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प्रश्न
Determine the order and degree of the following differential equation:
(y''')2 + 3y'' + 3xy' + 5y = 0
उत्तर
The given D.E. is (y''')2 + 3y'' + 3xy' + 5y = 0
This can be written as:
`((d^3y)/dx^3)^2 + 3(d^2y)/dx^2 + 3x(dy/dx) + 5y` = 0
This D.E. has highest order derivative `(d^3y)/dx^3` with power 2.
∴ The given D.E. has order 3 and degree 2.
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