Find differential equation using variation principle and lagrangian

Question

I am given lagrangian $L=u_t^2+u_xu_y$

And need to find differential equation using variation principle. I know that solution should be $u_{tt}+u_{xy} = 0$

I tried to go from here with Euler-Lagrange equations:

$\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial x_t}=0.$

$\frac{\partial L}{\partial y}-\frac{d}{dt}\frac{\partial L}{\partial y_t}=0.$

But cannot get needed equation. Can you please show me where I can be wrong?

Is it true that $\frac{\partial L}{\partial x_t}==0$ ?

Answer 1

Solved it!
It's the case of single function of several variables and single derivative here: wiki Euler–Lagrange equation

$L=u_t^2+u_xu_y$

So ${\large \frac{\partial L}{\partial u} - \sum_{j=1}^{n} \frac{\partial}{\partial x_j}\left(\frac{\partial L}{\partial u_{x_j}}\right) = 0 }$ where in my case $x_1=x$, $x_2=y$, $x_3=t$

${\large \frac{\partial L}{\partial u} - \frac{\partial }{\partial x}\left(\frac{\partial L}{\partial u_x}\right) - \frac{\partial }{\partial y}\left(\frac{\partial L}{\partial u_y}\right) - \frac{\partial }{\partial t}\left(\frac{\partial L}{\partial u_t}\right) } = 0$

$0-2u_{tt}-u_{yx}-u_{xy}=0$

$u_{tt}+u_{xy}=0$

as needed