If x= a costheta + b sintheta and y= a sintheta - b costheta, then show that y^2 (d^2y/dx^2)-x(dy/dx)+y=0?

x = a*cos θ + b*sin θ
y = a*sin θ - b*cos θ

x² + y²
= ( a*cos θ + b*sin θ )² + ( a*sin θ - b*cos θ )²
= a²cos² θ + 2ab*sin θ*cos θ + b²sin² θ + a²sin² θ - 2ab*sin θ*cos θ + b²cos² θ
= ( a² + b² )cos² θ + ( a² + b² )sin² θ
= ( a² + b² )( sin² θ + cos² θ )
= a² + b²

Hence, x² + y² = a² + b²

2x*dx + 2y*dy = 0
y' = dy/dx = - 2x / ( 2y ) = - x / y

By the formula ( u / v )' = ( u' v - u v' )/ v² , we have
y'' = dy' / dx = - ( 1*y - xy' ) / y²
y²y'' = - ( 1*y - xy' ) = - y + xy'
y²y'' - xy' + y = 0
Q.E.D.