A) \[(1-{{x}^{2}}){{y}_{2}}+x{{y}_{1}}+{{p}^{2}}y=0\]
B) \[(1-{{x}^{2}}){{y}_{2}}+x{{y}_{1}}-{{p}^{2}}y=0\]
C) \[(1+{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{p}^{2}}y=0\]
D) \[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{p}^{2}}y=0\]
Correct Answer: D
Solution :
\[x=\sin t\], \[y=\cos pt\] \[\frac{dx}{dt}=\cos t\]; \[\frac{dy}{dt}=-p\sin pt\]; \[\frac{dy}{dx}=\frac{-p\sin pt}{\cos t}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-\cos t\,{{p}^{2}}\cos pt(dt/dx)-p\sin pt\sin t(dt/dx)}{{{\cos }^{2}}t}\] Þ \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}+{{p}^{2}}y=0\] or \[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{p}^{2}}y=0\].
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