Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading the book Ordinary Differential Equations and Dynamical Systems by Gerald Teschl. Or rather, I am reading the online edition: http://www.mat.univie.ac.at/~gerald/ftp/book-ode/index.html

On page 15 it says:

More generally, consider the differential equation $$ \dot{x} = f\left(\frac{ax + bt + c}{\alpha x + \beta t + \gamma}\right). $$ Two cases can occur. If $a \beta - \alpha b = 0$, our differential equation is of the form $$\dot{x} = \widetilde{f}(ax +bt) $$

I understand the rest of the page. But this step makes no sense to me. Would someone care to explain to me what the author might mean?

Thanks in advance

share|improve this question

2 Answers 2

up vote 3 down vote accepted

$\beta = \frac{\alpha b}{a}$

plug in to get

$\dot{x} = f\left(\frac{a x + bt + c}{\alpha x + \alpha b/a t + \gamma}\right) = f\left(\frac{a x + bt + c}{\alpha/a( a x + b t + a\gamma/\alpha)}\right)$

which is a function of $ax + bt$.

share|improve this answer
    
Thank you, that seems correct :) –  DoubleTrouble Feb 2 '13 at 20:33

Since $a\beta - \alpha b$ is a determinant, and in the special case $c=\gamma=0$ you can always write the argument of $f$ in a $ax+bt$ form (because the numerator is a multiple of the denominator). In general, if the numerator is a multiple of the denominator, you could do a change of coordinates and resort the argument to a $ax+bt$ form, but seems to me that the condition $a\beta -\alpha b=0$ is necessary but not sufficient.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.