If \[x=\sin t\], \[y=\cos pt\], then [Karnataka CET 2005]

\[(1-{{x}^{2}}){{y}_{2}}+x{{y}_{1}}+{{p}^{2}}y=0\]

\[(1-{{x}^{2}}){{y}_{2}}+x{{y}_{1}}-{{p}^{2}}y=0\]

\[(1+{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{p}^{2}}y=0\]

\[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{p}^{2}}y=0\]

If \[x=\sin t\], \[y=\cos pt\], then [Karnataka CET

1.	If \[x=\sin t\], \[y=\cos pt\], then [Karnataka CET 2005]
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.	\[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{p}^{2}}y=0\]
Answer» E.

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If \[x=\sin t\], \[y=\cos pt\], then [Karnataka CET 2005]

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