Question

If $x=\sin t,y=\sin kt$ satisfies $(1-x^2)y_2-x{y}_1+Ay=0$ then $A$ is Equal to

• A. $k$
• B. $1$
• C. $k^2$
• D. $1+k$

Answer 1

Answer: (C)

$k^2$

Given, $x=\sin t\Rightarrow \frac{dx}{dt}=\cos ty=\sin kt\Rightarrow \frac{dy}{dt}=k\cos kt\frac{dy}{dx}=\frac{k\cos kt}{\cos t}\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{kcoskt}{\cos t}\right)\times \frac{dt}{dx}$

we know $\left(\frac{f}{g}\right)=\frac{f^{\prime }g-f{g}^{\prime }}{g^2}{y}_2=\frac{(-k^2\sin kt)\left(\cos t\right)-(-\sin t)\left(k\cos kt\right)}{{\cos }^{2}t}\times \frac{1}{\cos t}({\cos }^{2}t{)y}_2=-k^2\sin kt+\frac{\left(\sin t\right)\left(k\cos kt\right)}{\cos t}(1-x^2{)y}_2=-k^{2}y+x{y}_1$

$\Rightarrow (1-x^2{)y}_2-x{y}_1+k^{2}y=0$