BACK TO BASICS
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here I am using the grey background to highlight functions, not MATLAB code.
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Misha and Basa,
the usage of 1/u in Misha's solution, in this context is erroneous, let me show you why:
1.- one of the possible solutions to dx/dy+y+x-1=0 is
x=1-exp(-y)
2.- inverting x(y)
x=1-exp(-y)
ln(1-x)=-y
so we have
y=-ln(1-x), or ln(|1-x|)
the 1st key point you need to understand is that d(y)/dx is not a division operator, it's a convention to express derivative, which is an operator that generally speaking does not comply with
if u=f(v) d(f(v),v)=k
let be v=g(u)
it's not generally correct assume d(g(u),u)=k/f(v)
2.- according to Misha,
dy/dx=1/(1-x-y)
but
dy/dx=d(-ln(1-x))/dx=1/(1-x)
yet
1/(1-x-y)=1/(1-x+ln(1-x)) or 1/(1-ln(|1-x|))
which obviously are not the same
Misha the 2nd reason why you shouldn't be using 1/u is that you may be losing poles while constraining the function domain to that of u=f(v), not the variable intended to be the domain.
For these reasons I consider my answer deserves to be the accepted answer.
Additionally, Basa, I gave the Simulink diagram first, Misha had just given some general indications, Misha saw my diagram, he builds an erroneous Simulink equation, and you Basa mark as accepted an erroneous Simulink circuit for the above reasons.
Please reconsider marking my answer as accepted.
John BG
