Question

If u=xy/√x²+y²+1 then verify that d²u/dx dy=d²u/dy dx

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{U=\dfrac{xy}{\sqrt{x^2+y^2+1}}}$

$\tt{\implies\,U^2=\dfrac{x^2\,y^2}{x^2+y^2+1}}$

$\tt{\implies\,U^2\left(x^2+y^2+1\right)=x^2\,y^2}$

Now,

$\tt{\implies\,\left(x^2+y^2+1\right)\cdot\,2\,U\,\dfrac{\partial\,U}{\partial\,x}+U^2\cdot2x=2\,y^2\,x}$

$\tt{\implies\,\left(x^2+y^2+1\right)\cdot\,U\,\dfrac{\partial\,U}{\partial\,x}+U^2\cdot\,x=y^2\,x}$

$\tt{\implies\,\left(x^2+y^2+1\right)\cdot\,\dfrac{x\,y}{\sqrt{x^2+y^2+1}}\cdot\dfrac{\partial\,U}{\partial\,x}+\dfrac{x^2\,y^2}{x^2+y^2+1}\cdot\,x=y^2\,x}$

$\tt{\implies\,\dfrac{x^2+y^2+1}{\sqrt{x^2+y^2+1}}\cdot\dfrac{\partial\,U}{\partial\,x}+\dfrac{x\,y}{x^2+y^2+1}\cdot\,x=y}$

$\tt{\implies\,\dfrac{x^2+y^2+1}{\sqrt{x^2+y^2+1}}\cdot\dfrac{\partial\,U}{\partial\,x}=y-\dfrac{x^2\,y}{x^2+y^2+1}}$

$\tt{\implies\,\sqrt{x^2+y^2+1}\cdot\dfrac{\partial\,U}{\partial\,x}=\dfrac{x^2\,y-y^3-y-x^2\,y}{x^2+y^2+1}}$

$\tt{\implies\,\sqrt{x^2+y^2+1}\cdot\dfrac{\partial\,U}{\partial\,x}=\dfrac{-y^3-y}{x^2+y^2+1}}$

$\tt{\displaystyle\implies\,\left(x^2+y^2+1\right)^{\frac{3}{2}}\cdot\dfrac{\partial\,U}{\partial\,x}=-y^3-y\,\,\,\,\,...(1)}$

$\tt{\displaystyle\implies\,\dfrac{\partial}{\partial\,y}\left\{\left(x^2+y^2+1\right)^{\frac{3}{2}}\cdot\dfrac{\partial\,U}{\partial\,x}\right\}=\dfrac{\partial}{\partial\,y}\left(-y^3-y\right)}$

$\tt{\displaystyle\implies\,\left(x^2+y^2+1\right)^{\frac{3}{2}}\cdot\dfrac{\partial^2\,U}{\partial\,x\,\partial\,y}+\dfrac{3}{2}\left(x^2+y^2+1\right)^{\frac{1}{2}}\cdot(2\,y)\cdot\dfrac{\partial\,U}{\partial\,x}=-3y^2-1}$

$\tt{\displaystyle\implies\,\left(x^2+y^2+1\right)^{\frac{3}{2}}\cdot\dfrac{\partial^2\,U}{\partial\,x\,\partial\,y}+3y\left(x^2+y^2+1\right)^{\frac{1}{2}}\cdot\dfrac{\partial\,U}{\partial\,x}=-3y^2-1}$

From (1), we get,

$\tt{\displaystyle\implies\,\left(x^2+y^2+1\right)^{\frac{3}{2}}\cdot\dfrac{\partial^2\,U}{\partial\,x\,\partial\,y}-3y\cdot\dfrac{y^3+y}{x^2+y^2+1}=-3y^2-1}$