Lagrange equation

Definition

A Lagrange equation' is a first-order differential equation that is linear in both the dependent and independent variable, but not in terms of the derivative of the dependent variable. Explicitly, if the independent variable is $x$ and the dependent variable is $y$, the Lagrange equation has the form:

$\!F(y')x+G(y')y=H(y')$

We first solve for $y$ (dividing both sides by $G(y')$) to get an equation of the form:

$\!y=f(y')x+g(y')$

where $f=-F/G$ and $g=H/G$. Note that this process may involve some loss of solutions, since it excludes the possibility $G(y')=0$. Those solution cases can be considered separately. For the rest of the discussion, we assume that the equation is in the "solved for $y$" form.

Solution method

We differentiate both sides with respect to $x$ to obtain:

$\!y'=f'(y')y''x+f(y')+g'(y')y''$

We now see that the differential equation involves only $y'$ and higher derivatives, so set $p=y'$ to get:

$\!p=f'(p)p'x+f(p)+g'(p)p'$

This becomes:

$\!p=f(p)+(xf'(p)+g'(p))p'$

so that:

$\!p-f(p)=(xf'(p)+g'(p))p'$

We switch the roles of dependent and independent variable, thinking of $x$ as the dependent variable now. We can rewrite the above differential equation as:

$\!(p-f(p)){\frac {dx}{dp}}=xf'(p)+g'(p)$

Rearranging:

$\!(p-f(p)){\frac {dx}{dp}}-f'(p)x=g'(p)$

Now, we separate out the solution possibility $f(p)=p$. For any other solution, we divide by $p-f(p)$ to get a first-order linear differential equation which we can solve for $x$ in terms of $p$. Suppose the general solution is of the form:

$x=\Phi (p,C)$

Then, the overall general solution is given by the following parametric curve:

$x=\Phi (p,C),y=f(p)\Phi (p,C)+g(p)$

In addition, there may be special solutions corresponding to the $p=f(p)$ case. Specifically, for all $p$ satisfying $p=f(p)$ (hopefully, a discrete set of values), we have straight line solutions $y=f(p)x+g(p)$.