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Question
Determine the order and degree of the following differential equation:
`[1 + (dy/dx)^2]^(3/2) = 8(d^2y)/dx^2`
Solution
The given D.E. is `[1 + (dy/dx)^2]^(3/2) = 8(d^2y)/dx^2`
On squaring both sides, we get
`[1 + (dy/dx)^2]^3 = 8^2.((d^2y)/dx^2)^2`
This D.E. has highest order derivative `(d^2y)/dx^2` with power 2.
∴ The given D.E. has order 2 and degree 2.
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