Differential Equations Question 140
Question: If $ x=\sin t $ , $ y=\cos pt $ , then
[Karnataka CET 2005]
Options:
A) $ (1-x^{2})y_2+xy_1+p^{2}y=0 $
B) $ (1-x^{2})y_2+xy_1-p^{2}y=0 $
C) $ (1+x^{2})y_2-xy_1+p^{2}y=0 $
D) $ (1-x^{2})y_2-xy_1+p^{2}y=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ x=\sin t $ , $ y=\cos pt $
$ \frac{dx}{dt}=\cos t $ ; $ \frac{dy}{dt}=-p\sin pt $ ; $ \frac{dy}{dx}=\frac{-p\sin pt}{\cos t} $
$ \frac{d^{2}y}{dx^{2}}=\frac{-\cos tp^{2}\cos pt(dt/dx)-p\sin pt\sin t(dt/dx)}{{{\cos }^{2}}t} $
Therefore $ (1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+p^{2}y=0 $
or $ (1-x^{2})y_2-xy_1+p^{2}y=0 $ .
BETA
