Solving 1st Order PDE: Finding L with Ax^2 Form

[PhyAmateur]

How to find L if the form is:

$$ (\frac{\partial L}{\partial x})^2 - (\frac{\partial L}{\partial y})^2 = -1$$

The author wrote, $$L = y + ax^2 + ..$$
but I didn't get how?

 

[Mark44]

How to find L if the form is:

$$ (\frac{\partial L}{\partial x})^2 - (\frac{\partial L}{\partial y})^2 = -1$$

The author wrote, $$L = y + ax^2 + ..$$
but I didn't get how?

This is actually a first order PDE, since both partials are first partials. Since you have "second order" in the title, I'm wondering if you have written the equation correctly.

Did you mean this equation?
$$\frac{\partial^2 L}{\partial x^2} - \frac{\partial^2 L}{\partial y^2} = -1 $$

For the equation as you wrote it, L = ax + by + c is a solution, with a² - b² = -1. I don't see how L = y + ax² could be a solution.

 

[PhyAmateur]

Oh I have written the title wrongly! I will edit it right away. Can you please tell me how did you get L=ax+by+c?

 

[PhyAmateur]

I can't edit the title :(, I would really appreciate if anyone from the mentors could edit my title!

 

[Mark44]

Oh I have written the title wrongly! I will edit it right away. Can you please tell me how did you get L=ax+by+c?

My approach was pretty simpleminded - it's been many years since I had a class on PDEs. My reasoning was that since the difference of the squares of the first partials was a constant (-1), then L(x, y) must be first-degree in both x and y, so that L(x, y) = ax + by + c.

You can check that this is a solution, provided that a² - b² = -1. If there are other solutions, my recollections from the class I had are too vague, and maybe someone else can weigh in.

I can't edit the title :(, I would really appreciate if anyone from the mentors could edit my title!

Done