1.5 Finding xi and eta knowing the first order ode type. Table lookup method.

There is a short cut to obtaining \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) if the first order ode type is known or can be determined. (of course, if we know the ode type, then a direct method for solving the ode can be used which is much simpler, since the type is known and there is no need to use Lie symmetry), but still Lie symmetry can be useful in this case, and also it allows us to find the integrating factor quickly, which provides one more method to solve the ode. An example of a first order ode which does not have known type is

\[ \left ( x\cos y-e^{-\sin y}\right ) y^{\prime }+1=0 \]

The above can be solved using Lie symmetry but with functional form of ansatz \(\xi =f\left ( x\right ) g\left ( y\right ) ,\eta =0\). which gives \(\xi =e^{-\sin y},\eta =0\).

I am in the process of building table for ready to use infinitesimal based on the first ode type. The following small list is the current ones determined. For some first order ode such as linear \(y^{\prime }=f\left ( x\right ) y\left ( x\right ) +g\left ( x\right ) \) or separable \(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \) the infinitesimals can be written directly (but again, for these simple ode’s Lie method is not really needed but it provides good illustration on how to use it. Lie method is meant to be used for ode’s which have no known type or difficult to solve otherwise). For an ode type not given in this list, an ansatz have to be used to solve the similarity PDE.

ode type	form	\(\xi \)	\(\eta \)	notes
linear ode	\(y'=f(x) y(x) +g(x)\)	\(0\)	\(e^{\int fdx}\)	Notice that \(g\left ( x\right ) \) does not affect the result
separable ode	\(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \)	\(\frac {1}{f}\)	\(0\)	This works for any \(g\) function that depends on \(y\) only
		\(0\)	\(g\)

Currently the above are the ones I am able to determine for known first order ode’s. If I find more, will add them. The table lookup is much faster to use than having to solve the similarity PDE each time using ansatz in order to find \(\xi ,\eta \).