1. ## Lagrange Equation

$
y=xy'+\sqrt{b^{2}+a^{2}(y')^{2}}
$

2. Originally Posted by sinichko

$
y=xy'+\sqrt{b^{2}+a^{2}(y')^{2}}
$
If you differentiate both sides you get

$
y' = y' + x y'' + \frac{a^2 y' y''}{\sqrt{b^{2}+a^{2}(y')^{2}}}
$

or

$
\left(\frac{a^2 y'}{\sqrt{b^{2}+a^{2}(y')^{2}}} + x\right) y'' = 0
$
.

You now have two cases which can be solved. Remember, sub in your answer into the original ODE to make sure both (if any) are actual solutions.