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Q14. If x = a, cos $\mathrm{\theta }$ + b sin ​$\mathrm{\theta }$ and y = a sin $\mathrm{\theta }$ – b cos $\mathrm{\theta }$, show that $y^2\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+y=0$.

Answer 1

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$$\begin{gathered} Given \\ x=a\cos \theta +b\sin \theta \\ y=a\sin \theta -b\cos \theta \\ Now \\ {x}^2=a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\sin \theta \cos \theta \\ {y}^2=a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\sin \theta \cos \theta \\ Adding{x}^2+y^{2}weget \\ {x}^2+y^2=a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\sin \theta \cos \theta +a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\sin \theta \cos \theta \\ =a^2\left({\cos }^2\theta +{\sin }^2\theta \right)+b^2\left({\cos }^2\theta +{\sin }^2\theta \right) \\ =a^2+b^2 \\ Nowdifferentiatingbothsidesweget \\ 2x+2y\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-x}{y} \\ againdifferentiatingweget \\ \frac{d^{2}y}{d{x}^2}=-\frac{y-x{\displaystyle \frac{dy}{dx}}}{y^2} \\ \Rightarrow y^2\frac{d^{2}y}{d{x}^2}=-y+x\frac{dy}{dx} \\ \Rightarrow y^2\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+y=0 \\ Henceproved \end{gathered}$$

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