.