How do I prove an equation in partial derivative?

Question

When x = e^s cost and y = e^s sint how to get the below equation correctly? Everytime I get extra things in my answer. Can someone help me to get the correct answer. But I get a different answer. I want to identify the wrong point in my answer.

Answer 1

I use subscripts for derivatives. Note these identities: $$ x=x_s, \;\;\;y=y_s,\;\;\; x_t = -y,\;\;\; y_t = x,\;\;\; x^2+y^2=e^{2s} $$ Then you get: \begin{align} u_s &= u_xx_s + u_yy_s\\ u_{ss} &= x_s\partial_s u_x + u_x x_{ss} + y_s\partial_s u_y + u_y y_{ss}\\ &= u_{xx}x^2 + u_{xy}xy + u_{x}x + u_{yx}xy + u_{yy}y^2 + u_yy\\ &= u_{xx}x^2 + u_{yy}y^2 + 2u_{yx}xy + u_yy + u_{x}x\\ u_t &= u_xx_t+u_yy_t \\ &= -yu_x+u_yx\\ u_{tt} &= -y_tu_x-y\partial_tu_x + x_tu_y + x\partial_t u_y\\ &= -xu_x + y^2u_{xx} - 2xyu_{xy} - yu_y-u_{yy}x^2 \\ u_{tt}+u_{ss} &= [y^2+x^2](u_{xx}+u_{yy})- 2xyu_{xy} - yu_y-xu_x+ 2u_{yx}xy + u_yy + u_{x}x\\ &= e^{2s}(u_{xx}+u_{yy})\\ \therefore\; u_{xx}+u_{yy} &= e^{-2s}( u_{tt} + u_{ss} ) \end{align}