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Differential equation of \[y=\sec ({{\tan }^{-1}}x)\] is   [UPSEAT 2002]

A.                 \[(1+{{x}^{2}})\frac{dy}{dx}=y+x\]           
B.                 \[(1+{{x}^{2}})\frac{dy}{dx}=y-x\]
C.                 \[(1+{{x}^{2}})\frac{dy}{dx}=xy\]               
D.                 \[(1+{{x}^{2}})\frac{dy}{dx}=\frac{x}{y}\]
Answer» D.                 \[(1+{{x}^{2}})\frac{dy}{dx}=\frac{x}{y}\]


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