# Given that u(s,t)=v(s^2-t^2,t^2-s^2) and v is differentiable prove that u verifies equation t*du/ds+s*du/dt=0

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We'll note x = s^2 - t^2 and y = t^2 - s^2

The constraint form enunciation u(s,t) = v(s^2-t^2,t^2-s^2) is turning into u(s,t) = v(x,y).

We'll apply the chain rule and we'll get:

du/ds = (dv/dx)*(dx/ds) + (dv/dy)*(dy/ds)

du/ds = (dv/dx)*(2s) + (dv/dy)*(-2s) (1)

du/dt = (dv/dx)*(dx/dt) + (dv/dy)*(dy/dt)

du/dt = (dv/dx)*(-2t) + (dv/dy)*(2t) (2)

Now, we'll substitute (1) and (2) in the identity that has to be demonstrated:

t*(du/ds)+s*(du/dt) = [2st*(dv/dx) - 2st(dv/dy)] + [-2st(dv/dx) + 2st(dv/dy)]

**If we'll remove the brackets, we'll cancel out the like terms and the relation t*(du/ds)+s*(du/dt) yields 0.**