Question

Verify that u=2x+a−−−−−√+2y+b−−−−−√ is a solution of the equation ∂x∂u+∂y∂u=u.

Answer 1

∂x∂u+∂y∂u=u.

Divide throughout by ∂x∂y, we get

$\frac{\partial u}{\partial \:y}+\frac{\partial u}{\partial \:x}= \frac{\partial{u}}{\partial \:x\partial \:y}$

Given u= $\left(2x+a\right)-\sqrt{2y+b}$

Left side:

$\frac{\partial }{\partial \:x}\left(\left(2x+a\right)-\sqrt{2y+b}\right)+\frac{\partial }{\partial \:y}\left(\left(2x+a\right)-\sqrt{2y+b}\right)$

$\frac{\partial }{\partial \:x}\left(\left(2x+a\right)-\sqrt{2y+b}\right)= 2$

$\frac{\partial }{\partial \:y}\left(\left(2x+a\right)-\sqrt{2y+b}\right)= \frac{1}{-\sqrt{2y+b}}$

Hence left side is $2-\frac{1}{\sqrt{2y+b}}$

Right side:

$\frac{\partial }{\partial \:x\partial \:y}\left(\left(2x+a\right)-\sqrt{2y+b}\right)$ = $\frac{\partial }{\partial \:y}(2)$ =0

Hence both sides are not equal and hence u is not the solution for given equation.