The degree of the differential equation 3d2ydx2={1+(dydx)2}3/2 is
A
1
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B
2
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C
3
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D
6
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Solution
The correct option is B 2 3d2ydx2={1+(dydx)2}3/2 On squaring, we get 9(d2ydx2)2={1+(dydx)2}3 Obviously the highest derivative d2ydx2 contains a degree 2.