If x = a sin t - b cos t and y = a cos t + b | vedclass

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(A) Given: $x = a \sin t - b \cos t$ and $y = a \cos t + b \sin t$. Squaring and adding both equations: $x^{2} + y^{2} = (a \sin t - b \cos t)^{2} + (

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(A) Given: $x = a \sin t - b \cos t$ and $y = a \cos t + b \sin t$. Squaring and adding both equations: $x^{2} + y^{2} = (a \sin t - b \cos t)^{2} + (a \cos t + b \sin t)^{2}$ $x^{2} + y^{2} = a^{2} \sin^{2} t + b^{2} \cos^{2} t - 2ab \sin t \cos t + a^{2} \cos^{2} t + b^{2} \sin^{2} t + 2ab \sin t \cos t$ $x^{2} + y^{2} = a^{2}(\sin^{2} t + \cos^{2} t) + b^{2}(\cos^{2} t + \sin^{2} t) = a^{2} + b^{2}$. Differentiating $x^{2} + y^{2} = a^{2} + b^{2}$ with respect to $x$: $2x + 2y \frac{dy}{dx} = 0 \implies y \frac{dy}{dx} = -x$. Differentiating again with respect to $x$: $y \frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{2} = -1$. Substitute $\frac{dy}{dx} = -\frac{x}{y}$: $y \frac{d^{2}y}{dx^{2}} + \left(-\frac{x}{y}\right)^{2} = -1$ $y \frac{d^{2}y}{dx^{2}} + \frac{x^{2}}{y^{2}} = -1$. Multiplying by $y^{2}$: $y^{3} \frac{d^{2}y}{dx^{2}} + x^{2} = -y^{2}$ $y^{3} \frac{d^{2}y}{dx^{2}} + x^{2} + y^{2} = 0$.

If $x = a \sin t - b \cos t$ and $y = a \cos t + b \sin t$,then $y^{3} \frac{d^{2} y}{d x^{2}} + x^{2} + y^{2} = $

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