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The degree of the differential equation dddd[1+(dydx)2]32=d2ydx2 is ______.

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Question

The degree of the differential equation `[1 + (("d"y)/("d"x))^2]^(3/2) = ("d"^2y)/("d"x^2)` is ______.

Options

  • 4

  • `3/2`

  • not defined

  • 2

MCQ
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Solution

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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Chapter 9: Differential Equations - Exercise [Page 195]

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NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 9 Differential Equations
Exercise | Q 35 | Page 195

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