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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`
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Solution
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`
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