Differential equation C4

MEPS1996

The points (x,y) on the curve C satisfy:
(x + 1)(x + 2) dy/dx=xy
A point on C is (-4,-3)

Find the equation of C in the form y=f(x)

I rearraged the equtions to get all the xs on the RHS and the ys on the LHS, then integrated both sides and found the arbitrary constant to be 0.
So the gives: ln(mod(y))=2ln(mod(x+1))-ln(mod(x+1))

So mod(y)=(x+1)^2/mod(x+1)

Im struggling to see how this could be put in the form y=f(x) without defining the function piecewise for x>=1 and x<1.
In the markscheme the modulus signs have simply been omitted without justification. Is this justified? Or ar C4 level is it acceptable to just remove modulus signs?

Reply 1

ghostwalker

Original post by MEPS1996

So mod(y)=(x+1)^2/mod(x+1)

Im struggling to see how this could be put in the form y=f(x) without defining the function piecewise for x>=1 and x<1.
In the markscheme the modulus signs have simply been omitted without justification. Is this justified? Or ar C4 level is it acceptable to just remove modulus signs?

The graph of this function is disconnected, as you realised. The point (-4,-3) is sufficient of fix the graph for x<-1, but it doesn't fix the graph for x>-1. I can only assume that you're meant to restrict the domain to x<-1.

As such x+1, and y are both <=0, removing the mod signs you'd need the negative of each, which cancel, to give the effect of "just omitting the mod signs".

It does really need a bit of justification, but is valid here.