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If x = a cos θ + b sin θ, y = a sin θ − b cos θ, show that y2 (d2y)/(dx2)-xdy/dx+y=0 - Mathematics

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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`

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Solution

We have

x=acosθ+bsinθ      .....(1)

y=asinθbcosθ      .....(2)

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

x2+y2=(acosθ+bsinθ)2+(asinθbcosθ)2

=a2cos2θ+b2sin2θ+2abcosθsinθ + a2sin2θ+b2cos2θ2abcosθsinθ

=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`

Concept: Second Order Derivative
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