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प्रश्न
The degree of the differential equation `[1 + (("d"y)/("d"x))^2]^(3/2) = ("d"^2y)/("d"x^2)` is ______.
विकल्प
4
`3/2`
not defined
2
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उत्तर
The degree of the differential equation `[1 + (("d"y)/("d"x))^2]^(3/2) = ("d"^2y)/("d"x^2)` is 2.
Explanation:
The given differential equation is
`[1 + (("d"y)/("d"x))^2]^(3/2) = (("d"^2y)/("d"x^2))`
Squaring both sides, we have
`[ 1 + (("d"y)/("d"x))^2]^3 = (("d"^2y)/("d"x^2))^2`
So, the degree of the given differential equation is 2.
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