Question

If x = a cos θ + b sin θ, y = a sin θ − b cos θ, show that `y^2 (d^2y)/(dx^2)-xdy/dx+y=0`

Answer 1

We have

x=acosθ+bsinθ      .....(1)

y=asinθ−bcosθ      .....(2)

Squaring and adding (1) and (2), we get

x²+y²=(acosθ+bsinθ)²+(asinθ−bcosθ)²

=a²cos²θ+b²sin²θ+2abcosθsinθ + a²sin²θ+b²cos²θ−2abcosθsinθ

=a²(cos²θ+sin²θ)+b2(sin²θ+cos²θ)

⇒x²+y²=a²+b²        .....(3)

Differentiating both sides of (3) w.r.t. x, we get

`2x+2ydy/dx=0`

`⇒2ydy/dx=−2x`

`⇒dy/dx=−x/y                 .....(4)`

Differentiating both sides of (4) w.r.t. x, we get

`Y^2 (d^2y)/(dx^2)-x dy/dx+y`

`=y^2(-(x^2+y^2)/Y63)-x(-x/y)+y`  [From (4) and (5)]

`=-(x^2+y^2)/y+x^2/y+y`

`=(-x^2-^2+x^2+Y^2)/y`

`=0`