MHB Non-Dimensionalisation of Two Differential Equations: Working Out and Struggles

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I've got two differential equations I need to non-dimensionalise

I've managed to do the \[ dc/dT=α- μc \] with the following working out:

By letting:
\[ N=N0n \]
\[ C=C0c \]
\[ t=t0T \]

1596461095620.png


However I'm struggling with the first equation.

1596461128387.png


I'm up to here, it's just the left hand side of the equation I'm struggling to work out. Any help would be great, cheers
 
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Looking at $\frac{dN}{dt}= \frac{rN}{h+ rN}- bNC$ and comparing that to the desired $\frac{dr}{dt}= \frac{n}{n+1}- \beta nc$ my first thought would be to divide the numerator and denominator of $\frac{rN}{h+ rN}$ by h. That gives $\frac{\frac{rN}{h}}{1+ \frac{rN}{h}}$. So we should let $n= \frac{rN}{h}$. With that substitution, the first equation becomes $\frac{h}{r}\frac{dn}{dt}= \frac{n}{1+ n}- \left(\frac{br}{h}\right)nC$.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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