If m and n are the order and degree of the differential equation d2ydx25+4d2y3dx223+d3ydx3=x2−1 then

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m=3,n=3

m=3,n=2

m=3,n=5

m=3,n=1

answer is B.

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Detailed Solution

The given differential equation can be written asd3ydx32−x2−1d3ydx3+d3ydx3d2ydx25+4d2ydx23=0Clearly, its order is 3 and degree is 2.∴ m=3 and n=2

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