`x=acostheta+bsintheta` and `y=asintheta-bcostheta,` show that `y^2(d^2y)/(dx^2)-x(dy)/(dx)+y=0`

Explanation

`y = asintheta-bcostheta`

`x = acostheta+bsintheta`

`=>dx/(d theta) = -asintheta+bcostheta`

`=>dy/(d theta) = `

`:.dy/dx = dy/(d theta)*(d theta)/dx = (acostheta+bsintheta)/(-asintheta+bcostheta)`

`=>dy/dx = x/(-y)`

`=>(d^2y)/dx^2 = (-y + xdy/dx)/y^2`

`=>y^2(d^2y)/dx^2-xdy/dx+y =0`