u = tan^{-1}(y^2/x) Prove that x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = -sin(2u) sin^2(u) = -2cos(u) sin(3u)

Steps to Solve

1. Define the variable and its derivatives

Start by defining the variable ( u ): $$ u = \tan^{-1}\left(\frac{y^2}{x}\right) $$

Now compute the first and second derivatives ( u_x ), ( u_y ), ( u_{xx} ), ( u_{xy} ), and ( u_{yy} ).

2. First Derivative with respect to ( x )

Using implicit differentiation: $$ u_x = \frac{\partial}{\partial x} \tan^{-1}\left(\frac{y^2}{x}\right) = \frac{1}{1 + \left(\frac{y^2}{x}\right)^2} \cdot \left(-\frac{y^2}{x^2}\right) $$ Simplifying gives: $$ u_x = -\frac{y^2}{x^2 + y^4} $$

3. First Derivative with respect to ( y )

Now compute ( u_y ): $$ u_y = \frac{\partial}{\partial y} \tan^{-1}\left(\frac{y^2}{x}\right) = \frac{1}{1 + \left(\frac{y^2}{x}\right)^2} \cdot \left(\frac{2y}{x}\right) $$ This simplifies to: $$ u_y = \frac{2y}{x + \frac{y^4}{x}} = \frac{2y}{x + y^4/x} $$

4. Second derivatives ( u_{xx}, u_{xy}, u_{yy} )

Compute the second derivatives:

• For ( u_{xx} ): $$ u_{xx} = \frac{\partial}{\partial x} u_x = -\frac{\partial}{\partial x} \left( \frac{y^2}{x^2 + y^4} \right) $$ Use the quotient rule to find this derivative.

• For ( u_{yy} ): $$ u_{yy} = \frac{\partial}{\partial y} u_y = ...$$ Again use appropriate differentiation techniques.

• For ( u_{xy} ): $$ u_{xy} = \frac{\partial}{\partial y} u_x = ...$$ Similar steps apply with implicit differentiation.

5. Combine Second Derivatives

Using the terms found for ( u_{xx}, u_{xy}, u_{yy} ): $$ x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} $$

6. Prove the identity

Establish that this combination equals ( -\sin(2u) \sin^2(u) ) or ( -2\cos(u) \sin(3u) ).