Question

If $x=a\cos \theta +b\sin \theta ,y=a\sin \theta -b\cos \theta ,$ show that
$\frac{y^2{d}^{2}y}{d{x}^2}-\frac{xdy}{dx}+y=0$.

Answer 1

$y^2=a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\cos \theta \sin \theta x^2=b^2{\sin }^2\theta +a^2{\cos }^2\theta +2ab\cos \theta \sin \theta$

Now adding both the equations we get,

$y^2+x^2=a^2+b^2$

Now differentiating with respect to $x$ we get,

$2y{y}^{{}^{\prime }}+2x=0$

Now again differentiating with respect to $x$ we get,

$2y{y}^{{}^{\prime \prime }}+2{\left(y^{{}^{\prime }}\right)}^2+2=0$

Multiplying both the sides with $y$ and writing $y^{{}^{\prime }}=\frac{-x}{y}$

We get $y^2\frac{d^{2}y}{d{x}^2}-\frac{xdy}{dx}+y=0.$