Submitted by TimHargreaves on Fri, 04/07/2017 - 15:41
Forums:
In question 5 of the STEP II Calculus support module, the solutions jump from:
$$\ln |v| = -x + \ln |x-1| +c$$
to
$$v=Ae^{-x}(x-1)$$
$c$ and $A$ are arbitrary constants with $c=\ln A$
There is no obvious domain on v or x so I'm struggling to see how the removable of the absolute value signs was justified. Is this just a simplification to make the question appropriate for STEP I (that's the paper it came from) or am I missing the reason why this is okay to do?
Generally, when solving
Generally, when solving differential equations and manipulating log signs, we tend to be quite cavalier with modulus signs. Let's consider this question and see why it might be ok.
You could keep them and get something like $$|v|=A\mathrm{e}^{-x}|x-1|.$$To get a single valued function $v=v(x)$, we should choose a branch as this equation currently defines two branches (have a look at a sketch: https://www.desmos.com/calculator/r6dprrnh1o - not sure if this link will work). Since we want a nice smooth function, rather than something with a "cusp" at $x=1$, we'll choose either $$v=A\mathrm{e}^{-x}(x-1)\quad \text{or} \quad v=-A\mathrm{e}^{-x}(x-1).$$At this point, you realise that the sign in front of $A$ can really just be incorporated into $A$ so finally we get $$v=A\mathrm{e}^{-x}(x-1).$$Now this is the same as if we'd just ignored the modulus signs in the first place which is why we generally don't worry about them.
You can check that your solution is valid by substituting back into the differential equation.
*manipulating logs
*manipulating logs