Question

If $x=\sin t$ and $y=\sin pt,$ prove that: $(1-x^2)y_2-x{y}_1+p^{2}y=0$

Answer 1

$x=\sin t\to \frac{dx}{dt}=\cos t$

$y=\sin pt\to \frac{dy}{dt}=p\cos pt$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{p\cos pt}{\cos t}$

$y_1=\frac{p\cos pt}{\cos t}$

$y_1\cos t=p\cos pt$

squaring on both sides

$y_1^2{\cos }^{2}t=p^2{\cos }^{2}pt$

$y_1^2(1-{\sin }^{2}t)=p^2(1-{\sin }^{2}pt)$

$y_1^2(1-x^2)=p^2(1-y^2)$

differentiating on both sides

$2y_1{y}_2(1-x^2)-2x{y}_1^2=p^2(-2y{y}_1)$

$y_2(1-x^2)-x{y}_1=-p^{2}y$

$y_2(1-x^2)-x{y}_1+p^{2}y=0$