Question

If $x=a\cos \theta +b\sin \theta$ and $y=a\sin \theta -b\cos \theta$ then prove that $y^2\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+y=0$

Answer 1

We are given,

$\begin{array}{c}a\cos \theta +b\sin \theta \\ a\sin \theta -b\cos \theta \end{array}\}---\left(1\right)$

Now, differentiate $x$ and $y$ wrt $\theta$ we get,

$\frac{dx}{d\theta }=a(-\sin \theta )+b\left(\cos \theta \right)=-a\sin \theta +b\cos \theta$

& $\frac{dy}{d\theta }=a\left(\cos \theta \right)-b\left(\sin \theta \right)=a\cos \theta +b\sin \theta$

So, $\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}$

$\frac{dy}{dx}=\frac{a\cos \theta +b\sin \theta }{-a\sin \theta +b\cos \theta }$

$\frac{dy}{dx}=\frac{x}{-y}$ ( From $\left(1\right)$)

Now, differentiate above equation wrt $x$ we get

$\frac{d^{2}y}{d{x}^2}=\frac{1}{-y}+\frac{-(-x)}{y^2}.\frac{dy}{dx}$

$\frac{d^{2}y}{d{x}^2}=\frac{1}{-y}+\frac{x}{y^2}\frac{dy}{dx}$

$y^2\frac{d^{2}y}{d{x}^2}=-y+x\frac{dy}{dx}$

So, $y^2\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+y=0$

Hence proved that...