If `x=t-sin t` and `y=1-cos t` , show that `y(d^3y)/(dx^3)+2(dy/dx)(d^2y)/(dx^2)=0 ` for `t != 2n(pi)` , `n(in)ZZ` .
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`x=t-sin t`
`y=1-cos t`
`dx/dt =1-cos t`
`dy/dt =sin t`
Using chain rule
`dy/dx `
`= dy/dt xx dt/dx `
`= (sin t)/(1-cos t)`
`=(2 sin (t/2) xx cos (t/2))/ (2 sin^2 (t/2))` Where `sin (t/2) != 0`
`=cot (t/2)`
`(d^2y)/dx^2 `
`= (d(dy/dx))/dxxxdt/dx`
`=(d(cot(t/2)))/dtxx 1/(1-cost)`
`=-cosec^2 (t/2) xx (1/2) xx 1/(2 sin^2 (t/2))`
`= (-1/4) cosec^4 (t/2)`
`(d^3y)/dx^3`
`=d/dt((d^2y)/dx^2)xxdt/dx`
`= (-4/4)cosec^3 (t/2) xx (-cosec (t/2) xx cot (t/2)) xx 1/2xx1/(1-cost)`
`=(1/2)cosec^4 (t/2) xx cot (t/2) xx 1/(2 sin^2 (t/2))`
`= -[-(1/4)cosec^4 (t/2)] cot (t/2) xx 1/(sin^2 (t/2))`
`= -(d^2y)/dx^2dy/dxxx1/(sin^2(t/2))`
`y (d^3y)/dx^3 = -(d^2y)/dx^2 y cot (t/2) xx 1/(sin^2 (t/2))`
`y(dy^3)/dx^3 = -(d^2y)/dx^2 xx (1-cost) cot (t/2) xx 2/(1-cos t)`
`y(dy^3)/dx^3 = -2(d^2y/dx^2)(dy/dx)`
`y (d^3y)/dx^3+2((d^2y)/dx^2)(dy/dx)=0`
So the answer is obtained as required.
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