If x = acosθ + bsinθ, y = asinθ - bcosθ, show that y²d²y/dx² - xdy/dx + y = 0

y²=a²sin²θ+b²cos²θ-2abcosθsinθ
x²=b²sin²θ+a²cos²θ+2abcosθsinθ
Now adding both the equations we get,
y²+x²=a²+b²
Now differentiating with respect to x we get,
2yy'+2x=0
Now again differentiating with respect to x we get,
2yy''+2(y')²+2=0
Multiplying both the sides with y and writing y'=-x/y
We get y²d²y/dx² - xdy/dx + y = 0.